Related
What is array to pointer decay? Is there any relation to array pointers?
It's said that arrays "decay" into pointers. A C++ array declared as int numbers [5] cannot be re-pointed, i.e. you can't say numbers = 0x5a5aff23. More importantly the term decay signifies loss of type and dimension; numbers decay into int* by losing the dimension information (count 5) and the type is not int [5] any more. Look here for cases where the decay doesn't happen.
If you're passing an array by value, what you're really doing is copying a pointer - a pointer to the array's first element is copied to the parameter (whose type should also be a pointer the array element's type). This works due to array's decaying nature; once decayed, sizeof no longer gives the complete array's size, because it essentially becomes a pointer. This is why it's preferred (among other reasons) to pass by reference or pointer.
Three ways to pass in an array1:
void by_value(const T* array) // const T array[] means the same
void by_pointer(const T (*array)[U])
void by_reference(const T (&array)[U])
The last two will give proper sizeof info, while the first one won't since the array argument has decayed to be assigned to the parameter.
1 The constant U should be known at compile-time.
Arrays are basically the same as pointers in C/C++, but not quite. Once you convert an array:
const int a[] = { 2, 3, 5, 7, 11 };
into a pointer (which works without casting, and therefore can happen unexpectedly in some cases):
const int* p = a;
you lose the ability of the sizeof operator to count elements in the array:
assert( sizeof(p) != sizeof(a) ); // sizes are not equal
This lost ability is referred to as "decay".
For more details, check out this article about array decay.
Here's what the standard says (C99 6.3.2.1/3 - Other operands - Lvalues, arrays, and function designators):
Except when it is the operand of the sizeof operator or the unary & operator, or is a
string literal used to initialize an array, an expression that has type ‘‘array of type’’ is
converted to an expression with type ‘‘pointer to type’’ that points to the initial element of
the array object and is not an lvalue.
This means that pretty much anytime the array name is used in an expression, it is automatically converted to a pointer to the 1st item in the array.
Note that function names act in a similar way, but function pointers are used far less and in a much more specialized way that it doesn't cause nearly as much confusion as the automatic conversion of array names to pointers.
The C++ standard (4.2 Array-to-pointer conversion) loosens the conversion requirement to (emphasis mine):
An lvalue or rvalue of type “array of N T” or “array of unknown bound of T” can be converted to an rvalue
of type “pointer to T.”
So the conversion doesn't have to happen like it pretty much always does in C (this lets functions overload or templates match on the array type).
This is also why in C you should avoid using array parameters in function prototypes/definitions (in my opinion - I'm not sure if there's any general agreement). They cause confusion and are a fiction anyway - use pointer parameters and the confusion might not go away entirely, but at least the parameter declaration isn't lying.
"Decay" refers to the implicit conversion of an expression from an array type to a pointer type. In most contexts, when the compiler sees an array expression it converts the type of the expression from "N-element array of T" to "pointer to T" and sets the value of the expression to the address of the first element of the array. The exceptions to this rule are when an array is an operand of either the sizeof or & operators, or the array is a string literal being used as an initializer in a declaration.
Assume the following code:
char a[80];
strcpy(a, "This is a test");
The expression a is of type "80-element array of char" and the expression "This is a test" is of type "15-element array of char" (in C; in C++ string literals are arrays of const char). However, in the call to strcpy(), neither expression is an operand of sizeof or &, so their types are implicitly converted to "pointer to char", and their values are set to the address of the first element in each. What strcpy() receives are not arrays, but pointers, as seen in its prototype:
char *strcpy(char *dest, const char *src);
This is not the same thing as an array pointer. For example:
char a[80];
char *ptr_to_first_element = a;
char (*ptr_to_array)[80] = &a;
Both ptr_to_first_element and ptr_to_array have the same value; the base address of a. However, they are different types and are treated differently, as shown below:
a[i] == ptr_to_first_element[i] == (*ptr_to_array)[i] != *ptr_to_array[i] != ptr_to_array[i]
Remember that the expression a[i] is interpreted as *(a+i) (which only works if the array type is converted to a pointer type), so both a[i] and ptr_to_first_element[i] work the same. The expression (*ptr_to_array)[i] is interpreted as *(*a+i). The expressions *ptr_to_array[i] and ptr_to_array[i] may lead to compiler warnings or errors depending on the context; they'll definitely do the wrong thing if you're expecting them to evaluate to a[i].
sizeof a == sizeof *ptr_to_array == 80
Again, when an array is an operand of sizeof, it's not converted to a pointer type.
sizeof *ptr_to_first_element == sizeof (char) == 1
sizeof ptr_to_first_element == sizeof (char *) == whatever the pointer size
is on your platform
ptr_to_first_element is a simple pointer to char.
Arrays, in C, have no value.
Wherever the value of an object is expected but the object is an array, the address of its first element is used instead, with type pointer to (type of array elements).
In a function, all parameters are passed by value (arrays are no exception). When you pass an array in a function it "decays into a pointer" (sic); when you compare an array to something else, again it "decays into a pointer" (sic); ...
void foo(int arr[]);
Function foo expects the value of an array. But, in C, arrays have no value! So foo gets instead the address of the first element of the array.
int arr[5];
int *ip = &(arr[1]);
if (arr == ip) { /* something; */ }
In the comparison above, arr has no value, so it becomes a pointer. It becomes a pointer to int. That pointer can be compared with the variable ip.
In the array indexing syntax you are used to seeing, again, the arr is 'decayed to a pointer'
arr[42];
/* same as *(arr + 42); */
/* same as *(&(arr[0]) + 42); */
The only times an array doesn't decay into a pointer are when it is the operand of the sizeof operator, or the & operator (the 'address of' operator), or as a string literal used to initialize a character array.
It's when array rots and is being pointed at ;-)
Actually, it's just that if you want to pass an array somewhere, but the pointer is passed instead (because who the hell would pass the whole array for you), people say that poor array decayed to pointer.
Array decaying means that, when an array is passed as a parameter to a function, it's treated identically to ("decays to") a pointer.
void do_something(int *array) {
// We don't know how big array is here, because it's decayed to a pointer.
printf("%i\n", sizeof(array)); // always prints 4 on a 32-bit machine
}
int main (int argc, char **argv) {
int a[10];
int b[20];
int *c;
printf("%zu\n", sizeof(a)); //prints 40 on a 32-bit machine
printf("%zu\n", sizeof(b)); //prints 80 on a 32-bit machine
printf("%zu\n", sizeof(c)); //prints 4 on a 32-bit machine
do_something(a);
do_something(b);
do_something(c);
}
There are two complications or exceptions to the above.
