I am just learning C++ and a little confused about arrays and references. I wrote the following program:
void printArray(int arr[]) {
cout << arr[0] << arr[1] << arr[2] << endl;
}
int main() {
int arr[3] = {5, 7, 9};
int *aPtr = &arr[0];
cout << aPtr[0] << aPtr[1] << aPtr[2] << endl;
int *bPtr = arr;
cout << bPtr[0] << bPtr[1] << bPtr[2] << endl;
printArray(arr);
}
The output is :
579
579
579
And I have two questions:
Is array in C++ a reference? I mean, can I state that i = &i[0]?
If answer to the first question is yes, can I say that as array is reference we don't need to use & in arguments of the function printArray. I mean, we don't declare the function this way printArray(int &arr[])?
No, an array is not a reference in C++. It is an array, the length of which forms part of the type (so for example the type of int[3] is not the same as that of int[42]).
What can be confusing is that C++ inherits from C the strange features that
array function parameters have their type "adjusted" to pointer
array names can "decay" to pointers very easily. That makes it possible to assign an array to a pointer.
Point 1 above means that these two function declarations are completely equivalent:
// two ways do declaring the same function
void foo(int a[42]);
void foo(int* a);
and point 2 means you can call the function passing it an array:
int a[3] = {};
int b[100] = {};
foo(a);
foo(b);
and other funny stuff, for example, the type of expression +a being int*.
Related
int q[10]={0};
cout << q << endl;
cout << &q << endl;
cout << &q[0] << endl;
output is
0x7fffd4d2f860
0x7fffd4d2f860
0x7fffd4d2f860
Now when i do this->
int *r=q; // allowed
int *r=&q[0] // allowed
int *r=&q // not allowed
Why is the third assignment not allowed when it is essentially the same thing?
If you have an array declared like
T a[N];
where T is some type specifier then a pointer to the array will be declared like
T ( *p )[N] = &a;
A general rule is the following. If you have a multidimensional array (including one-dimensional arrays) like for example
T a[N1][N2][N3];
then this declaration you may rewrite like
T ( a[N1] )[N2][N3];
To get a pointer to the first element of the array just substitute the content in the parentheses the following way
T ( *p )[N2][N3] = a;
If you want to get a pointer to the whole array then rewrite the declaration of the array like
T ( a )[N1][N2][N3];
and make the substitution
T ( *p )[N1][N2][N3] = &a;
Compare this with a declaration of a scalar object and a pointer to it.
For example
T obj;
You may rewrite the declaration like
T ( obj );
Now to get a pointer to the object you can write
T ( *p ) = &obj;
Of course in this case the parentheses are redundant and the above declaration is equivalent to
T *p = &obj;
As for this code snippet
int q[10]={0};
cout << q << endl;
cout << &q << endl;
cout << &q[0] << endl;
and its output
0x7fffd4d2f860
0x7fffd4d2f860
0x7fffd4d2f860
then array designators used in expressions with rare exceptions are converted to pointers to their first elements.
So in fact the two expression q and &q[0] in these statements
cout << q << endl;
cout << &q[0] << endl;
are equivalent. On the other hand, the address of the array itself is the address of the memory extent that the array occupies. And in the beginning of the extent there is the first element of the array. So the three expressions give the same result: the address of the extent of memory occupied by the array.
Why is the third assignment not allowed when it is essentially the same thing?
Because The C++ language has a feature called "type safety". There is a type system that helps you keep the logic of your program sound.
One particular rule is that arbitrary pointer types can not be used to initialise pointers of other, incompatible types. In this case, you have a pointer to type int (.i.e. int*) that you try to initalise with an expression of type pointer to type array of 10 int (i.e. int(*)[10]). One type is not implicitly convertible to the other, hence the program is ill-formed.
Then why does cout print same things in all of the three cases?
Because all of the pointers have the same value. The first byte of the first element of the array is the same byte as the first byte of the entire array.
It just so happens that the stream insertion operator handles all pointer types1 exactly the same, so pointers with same value but different type produce the same output.
1 Pointers to character types are an exception. They are treated entirely differently.
Why can't we assign address of array to pointer?