First, when dealing with multidimensional arrays in C and C++, only the first dimension is lost. This is because arrays are layed out contiguously in memory, so the compiler must know all but the first dimension to be able to calculate offsets into that block of memory.
void do_something(int array[][10])
{
// We don't know how big the first dimension is.
}
int main(int argc, char *argv[]) {
int a[5][10];
int b[20][10];
do_something(a);
do_something(b);
return 0;
}
Second, in C++, you can use templates to deduce the size of arrays. Microsoft uses this for the C++ versions of Secure CRT functions like strcpy_s, and you can use a similar trick to reliably get the number of elements in an array.
tl;dr: When you use an array you've defined, you'll actually be using a pointer to its first element.
Thus:
When you write arr[idx] you're really just saying *(arr + idx).
functions never really take arrays as parameters, only pointers - either directly, when you specify an array parameter, or indirectly, if you pass a reference to an array.
Sort-of exceptions to this rule:
You can pass fixed-length arrays to functions within a struct.
sizeof() gives the size taken up by the array, not the size of a pointer.
Try this code
void f(double a[10]) {
printf("in function: %d", sizeof(a));
printf("pointer size: %d\n", sizeof(double *));
}
int main() {
double a[10];
printf("in main: %d", sizeof(a));
f(a);
}
and you will see that the size of the array inside the function is not equal to the size of the array in main, but it is equal to the size of a pointer.
You probably heard that "arrays are pointers", but, this is not exactly true (the sizeof inside main prints the correct size). However, when passed, the array decays to pointer. That is, regardless of what the syntax shows, you actually pass a pointer, and the function actually receives a pointer.
In this case, the definition void f(double a[10] is implicitly transformed by the compiler to void f(double *a). You could have equivalently declared the function argument directly as *a. You could have even written a[100] or a[1], instead of a[10], since it is never actually compiled that way (however, you shouldn't do it obviously, it would confuse the reader).
Arrays are automatically passed by pointer in C. The rationale behind it can only be speculated.
int a[5], int *a and int (*a)[5] are all glorified addresses meaning that the compiler treats arithmetic and deference operators on them differently depending on the type, so when they refer to the same address they are not treated the same by the compiler. int a[5] is different to the other 2 in that the address is implicit and does not manifest on the stack or the executable as part of the array itself, it is only used by the compiler to resolve certain arithmetic operations, like taking its address or pointer arithmetic. int a[5] is therefore an array as well as an implicit address, but as soon as you talk about the address itself and place it on the stack, the address itself is no longer an array, and can only be a pointer to an array or a decayed array i.e. a pointer to the first member of the array.
For instance, on int (*a)[5], the first dereference on a will produce an int * (so the same address, just a different type, and note not int a[5]), and pointer arithmetic on a i.e. a+1 or *(a+1) will be in terms of the size of an array of 5 ints (which is the data type it points to), and the second dereference will produce the int. On int a[5] however, the first dereference will produce the int and the pointer arithmetic will be in terms of the size of an int.
To a function, you can only pass int * and int (*)[5], and the function casts it to whatever the parameter type is, so within the function you have a choice whether to treat an address that is being passed as a decayed array or a pointer to an array (where the function has to specify the size of the array being passed). If you pass a to a function and a is defined int a[5], then as a resolves to an address, you are passing an address, and an address can only be a pointer type. In the function, the parameter it accesses is then an address on the stack or in a register, which can only be a pointer type and not an array type -- this is because it's an actual address on the stack and is therefore clearly not the array itself.
You lose the size of the array because the type of the parameter, being an address, is a pointer and not an array, which does not have an array size, as can be seen when using sizeof, which works on the type of the value being passed to it. The parameter type int a[5] instead of int *a is allowed but is treated as int * instead of disallowing it outright, though it should be disallowed, because it is misleading, because it makes you think that the size information can be used, but you can only do this by casting it to int (*a)[5], and of course, the function has to specify the size of the array because there is no way to pass the size of the array because the size of the array needs to be a compile-time constant.
I might be so bold to think there are four (4) ways to pass an array as the function argument. Also here is the short but working code for your perusal.
#include <iostream>
#include <string>
#include <vector>
#include <cassert>
using namespace std;
// test data
// notice native array init with no copy aka "="
// not possible in C
const char* specimen[]{ __TIME__, __DATE__, __TIMESTAMP__ };
// ONE
// simple, dangerous and useless
template<typename T>
void as_pointer(const T* array) {
// a pointer
assert(array != nullptr);
} ;
// TWO
// for above const T array[] means the same
// but and also , minimum array size indication might be given too
// this also does not stop the array decay into T *
// thus size information is lost
template<typename T>
void by_value_no_size(const T array[0xFF]) {
// decayed to a pointer
assert( array != nullptr );
}
// THREE
// size information is preserved
// but pointer is asked for
template<typename T, size_t N>
void pointer_to_array(const T (*array)[N])
{
// dealing with native pointer
assert( array != nullptr );
}
// FOUR
// no C equivalent
// array by reference
// size is preserved
template<typename T, size_t N>
void reference_to_array(const T (&array)[N])
{
// array is not a pointer here
// it is (almost) a container
// most of the std:: lib algorithms
// do work on array reference, for example
// range for requires std::begin() and std::end()
// on the type passed as range to iterate over
for (auto && elem : array )
{
cout << endl << elem ;
}
}
int main()
{
// ONE
as_pointer(specimen);
// TWO
by_value_no_size(specimen);
// THREE
pointer_to_array(&specimen);
// FOUR
reference_to_array( specimen ) ;
}
I might also think this shows the superiority of C++ vs C. At least in reference (pun intended) of passing an array by reference.
Of course there are extremely strict projects with no heap allocation, no exceptions and no std:: lib. C++ native array handling is mission critical language feature, one might say.
What is array to pointer decay? Is there any relation to array pointers?
It's said that arrays "decay" into pointers. A C++ array declared as int numbers [5] cannot be re-pointed, i.e. you can't say numbers = 0x5a5aff23. More importantly the term decay signifies loss of type and dimension; numbers decay into int* by losing the dimension information (count 5) and the type is not int [5] any more. Look here for cases where the decay doesn't happen.
If you're passing an array by value, what you're really doing is copying a pointer - a pointer to the array's first element is copied to the parameter (whose type should also be a pointer the array element's type). This works due to array's decaying nature; once decayed, sizeof no longer gives the complete array's size, because it essentially becomes a pointer. This is why it's preferred (among other reasons) to pass by reference or pointer.
Three ways to pass in an array1:
void by_value(const T* array) // const T array[] means the same
void by_pointer(const T (*array)[U])
void by_reference(const T (&array)[U])
The last two will give proper sizeof info, while the first one won't since the array argument has decayed to be assigned to the parameter.
1 The constant U should be known at compile-time.
Arrays are basically the same as pointers in C/C++, but not quite. Once you convert an array:
const int a[] = { 2, 3, 5, 7, 11 };
into a pointer (which works without casting, and therefore can happen unexpectedly in some cases):
const int* p = a;
you lose the ability of the sizeof operator to count elements in the array:
assert( sizeof(p) != sizeof(a) ); // sizes are not equal
This lost ability is referred to as "decay".
For more details, check out this article about array decay.
Here's what the standard says (C99 6.3.2.1/3 - Other operands - Lvalues, arrays, and function designators):
Except when it is the operand of the sizeof operator or the unary & operator, or is a
string literal used to initialize an array, an expression that has type ‘‘array of type’’ is
converted to an expression with type ‘‘pointer to type’’ that points to the initial element of
the array object and is not an lvalue.
This means that pretty much anytime the array name is used in an expression, it is automatically converted to a pointer to the 1st item in the array.
Note that function names act in a similar way, but function pointers are used far less and in a much more specialized way that it doesn't cause nearly as much confusion as the automatic conversion of array names to pointers.
The C++ standard (4.2 Array-to-pointer conversion) loosens the conversion requirement to (emphasis mine):
An lvalue or rvalue of type “array of N T” or “array of unknown bound of T” can be converted to an rvalue
of type “pointer to T.”
So the conversion doesn't have to happen like it pretty much always does in C (this lets functions overload or templates match on the array type).
This is also why in C you should avoid using array parameters in function prototypes/definitions (in my opinion - I'm not sure if there's any general agreement). They cause confusion and are a fiction anyway - use pointer parameters and the confusion might not go away entirely, but at least the parameter declaration isn't lying.
"Decay" refers to the implicit conversion of an expression from an array type to a pointer type. In most contexts, when the compiler sees an array expression it converts the type of the expression from "N-element array of T" to "pointer to T" and sets the value of the expression to the address of the first element of the array. The exceptions to this rule are when an array is an operand of either the sizeof or & operators, or the array is a string literal being used as an initializer in a declaration.
Assume the following code:
char a[80];
strcpy(a, "This is a test");
The expression a is of type "80-element array of char" and the expression "This is a test" is of type "15-element array of char" (in C; in C++ string literals are arrays of const char). However, in the call to strcpy(), neither expression is an operand of sizeof or &, so their types are implicitly converted to "pointer to char", and their values are set to the address of the first element in each. What strcpy() receives are not arrays, but pointers, as seen in its prototype:
char *strcpy(char *dest, const char *src);
This is not the same thing as an array pointer. For example:
char a[80];
char *ptr_to_first_element = a;
char (*ptr_to_array)[80] = &a;
Both ptr_to_first_element and ptr_to_array have the same value; the base address of a. However, they are different types and are treated differently, as shown below:
a[i] == ptr_to_first_element[i] == (*ptr_to_array)[i] != *ptr_to_array[i] != ptr_to_array[i]
Remember that the expression a[i] is interpreted as *(a+i) (which only works if the array type is converted to a pointer type), so both a[i] and ptr_to_first_element[i] work the same. The expression (*ptr_to_array)[i] is interpreted as *(*a+i). The expressions *ptr_to_array[i] and ptr_to_array[i] may lead to compiler warnings or errors depending on the context; they'll definitely do the wrong thing if you're expecting them to evaluate to a[i].
sizeof a == sizeof *ptr_to_array == 80
Again, when an array is an operand of sizeof, it's not converted to a pointer type.
sizeof *ptr_to_first_element == sizeof (char) == 1
sizeof ptr_to_first_element == sizeof (char *) == whatever the pointer size
is on your platform
ptr_to_first_element is a simple pointer to char.
Arrays, in C, have no value.
Wherever the value of an object is expected but the object is an array, the address of its first element is used instead, with type pointer to (type of array elements).
In a function, all parameters are passed by value (arrays are no exception). When you pass an array in a function it "decays into a pointer" (sic); when you compare an array to something else, again it "decays into a pointer" (sic); ...
void foo(int arr[]);
Function foo expects the value of an array. But, in C, arrays have no value! So foo gets instead the address of the first element of the array.
int arr[5];
int *ip = &(arr[1]);
if (arr == ip) { /* something; */ }
In the comparison above, arr has no value, so it becomes a pointer. It becomes a pointer to int. That pointer can be compared with the variable ip.
In the array indexing syntax you are used to seeing, again, the arr is 'decayed to a pointer'
arr[42];
/* same as *(arr + 42); */
/* same as *(&(arr[0]) + 42); */
The only times an array doesn't decay into a pointer are when it is the operand of the sizeof operator, or the & operator (the 'address of' operator), or as a string literal used to initialize a character array.
It's when array rots and is being pointed at ;-)
Actually, it's just that if you want to pass an array somewhere, but the pointer is passed instead (because who the hell would pass the whole array for you), people say that poor array decayed to pointer.
Array decaying means that, when an array is passed as a parameter to a function, it's treated identically to ("decays to") a pointer.
void do_something(int *array) {
// We don't know how big array is here, because it's decayed to a pointer.
printf("%i\n", sizeof(array)); // always prints 4 on a 32-bit machine
}
int main (int argc, char **argv) {
int a[10];
int b[20];
int *c;
printf("%zu\n", sizeof(a)); //prints 40 on a 32-bit machine
printf("%zu\n", sizeof(b)); //prints 80 on a 32-bit machine
printf("%zu\n", sizeof(c)); //prints 4 on a 32-bit machine
do_something(a);
do_something(b);
do_something(c);
}
There are two complications or exceptions to the above.
First, when dealing with multidimensional arrays in C and C++, only the first dimension is lost. This is because arrays are layed out contiguously in memory, so the compiler must know all but the first dimension to be able to calculate offsets into that block of memory.
void do_something(int array[][10])
{
// We don't know how big the first dimension is.