Actually, we can assign address of an array to a pointer. We just cannot assign address of a array (or any other object for that matter) to a pointer of wrong type. We need a pointer to an array in this case:
int (*r)[10] = &q;
You cannot do the third assignment because &q's type is an int (*)[10], which is incompatible with the type of int* r.
The output of cout << &q does not reveal the type of &q. See this documentation link.
q is a fixed-length array. Specifying q by itself in an expression decays into a pointer to the 1st element of q. Thus, q decays to the same pointer value that &q[0] returns. &q, on the other hand, returns the memory address of the q variable itself, and for an array, its 1st element occupies that same memory address.
There is an operator<<(void*) defined for std::ostream, and void* can accept (almost) ANY type of pointer. Since all three of your cout calls resolve to the same memory address, and there is an operator<< that accepts all three types of pointers, that is why all three calls print the same number.
As for your assignments:
q is an int[10], which decays into an int*, which is why int *r=q; works.
&q[0] dereferences q to access its 1st element, which is an int, and then takes the address of that element, producing an int*, which is why int *r=&q[0]; works.
since q is an int[10], &q is an int(*)[10], which DOES NOT decay into an int*, which is why int *r=&q; does not work. You would have to declare r using the correct type:
int (*r)[10] = &q;
The following code demonstrates the differences between arr, &arr and &arr[0] where arr is an array of integers.
#include <iostream>
#include <string_view>
// Reference: https://stackoverflow.com/questions/81870/is-it-possible-to-print-a-variables-type-in-standard-c/56766138#56766138
// Type finding code start
template <typename T>
constexpr auto type_name()
{
std::string_view name, prefix, suffix;
#ifdef __clang__
name = __PRETTY_FUNCTION__;
prefix = "auto type_name() [T = ";
suffix = "]";
#elif defined(__GNUC__)
name = __PRETTY_FUNCTION__;
prefix = "constexpr auto type_name() [with T = ";
suffix = "]";
#elif defined(_MSC_VER)
name = __FUNCSIG__;
prefix = "auto __cdecl type_name<";
suffix = ">(void)";
#endif
name.remove_prefix(prefix.size());
name.remove_suffix(suffix.size());
return name;
}
// Type finding code end
int main()
{
int arr[5] = {1, 2, 3, 4, 5};
std::cout << "Value: " << arr << "\tType: " << type_name<decltype(arr)>() << std::endl;
std::cout << "Value: " << &arr << "\tType: " << type_name<decltype(&arr)>() << std::endl;
std::cout << "Value: " << &arr[0] << "\tType: " << type_name<decltype(&arr[0])>() << std::endl;
return 0;
}
See it in action.
Sample Output:
Value: 0x7ffcdadd22d0 Type: int [5]
Value: 0x7ffcdadd22d0 Type: int (*)[5]
Value: 0x7ffcdadd22d0 Type: int*
While all three are associated with the same memory location, they are of different types.
Reference: Is it possible to print a variable's type in standard C++?
This question already has answers here:
Passing Arrays to Function in C++
(5 answers)
What is array to pointer decay?
(11 answers)
Closed 4 months ago.
I'm sorry for my bad English first.
I've encountered a strange problem when coding in C++.
using namespace std;
void Func(int a[2][3])
{
cout <<(int) &a;
}
int main()
{
int a[2][3] =
{
{1,2,3},
{4,5,6}
};
cout << (int)&a << endl;
Func(a);
return 0;
}
I was confused that &a in main() and in function Func() returned different values. And strangely, the difference between them always is 212.
Can anyone explain please? Thank you for your help.
P/s:Thank you all for your answer .My teacher says that C++ doesn't allow passing an array by value, because if the array has 1 million elements, that would decrease the performance a lot only for copying all of them, so he says only pass by reference is allowed. That's what make me think those two &a should be the same. Now I get it, thank you everyone!
Your function declaration
void Func(int a[2][3])
is completely equivalent and interchangeable with:
void Func(int (*a)[3]).
As you can see, you are passing a pointer to an array of three ints by value. Therefore the address of the local function parameter is different from the address of the variable in main, even if they may hold the same value.