}
int main(int argc, char *argv[]) {
int a[5][10];
int b[20][10];
do_something(a);
do_something(b);
return 0;
}
Second, in C++, you can use templates to deduce the size of arrays. Microsoft uses this for the C++ versions of Secure CRT functions like strcpy_s, and you can use a similar trick to reliably get the number of elements in an array.
tl;dr: When you use an array you've defined, you'll actually be using a pointer to its first element.
Thus:
When you write arr[idx] you're really just saying *(arr + idx).
functions never really take arrays as parameters, only pointers - either directly, when you specify an array parameter, or indirectly, if you pass a reference to an array.
Sort-of exceptions to this rule:
You can pass fixed-length arrays to functions within a struct.
sizeof() gives the size taken up by the array, not the size of a pointer.
Try this code
void f(double a[10]) {
printf("in function: %d", sizeof(a));
printf("pointer size: %d\n", sizeof(double *));
}
int main() {
double a[10];
printf("in main: %d", sizeof(a));
f(a);
}
and you will see that the size of the array inside the function is not equal to the size of the array in main, but it is equal to the size of a pointer.
You probably heard that "arrays are pointers", but, this is not exactly true (the sizeof inside main prints the correct size). However, when passed, the array decays to pointer. That is, regardless of what the syntax shows, you actually pass a pointer, and the function actually receives a pointer.
In this case, the definition void f(double a[10] is implicitly transformed by the compiler to void f(double *a). You could have equivalently declared the function argument directly as *a. You could have even written a[100] or a[1], instead of a[10], since it is never actually compiled that way (however, you shouldn't do it obviously, it would confuse the reader).
Arrays are automatically passed by pointer in C. The rationale behind it can only be speculated.
int a[5], int *a and int (*a)[5] are all glorified addresses meaning that the compiler treats arithmetic and deference operators on them differently depending on the type, so when they refer to the same address they are not treated the same by the compiler. int a[5] is different to the other 2 in that the address is implicit and does not manifest on the stack or the executable as part of the array itself, it is only used by the compiler to resolve certain arithmetic operations, like taking its address or pointer arithmetic. int a[5] is therefore an array as well as an implicit address, but as soon as you talk about the address itself and place it on the stack, the address itself is no longer an array, and can only be a pointer to an array or a decayed array i.e. a pointer to the first member of the array.
For instance, on int (*a)[5], the first dereference on a will produce an int * (so the same address, just a different type, and note not int a[5]), and pointer arithmetic on a i.e. a+1 or *(a+1) will be in terms of the size of an array of 5 ints (which is the data type it points to), and the second dereference will produce the int. On int a[5] however, the first dereference will produce the int and the pointer arithmetic will be in terms of the size of an int.
To a function, you can only pass int * and int (*)[5], and the function casts it to whatever the parameter type is, so within the function you have a choice whether to treat an address that is being passed as a decayed array or a pointer to an array (where the function has to specify the size of the array being passed). If you pass a to a function and a is defined int a[5], then as a resolves to an address, you are passing an address, and an address can only be a pointer type. In the function, the parameter it accesses is then an address on the stack or in a register, which can only be a pointer type and not an array type -- this is because it's an actual address on the stack and is therefore clearly not the array itself.
You lose the size of the array because the type of the parameter, being an address, is a pointer and not an array, which does not have an array size, as can be seen when using sizeof, which works on the type of the value being passed to it. The parameter type int a[5] instead of int *a is allowed but is treated as int * instead of disallowing it outright, though it should be disallowed, because it is misleading, because it makes you think that the size information can be used, but you can only do this by casting it to int (*a)[5], and of course, the function has to specify the size of the array because there is no way to pass the size of the array because the size of the array needs to be a compile-time constant.
I might be so bold to think there are four (4) ways to pass an array as the function argument. Also here is the short but working code for your perusal.
#include <iostream>
#include <string>
#include <vector>
#include <cassert>
using namespace std;
// test data
// notice native array init with no copy aka "="
// not possible in C
const char* specimen[]{ __TIME__, __DATE__, __TIMESTAMP__ };
// ONE
// simple, dangerous and useless
template<typename T>
void as_pointer(const T* array) {
// a pointer
assert(array != nullptr);
} ;
// TWO
// for above const T array[] means the same
// but and also , minimum array size indication might be given too
// this also does not stop the array decay into T *
// thus size information is lost
template<typename T>
void by_value_no_size(const T array[0xFF]) {
// decayed to a pointer
assert( array != nullptr );
}
// THREE
// size information is preserved
// but pointer is asked for
template<typename T, size_t N>
void pointer_to_array(const T (*array)[N])
{
// dealing with native pointer
assert( array != nullptr );
}
// FOUR
// no C equivalent
// array by reference
// size is preserved
template<typename T, size_t N>
void reference_to_array(const T (&array)[N])
{
// array is not a pointer here
// it is (almost) a container
// most of the std:: lib algorithms
// do work on array reference, for example
// range for requires std::begin() and std::end()
// on the type passed as range to iterate over
for (auto && elem : array )
{
cout << endl << elem ;
}
}
int main()
{
// ONE
as_pointer(specimen);
// TWO
by_value_no_size(specimen);
// THREE
pointer_to_array(&specimen);
// FOUR
reference_to_array( specimen ) ;
}
I might also think this shows the superiority of C++ vs C. At least in reference (pun intended) of passing an array by reference.
Of course there are extremely strict projects with no heap allocation, no exceptions and no std:: lib. C++ native array handling is mission critical language feature, one might say.
What is array to pointer decay? Is there any relation to array pointers?
It's said that arrays "decay" into pointers. A C++ array declared as int numbers [5] cannot be re-pointed, i.e. you can't say numbers = 0x5a5aff23. More importantly the term decay signifies loss of type and dimension; numbers decay into int* by losing the dimension information (count 5) and the type is not int [5] any more. Look here for cases where the decay doesn't happen.
If you're passing an array by value, what you're really doing is copying a pointer - a pointer to the array's first element is copied to the parameter (whose type should also be a pointer the array element's type). This works due to array's decaying nature; once decayed, sizeof no longer gives the complete array's size, because it essentially becomes a pointer. This is why it's preferred (among other reasons) to pass by reference or pointer.
Three ways to pass in an array1:
void by_value(const T* array) // const T array[] means the same
void by_pointer(const T (*array)[U])
void by_reference(const T (&array)[U])
The last two will give proper sizeof info, while the first one won't since the array argument has decayed to be assigned to the parameter.
1 The constant U should be known at compile-time.
Arrays are basically the same as pointers in C/C++, but not quite. Once you convert an array:
const int a[] = { 2, 3, 5, 7, 11 };
into a pointer (which works without casting, and therefore can happen unexpectedly in some cases):
const int* p = a;
you lose the ability of the sizeof operator to count elements in the array:
assert( sizeof(p) != sizeof(a) ); // sizes are not equal
This lost ability is referred to as "decay".
For more details, check out this article about array decay.
Here's what the standard says (C99 6.3.2.1/3 - Other operands - Lvalues, arrays, and function designators):
Except when it is the operand of the sizeof operator or the unary & operator, or is a
string literal used to initialize an array, an expression that has type ‘‘array of type’’ is
converted to an expression with type ‘‘pointer to type’’ that points to the initial element of
the array object and is not an lvalue.
This means that pretty much anytime the array name is used in an expression, it is automatically converted to a pointer to the 1st item in the array.
Note that function names act in a similar way, but function pointers are used far less and in a much more specialized way that it doesn't cause nearly as much confusion as the automatic conversion of array names to pointers.
The C++ standard (4.2 Array-to-pointer conversion) loosens the conversion requirement to (emphasis mine):
An lvalue or rvalue of type “array of N T” or “array of unknown bound of T” can be converted to an rvalue
of type “pointer to T.”
So the conversion doesn't have to happen like it pretty much always does in C (this lets functions overload or templates match on the array type).
This is also why in C you should avoid using array parameters in function prototypes/definitions (in my opinion - I'm not sure if there's any general agreement). They cause confusion and are a fiction anyway - use pointer parameters and the confusion might not go away entirely, but at least the parameter declaration isn't lying.
"Decay" refers to the implicit conversion of an expression from an array type to a pointer type. In most contexts, when the compiler sees an array expression it converts the type of the expression from "N-element array of T" to "pointer to T" and sets the value of the expression to the address of the first element of the array. The exceptions to this rule are when an array is an operand of either the sizeof or & operators, or the array is a string literal being used as an initializer in a declaration.
Assume the following code:
char a[80];
strcpy(a, "This is a test");
The expression a is of type "80-element array of char" and the expression "This is a test" is of type "15-element array of char" (in C; in C++ string literals are arrays of const char). However, in the call to strcpy(), neither expression is an operand of sizeof or &, so their types are implicitly converted to "pointer to char", and their values are set to the address of the first element in each. What strcpy() receives are not arrays, but pointers, as seen in its prototype:
char *strcpy(char *dest, const char *src);
This is not the same thing as an array pointer. For example:
char a[80];
char *ptr_to_first_element = a;
char (*ptr_to_array)[80] = &a;
Both ptr_to_first_element and ptr_to_array have the same value; the base address of a. However, they are different types and are treated differently, as shown below:
a[i] == ptr_to_first_element[i] == (*ptr_to_array)[i] != *ptr_to_array[i] != ptr_to_array[i]
Remember that the expression a[i] is interpreted as *(a+i) (which only works if the array type is converted to a pointer type), so both a[i] and ptr_to_first_element[i] work the same. The expression (*ptr_to_array)[i] is interpreted as *(*a+i). The expressions *ptr_to_array[i] and ptr_to_array[i] may lead to compiler warnings or errors depending on the context; they'll definitely do the wrong thing if you're expecting them to evaluate to a[i].
sizeof a == sizeof *ptr_to_array == 80
Again, when an array is an operand of sizeof, it's not converted to a pointer type.
sizeof *ptr_to_first_element == sizeof (char) == 1
sizeof ptr_to_first_element == sizeof (char *) == whatever the pointer size
is on your platform
ptr_to_first_element is a simple pointer to char.
Arrays, in C, have no value.
Wherever the value of an object is expected but the object is an array, the address of its first element is used instead, with type pointer to (type of array elements).
In a function, all parameters are passed by value (arrays are no exception). When you pass an array in a function it "decays into a pointer" (sic); when you compare an array to something else, again it "decays into a pointer" (sic); ...
void foo(int arr[]);
Function foo expects the value of an array. But, in C, arrays have no value! So foo gets instead the address of the first element of the array.
int arr[5];
int *ip = &(arr[1]);
if (arr == ip) { /* something; */ }
In the comparison above, arr has no value, so it becomes a pointer. It becomes a pointer to int. That pointer can be compared with the variable ip.
In the array indexing syntax you are used to seeing, again, the arr is 'decayed to a pointer'
arr[42];
/* same as *(arr + 42); */
/* same as *(&(arr[0]) + 42); */
The only times an array doesn't decay into a pointer are when it is the operand of the sizeof operator, or the & operator (the 'address of' operator), or as a string literal used to initialize a character array.
It's when array rots and is being pointed at ;-)
Actually, it's just that if you want to pass an array somewhere, but the pointer is passed instead (because who the hell would pass the whole array for you), people say that poor array decayed to pointer.
Array decaying means that, when an array is passed as a parameter to a function, it's treated identically to ("decays to") a pointer.
void do_something(int *array) {
// We don't know how big array is here, because it's decayed to a pointer.
printf("%i\n", sizeof(array)); // always prints 4 on a 32-bit machine
}
int main (int argc, char **argv) {
int a[10];
int b[20];
int *c;
printf("%zu\n", sizeof(a)); //prints 40 on a 32-bit machine
printf("%zu\n", sizeof(b)); //prints 80 on a 32-bit machine
printf("%zu\n", sizeof(c)); //prints 4 on a 32-bit machine
do_something(a);
do_something(b);
do_something(c);
}
There are two complications or exceptions to the above.
First, when dealing with multidimensional arrays in C and C++, only the first dimension is lost. This is because arrays are layed out contiguously in memory, so the compiler must know all but the first dimension to be able to calculate offsets into that block of memory.
void do_something(int array[][10])
{
// We don't know how big the first dimension is.
}
int main(int argc, char *argv[]) {
int a[5][10];
int b[20][10];
do_something(a);
do_something(b);
return 0;
}
Second, in C++, you can use templates to deduce the size of arrays. Microsoft uses this for the C++ versions of Secure CRT functions like strcpy_s, and you can use a similar trick to reliably get the number of elements in an array.
tl;dr: When you use an array you've defined, you'll actually be using a pointer to its first element.