You're passing the a argument by value, so each function has its own copy of it (of the pointer, not the array data). The constant offset you're seeing comes from the distance between the stack frames of the two functions, and this is constant.
If you change the function to get a reference to the array (void Func(int (&a)[2][3]) you will get the same value in both cases
The parameter and the local variable are distinct objects and since their lifetimes overlap, they must have distinct memory addresses.
Please read about pass by value and pass by references.
So what happened here is:
you initialised an array in main function. &a will refer to the address of a.
you passed as a pass by value argument to another function. A copy of a is created to be consumed in Func and &a will refer to the memory location of a local to Func.
I hope the concept is clear.
Use the following syntax to pass arrays by (const) reference : const int (&a)[2][3]
#include <iostream>
void func(const int (&a)[2][3])
{
for (const auto& row : a)
{
for(const auto& value : row )
{
std::cout << value << " ";
}
std::cout << "\n";
}
}
int main()
{
int a[2][3] =
{
{1,2,3},
{4,5,6}
};
func(a);
return 0;
}
This is because C rules on how pointers and arrays work are a little weird. You're actually taking the address of a pointer to the array, not the actual address of the array. If you want to get the address to the array you need to take the address of the first element instead:
&a[0]
For starters it is a bad idea to cast an address to the type int like
cout << (int)&a << endl;
you could just write
cout << static_cast<void *>( a ) << endl;
or
cout << static_cast<void *>( &a ) << endl;
Or even like
cout << a << endl;
cout << &a << endl;
though with static_cast the code looks more readable.
The both statements will output the same value: the address of the extent of memory occupied by the array.
In this function call
Func(a);
the array designator is implicitly converted to pointer to its first element of the type int( * )[3].
The value of the pointer expression is assigned to the local variable (parameter) a of the function Func.
The function parameter of the pointer type a and the array a defined in main occupy different extents of memory.
If to rename the function parameter as for example
void Func(int b[2][3])
to distinguish it from the array with the same name defined in main then you may imagine the function call the following way
Func(a);
//...
void Func( /*int b[2][3] */ )
{
int ( *b )[3] = a;
cout << static_cast<void *>( &b );
}
Pay attention to that the function parameter declared as having the array type int[2][3] is implicitly adjusted by the compiler to pointer to the array element type that is int ( * )[3].
So as you can see this statement
cout << static_cast<void *>( &b );
outputs the address of the local variable (parameter) b.
If you want to get the address of the array a within the function Func then you should write
cout << static_cast<void *>( b );
In this case the addresses outputted in main and in the function will coincide because the parameter b stores the address of the first element of the array a.
Here is a demonstration program.
#include <iostream>
void Func( int a[2][3] )
{
std::cout << static_cast< void * >( a ) << '\n';
}
int main()
{
int a[2][3] =
{
{1,2,3},
{4,5,6}
};
std::cout << static_cast<void *>( a ) << '\n';
std::cout << static_cast<void *>( &a ) << '\n';
Func( a );
}
The program output might look like
010FFD08
010FFD08
010FFD08
As you can see the three values are equal each other.
But if you will write in the function Func
std::cout << static_cast< void * >( &a ) << '\n';
^^^^
you will get the address of the local variable (parameter) a of the function. It is evident that this address differs from the address of the extent of memory occupied by the array because for the parameter a there was allocated a separate extent of memory the size of which is equal tp the value of the sizeof( int( * )[3] ) and usually this value is equal to 4 or 8.
I am wondering how come the # number1 code not working
as I am trying to use increment operator to display the next following element in the array.
But the # number2 code works , and it was the same code but in a function
//# number 1 code
using namespace std;
int main(){
int arrays[5]={2,4,6,8,10};
for(int x=0;x<5;x++){
cout<<*arrays<<endl;
arrays++; //error: lvalue required as increment operand
}
}
//# number 2 code
using namespace std;
void display(int *arr,int size){
for(int x=0; x<5;x++){
cout<<*arr<<endl;
arr++; //This time no error!!!