Thus:
When you write arr[idx] you're really just saying *(arr + idx).
functions never really take arrays as parameters, only pointers - either directly, when you specify an array parameter, or indirectly, if you pass a reference to an array.
Sort-of exceptions to this rule:
You can pass fixed-length arrays to functions within a struct.
sizeof() gives the size taken up by the array, not the size of a pointer.
Try this code
void f(double a[10]) {
printf("in function: %d", sizeof(a));
printf("pointer size: %d\n", sizeof(double *));
}
int main() {
double a[10];
printf("in main: %d", sizeof(a));
f(a);
}
and you will see that the size of the array inside the function is not equal to the size of the array in main, but it is equal to the size of a pointer.
You probably heard that "arrays are pointers", but, this is not exactly true (the sizeof inside main prints the correct size). However, when passed, the array decays to pointer. That is, regardless of what the syntax shows, you actually pass a pointer, and the function actually receives a pointer.
In this case, the definition void f(double a[10] is implicitly transformed by the compiler to void f(double *a). You could have equivalently declared the function argument directly as *a. You could have even written a[100] or a[1], instead of a[10], since it is never actually compiled that way (however, you shouldn't do it obviously, it would confuse the reader).
Arrays are automatically passed by pointer in C. The rationale behind it can only be speculated.
int a[5], int *a and int (*a)[5] are all glorified addresses meaning that the compiler treats arithmetic and deference operators on them differently depending on the type, so when they refer to the same address they are not treated the same by the compiler. int a[5] is different to the other 2 in that the address is implicit and does not manifest on the stack or the executable as part of the array itself, it is only used by the compiler to resolve certain arithmetic operations, like taking its address or pointer arithmetic. int a[5] is therefore an array as well as an implicit address, but as soon as you talk about the address itself and place it on the stack, the address itself is no longer an array, and can only be a pointer to an array or a decayed array i.e. a pointer to the first member of the array.
For instance, on int (*a)[5], the first dereference on a will produce an int * (so the same address, just a different type, and note not int a[5]), and pointer arithmetic on a i.e. a+1 or *(a+1) will be in terms of the size of an array of 5 ints (which is the data type it points to), and the second dereference will produce the int. On int a[5] however, the first dereference will produce the int and the pointer arithmetic will be in terms of the size of an int.
To a function, you can only pass int * and int (*)[5], and the function casts it to whatever the parameter type is, so within the function you have a choice whether to treat an address that is being passed as a decayed array or a pointer to an array (where the function has to specify the size of the array being passed). If you pass a to a function and a is defined int a[5], then as a resolves to an address, you are passing an address, and an address can only be a pointer type. In the function, the parameter it accesses is then an address on the stack or in a register, which can only be a pointer type and not an array type -- this is because it's an actual address on the stack and is therefore clearly not the array itself.
You lose the size of the array because the type of the parameter, being an address, is a pointer and not an array, which does not have an array size, as can be seen when using sizeof, which works on the type of the value being passed to it. The parameter type int a[5] instead of int *a is allowed but is treated as int * instead of disallowing it outright, though it should be disallowed, because it is misleading, because it makes you think that the size information can be used, but you can only do this by casting it to int (*a)[5], and of course, the function has to specify the size of the array because there is no way to pass the size of the array because the size of the array needs to be a compile-time constant.
I might be so bold to think there are four (4) ways to pass an array as the function argument. Also here is the short but working code for your perusal.
#include <iostream>
#include <string>
#include <vector>
#include <cassert>
using namespace std;
// test data
// notice native array init with no copy aka "="
// not possible in C
const char* specimen[]{ __TIME__, __DATE__, __TIMESTAMP__ };
// ONE
// simple, dangerous and useless
template<typename T>
void as_pointer(const T* array) {
// a pointer
assert(array != nullptr);
} ;
// TWO
// for above const T array[] means the same
// but and also , minimum array size indication might be given too
// this also does not stop the array decay into T *
// thus size information is lost
template<typename T>
void by_value_no_size(const T array[0xFF]) {
// decayed to a pointer
assert( array != nullptr );
}
// THREE
// size information is preserved
// but pointer is asked for
template<typename T, size_t N>
void pointer_to_array(const T (*array)[N])
{
// dealing with native pointer
assert( array != nullptr );
}
// FOUR
// no C equivalent
// array by reference
// size is preserved
template<typename T, size_t N>
void reference_to_array(const T (&array)[N])
{
// array is not a pointer here
// it is (almost) a container
// most of the std:: lib algorithms
// do work on array reference, for example
// range for requires std::begin() and std::end()
// on the type passed as range to iterate over
for (auto && elem : array )
{
cout << endl << elem ;
}
}
int main()
{
// ONE
as_pointer(specimen);
// TWO
by_value_no_size(specimen);
// THREE
pointer_to_array(&specimen);
// FOUR
reference_to_array( specimen ) ;
}
I might also think this shows the superiority of C++ vs C. At least in reference (pun intended) of passing an array by reference.
Of course there are extremely strict projects with no heap allocation, no exceptions and no std:: lib. C++ native array handling is mission critical language feature, one might say.
Here is the code I'm having trouble to understand:
char* myPtr = "example";
myPtr[1] = 'x';
How am I allowed to use myPtr[1]? Why can I choose positions like a do on arrays? myPtr is not even an array.
Obs. I know about lookup table, literal pooling and string literals, my concern is just how this even compile. I don't use pointers that much.
Can anyone help?
Apparently you made an assumption that applicability of [] operator to something necessarily implies that that "something" is an array. This is not true. The built-in [] operator has no direct relation to arrays. The [] is just a shorthand for a combination of * and + operators: by definition a[b] means *(a + b), where one operand is required to be a pointer and another is required to be an integer.
Moreover, when you apply the [] operator to an actual array, that array gets implicitly converted to a pointer type first, and only then the resultant pointer can act as an operand of [] operator. This actually means the opposite of what you supposedly assumed initially: operator [] never works with arrays. By the time we get to the [] the array has already decayed to a pointer.
As a related side-note, this latter detail manifests itself in one obscure peculiarity of the first C language standard. In C89/90 the array-to-pointer conversion was not allowed for rvalue arrays, which also prevented the [] operator from working with such arrays
struct S { int a[10]; };
struct S foo(void) { struct S s = { 0 }; return s; }
int main()
{
foo().a[5];
/* ERROR: cannot convert array to pointer, and therefore cannot use [] */
return 0;
}
C99 expanded the applicability of that conversion thus making the above code valid.
It compiles according to §5.2.1/1 [expr.sub] of the C++ standard:
A postfix expression followed by an expression in square brackets is a postfix expression. One of the expressions shall have the type “array of T” or “pointer to T” and the other shall have unscoped enumeration or integral type. The result is of type “T”. The type “T” shall be a completely-defined object type.
The expression E1[E2] is identical (by definition) to *((E1)+(E2)), except that in the case of an array operand, the result is an lvalue if that operand is an lvalue and an xvalue otherwise.
Since "example" has type char const[8] it may decay to char const* (it used to decay to char* as well, but it's mostly a relict of the past) which makes it a pointer.
At which point the expression myPtr[1] becomes *(myPtr + 1) which is well defined.
Pointers hold the address of memory location of variables of specific data types they are assigned to hold. As others have pointed out its counter-intuitive approach take a bit of learning curve to understand.
Note that the string "example" itself is immutable however, the compiler doesn't prevent the manipulation of the pointer variable, whose new value is changed to address of string 'x' (this is not same as the address of x in 'example'),
char* myPtr = "example";
myPtr[1] = 'x';
Since myPtr is referencing immutable data when the program runs it will crash, though it compiles without issues.
From C perspective, here, you are dereferencing a mutable variable.
By default in C, the char pointer is defined as mutable, unless specifically stated as immutable through keyword const, in which case the binding becomes inseparable and hence you cannot assign any other memory address to the pointer variable after defining it.
Lets say your code looked like this,
const char *ptr ="example";
ptr[1] = 'x';
Now the compilation will fail and you cannot modify the value as this pointer variable is immutable.
You should use char pointer only to access the individual character in a string of characters.
If you want to do string manipulations then I suggest you declare an int to store each character's ASCII values from the standard input output like mentioned here,
#include<stdio.h>
int main()
{
int countBlank=0,countTab=0,countNewLine=0,c;
while((c=getchar())!=EOF)
{
if(c==' ')
++countBlank;
else if(c=='\t')
++countTab;
else if(c=='\n')
++countNewLine;
putchar(c);
}
printf("Blanks = %d\nTabs = %d\nNew Lines = %d",countBlank,countTab,countNewLine);
}
See how the integer takes ASCII values in order to get and print individual characters using getchar() and putchar().
A special thanks to Keith Thompson here learnt some useful things today.
The most important thing to remember is this:
Arrays are not pointers.
But there are several language rules in both C and C++ that can make it seem as if they're the same thing. There are contexts in which an expression of array type or an expression of pointer type is legal. In those contexts, the expression of array type is implicitly converted to yield a pointer to the array's initial element.
char an_array[] = "hello";
const char *a_pointer = "goodbye";
an_array is an array object, of type char[6]. The string literal "hello" is used to initialize it.
a_pointer is a pointer object, of type const char*. You need the const because the string literal used to initialize it is read-only.
When an expression of array type (usually the name of an array object) appears in an expression, it is usually implicitly converted to a pointer to its initial (0th) element. So, for example, we can write:
char *ptr = an_array;
an_array is an array expression; it's implicitly converted to a char* pointer. The above is exactly equivalent to:
char *ptr = &(an_array[0]); // parentheses just for emphasis
There are 3 contexts in which an array expression is not converted to a pointer value:
When it's the operand of the sizeof operator. sizeof an_array yields the size of the array, not the size of a pointer.
When it's the operand of the unary & operator. &an_array yields the address of the entire array object, not the address of some (nonexistent) char* pointer object. It's of type "pointer to array of 6 chars", or char (*)[6].
When it's a string literal used as an initializer for an array object. In the example above:
char an_array[] = "hello";
the contents of the string literal "hello" are copied into an_array; it doesn't decay to a pointer.
Finally, there's one more language rule that can make it seem as if arrays were "really" pointer: a parameter defined with an array type is adjusted so that it's really of pointer type. You can define a function like:
void func(char param[10]);
and it really means:
void func(char *param);
The 10 is silently ignored.
The [] indexing operator requires two operands, a pointer and an integer. The pointer must point to an element of an array object. (A standalone object is treated as a 1-element array.) The expression
arr[i]
is by definition equivalent to
*(arr + i)
Adding an integer to a pointer value yields a new pointer that's advanced i elements forward in the array.
Section 6 of the comp.lang.c FAQ has an excellent explanation of all this stuff. (It applies to C++ as well as to C; the two languages have very similar rules in this area.)
In C++, your code generates a warning during compile:
{
//char* myPtr = "example"; // ISO C++ forbids converting a string
// constant to ‘char*’ [-Wpedantic]
// instead you should use the following form
char myPtr[] = "example"; // a c-style null terminated string
// the myPtr symbol is also treated as a char*, and not a const char*
myPtr[1] = 'k'; // still works,
std::cout << myPtr << std::endl; // output is 'ekample'
}
On the other hand, std::string is much more flexible, and has many more features:
{
std::string myPtr = "example";
myPtr[1] = 'k'; // works the same
// then, to print the corresponding null terminated c-style string
std::cout << myPtr.c_str() << std::endl;
// ".c_str()" is useful to create input to system calls requiring
// null terminated c-style strings
}
The semantics of abc[x] is "Add x*sizeof(type)" to abc where abc is any memory pointer. Arrays variable behave like memory pointers and they just point to beginning of the memory location allocated to array.
Hence adding x to array or pointer variable both will point to memory which is same as variable pointing to + x*sizeof(type which array contains or pointer points to, e.g. in case of int pointers or int array it's 4)
Array variables are not same as pointer as said in comment by Keith as array declaration will create fix sized memory block and any arithmetic on that will use size of array not the element types in that array.
I recently embarrassed myself while explaining to a colleague why
char a[100];
scanf("%s", &a); // notice a & in front of 'a'
is very bad and that the slightly better way to do it is:
char a[100];
scanf("%s", a); // notice no & in front of 'a'
Ok. For everybody getting ready to tell me why scanf should not be used anyway for security reasons: ease up. This question is actually about the meaning of "&a" vs "a".
The thing is, after I explained why it shouldn't work, we tried it (with gcc) and it works =)). I ran a quick
printf("%p %p", a, &a);
and it prints the same address twice.