}
}
int main(){
int arrays[5]={2,4,6,8,10};
display(arrays,5);
return 0;
}
That's because you cannot change the address of an array.
In # number 1 code when you do array++, you are actually trying to operate directly on the variable which is storing the base address of the array.
What you can try instead is something like below:
int *p = array;
p++;
Whereas in the case when you are calling a function passing the array's base address # number 2, you are implicitly doing what has been shown in the above code snippet.
This is a common problem for beginners. Arrays are not pointers!. Arrays are implicitly converted to pointers. That is where the confusion lies. Consider this:
int array[] = {1, 2, 3};
std::cout << *array << '\n';
What do you think is happening when we do *array. Does it really make sense to dereference an array? The array is being implicitly converted to a int * and then dereferenced. What about this:
int array[] = {1, 2, 3};
array++;
std::cout << *array << '\n';
This doesn't compile (as you found out for yourself). In this statement array++, the array is not implicitly converted to a pointer.
Arrays are converted to pointers when you pass them to functions that accept pointers. That makes it possible to do this:
int array[3] = {1, 2, 3};
display(array, 3);
An array is a sequence of objects stored on the stack. You access this sequence of objects as a pointer to the first object. Both arrays and pointers can be subscripted. They share many similarities but are fundamentally different.
To make your first example compile, subscript the array with x:
for (int x = 0; x < 5; x++) {
std::cout << arrays[x] << '\n';
}
Use :
int *arr = arrays;
arr++;
in code #1. It will work. This is because you need first to create a pointer to the base of the array which you can increment as in the second code, you have the pointer in the form of the passed argument to the function.
I wrote a simple piece of C++ code to pass addresses by reference.
I am passing the address of a variable (say y) and an array (say arr) to a class. Both arr and y will get modified inside the class. I want to have the modified values in my main().
Please find my question in the below piece of code as it is easier that way. Thanks.
#include <iostream>
using namespace std;
class A
{
public:
// Assign values to the array and increment _x.
void increment()
{
(*_x)++;
(*_arr)[0] = 1; // Q1. Is it safe to directly access the array like this.
(*_arr)[1] = 2; // Don't I have to allocate memory to the array ?
(*_arr)[2] = 3;
}
// Get the address of the Variable that is passed in main. x will now have &y2.
A (int* &arr, int* &x):
_x(x)
{
*_arr = arr;
}
private:
int* _x;
int** _arr;
};
int main()
{
int y = 9;
int arr[5];
int *pY = &y;
int *pArr = arr;
A *obj1 = new A(pArr, pY);
// This gives a compile time error. warning: initialization of non-const reference int *&' from rvalue `int *'
// A *obj1 = new A(&y); <-- Q2. Why does this give a Compile Time Error ?
obj1->increment();
cout << "y : " << y << endl;
cout << "[0]: " << arr[0] << "; [1]: " << arr[1] << "; [2]: " << arr[2] << endl;
cout << endl;
return 0;
}
In A::increment() function, I am directly assigning values to the array without
allocating memory. Is it safe to do ? If not, how can I allocate memory so that
I can still get the modified array values in main() ?
Why do I get a compile time error whey I pass &y to A's constructor ?
Thanks in advance.
Question 1
In A::increment() function, I am directly assigning values to the array without allocating memory. Is it safe to do ? If not, how can I allocate memory so that I can still get the modified array values in main() ?
Answer
Yes, it is safe.
Question 2
Why do I get a compile time error whey I pass &y to A's constructor ?
Answer
&y is not an lvalue. Hence, it cannot be used where the argument type is int*&.
Problem in posted code
*_arr = arr;
That is a problem since _arr has not been initialized to point to a valid memory. Using *_arr when _arr has not been initialized causes undefined behavior. You can change the type of _arr to:
int* _arr;
and simplify your code a little bit.
class A
{
public:
// Assign values to the array and increment _x.
void increment()
{
(*_x)++;
_arr[0] = 1; // Q1. Is it safe to directly access the array like this.
_arr[1] = 2; // Don't I have to allocate memory to the array ?
_arr[2] = 3;
}
// Get the address of the Variable that is passed in main. x will now have &y2.