Can anybody explain to me what's going on?
Well, the &a case should be obvious. You take the address of the array, exactly as expected.
a is a bit more subtle, but the answer is that a is the array. And as any C programmer knows, arrays have a tendency to degenerate into a pointer at the slightest provocation, for example when passing it as a function parameter.
So scanf("%s", a) expects a pointer, not an array, so the array degenerates into a pointer to the first element of the array.
Of course scanf("%s", &a) works too, because that's explicitly the address of the array.
Edit: Oops, looks like I totally failed to consider what argument types scanf actually expects. Both cases yield a pointer to the same address, but of different types. (pointer to char, versus pointer to array of chars).
And I'll gladly admit I don't know enough about the semantics for ellipsis (...), which I've always avoided like the plague, so looks like the conversion to whichever type scanf ends up using may be undefined behavior. Read the comments, and litb's answer. You can usually trust him to get this stuff right. ;)
Well, scanf expects a char* pointer as the next argument when seeing a "%s". But what you give it is a pointer to a char[100]. You give it a char(*)[100]. It's not guaranteed to work at all, because the compiler may use a different representation for array pointers of course. If you turn on warnings for gcc, you will see also the proper warning displayed.
When you provide an argument object that is an argument not having a listed parameter in the function (so, as in the case for scanf when has the vararg style "..." arguments after the format string), the array will degenerate to a pointer to its first element. That is, the compiler will create a char* and pass that to printf.
So, never do it with &a and pass it to scanf using "%s". Good compilers, as comeau, will warn you correctly:
warning: argument is incompatible with corresponding format string conversion
Of course, the &a and (char*)a have the same address stored. But that does not mean you can use &a and (char*)a interchangeably.
Some Standard quotes to especially show how pointer arguments are not converted to void* auto-magically, and how the whole thing is undefined behavior.
Except when it is the operand of the sizeof operator or the unary & operator, or is a
string literal used to initialize an array, an expression that has type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points to the initial element of the array object. (6.3.2.1/3)
So, that is done always - it isn't mentioned below explicitly anymore when listening valid cases when types may differ.
The ellipsis notation in a function prototype declarator causes argument type conversion to stop after the last declared parameter. The default argument promotions are performed on trailing arguments. (6.5.2.2/7)
About how va_arg behaves extracting the arguments passed to printf, which is a vararg function, emphasis added by me (7.15.1.1/2):
Each invocation of the va_arg macro modifies ap so that the
values of successive arguments are returned in turn. The parameter type shall be a type
name specified such that the type of a pointer to an object that has the specified type can be obtained simply by postfixing a * to type. If there is no actual next argument, or if type is not compatible with the type of the actual next argument (as promoted according to the default argument promotions), the behavior is undefined, except for the following cases:
one type is a signed integer type, the other type is the corresponding unsigned integer
type, and the value is representable in both types;
one type is pointer to void and the other is a pointer to a character type.
Well, here is what that default argument promotion is:
If the expression that denotes the called function has a type that does not include a
prototype, the integer promotions are performed on each argument, and arguments that
have type float are promoted to double. These are called the default argument
promotions. (6.5.2.2/6)
It's been a while since I programmed in C but here's my 2c:
char a[100] doesn't allocate a separate variable for the address of the array, so the memory allocation looks like this:
---+-----+---
...|0..99|...
---+-----+---
^
a == &a
For comparison, if the array was malloc'd then there is a separate variable for the pointer, and a != &a.
char *a;
a = malloc(100);
In this case the memory looks like this:
---+---+---+-----+---
...| a |...|0..99|...
---+---+---+-----+---
^ ^
&a != a
K&R 2nd Ed. p.99 describes it fairly well:
The correspondence between indexing
and pointer arithmetic is very close.
By definition, the value of a variable
or expression of type array is the
address of element zero of the array.
Thus after the assignment pa=&a[0];
pa and a have identical values. Since
the name of the array is a synonym for
the location of the initial element,
the assignment pa=&a[0] can also be
written as pa=a;
A C array can be implicitly converted to a pointer to its first element (C99:TC3 6.3.2.1 §3), ie there are a lot of cases where a (which has type char [100]) will behave the same way as &a[0] (which has type char *). This explains why passing a as argument will work.
But don't start thinking this will always be the case: There are important differences between arrays and pointers, eg regarding assignment, sizeof and whatever else I can't think of right now...
&a is actually one of these pitfalls: This will create a pointer to the array, ie it has type char (*) [100] (and not char **). This means &a and &a[0] will point to the same memory location, but will have different types.
As far as I know, there is no implicit conversion between these types and they are not guaranteed to have a compatible representation as well. All I could find is C99:TC3 6.2.5 §27, which doesn't says much about pointers to arrays:
[...] Pointers to other types need not have the same representation or alignment requirements.
But there's also 6.3.2.3 §7:
[...] When a pointer to an object is converted to a pointer to a character type, the result points to the lowest addressed byte of the object. Successive increments of the result, up to the size of the object, yield pointers to the remaining bytes of the object.
So the cast (char *)&a should work as expected. Actually, I'm assuming here that the lowest addressed byte of an array will be the lowest addressed byte of its first element - not sure if this is guaranteed, or if a compiler is free to add arbitrary padding in front of an array, but if so, that would be seriously weird...
Anyway for this to work, &a still has to be cast to char * (or void * - the standard guarantees that these types have compatible representations). The problem is that there won't be any conversions applied to variable arguments aside from the default argument promotion, ie you have to do the cast explicitly yourself.
To summarize:
&a is of type char (*) [100], which might have a different bit-representation than char *. Therefore, an explicit cast must be done by the programmer, because for variable arguments, the compiler can't know to what it should convert the value. This means only the default argument promotion will be done, which, as litb pointed out, does not include a conversion to void *. It follows that:
scanf("%s", a); - good
scanf("%s", &a); - bad
scanf("%s", (char *)&a); - should be ok
Sorry, a tiny bit off topic:
This reminded me of an article I read about 8 years ago when I was coding C full time. I can't find the article but I think it was titled "arrays are not pointers" or something like that. Anyway, I did come across this C arrays and pointers FAQ which is interesting reading.
char [100] is a complex type of 100 adjacent char's, whose sizeof equals to 100.
Being casted to a pointer ((void*) a), this variable yields the address of the first char.
Reference to the variable of this type (&a) yields address of the whole variable, which, in turn, also happens to be the address of the first char