A (int* &arr, int* &x):
_x(x),
_arr(arr)
{
}
private:
int* _x;
int* _arr;
};
without changing anything in main.
This is very rarely what you want; a T** is generally an array of arrays or else a pointer value that you want to modify in the caller’s scope. However, neither seems to be what you’re doing here.
It is safe to modify *_arr[0] if and only if _arr has been initialized to a array of non-const arrays, and (*_arr)[0] if and only if it has been initialized as a pointer to a non-const array. Neither appears to be the case here, but if it is, you probably want to give the array length explicitly.
In this example, &y is a constant. You can’t modify it, so you can’t pass it as a non-const variable. You can declare a pointer int *py = &y; and pass that. But consider whether that’s what you want to do.
By the way, it’s not good style to use identifiers that start with underscores, because by the standard, they’re reserved for the compiler to use.
You should tell us what you are trying to do. In my opinion it's nonsense using raw pointers/arrays and naked new/(missing?) delete in C++ without good reason. I would also like to note that it is not considered good practice using the _prefix for class members. Usually leading _ are used for std implementations. I recommend using m_prefix if you insist on one. And why do you give _arr the type int**? Is is supposed to be a 2D-Array? Additionally, it doesn't really make sense passing a pointer by reference. A pointer is already a pointer, if you know what I mean, just pass the pointer around.
I'm just going to assume that you are doing this to understand manual memory management or pointer arithmetics or - wait, right: Tell us what you are trying to do and why. Nevertheless, I don't understand what you have the class for:
#include <iostream>
void increment(int& x, int *arr, int sz)
{
++x;
for (int i = 0; i != sz; ++i)
{
// this just numbers the values respectively (starting at 1)
arr[i] = i + 1;
}
}
int main()
{
using namespace std;
int y = 9;
const int sz = 5;
int arr[sz];
increment(y, arr, sz);
cout << "y : " << y << '\n'
<< "[0]: " << arr[0] << "; [1]: " << arr[1] << "; [2]: " << arr[2] << "\n\n";
}
To answer your questions:
2. First thing first: I don't see any constructor that only takes one argument.
Read up on "Undefined Behaviour (UB)" starting point: What are all the common undefined behaviours that a C++ programmer should know about?
I can't repeat enough that I don't understand what you are going for and that makes it hard to give solid advice.
I tried fixing your version.. well its still terrible... I highly recommend on reading up on std::array, std::vector. Maybe on pointers, C-Style Arrays and how to pass C-Style Arrays as function arguments (note: for regular C++ programming you wouldn't be doing/using that, usually).
#include <iostream>
class A {
public:
// Assign values to the array and increment m_x.
void increment()
{
++(*m_x);
m_arr[0] = 1;
m_arr[1] = 2;
m_arr[2] = 3;
}
A (int* arr, int* x):
m_x(x), m_arr(arr)
{
}
private:
int* m_x;
int* m_arr;
};
int main()
{
using namespace std;
int y = 9;
int arr[5];
A obj1(arr, &y);
obj1.increment();
cout << "y : " << y << '\n'
<< "[0]: " << arr[0] << "; [1]: " << arr[1] << "; [2]: " << arr[2] << "\n\n";
A obj2(arr, &y);
obj2.increment();
cout << "y : " << y << '\n'
<< "[0]: " << arr[0] << "; [1]: " << arr[1] << "; [2]: " << arr[2] << "\n\n";
}
You should also read up un pointers/references and their differences
I am actually trying to make your programming life easier. Sorry for long answer.
In answer to your first question
In A::increment() function, I am directly assigning values to the
array without allocating memory. Is it safe to do ? If not, how can I
allocate memory so that I can still get the modified array values in
main() ?
you allocated memory in main(), in the line
int arr[5];
In terms of class design, you defined your class constructor to accept reference arguments, which means that an existing int* must be passed to each argument:
A (int* &arr, int* &x)
and you do so when you invoke the constructor:
A *obj1 = new A(pArr, pY);
so in this program, what you are doing is safe. A potential danger if you expect to use this class in another context would be if your arr array in main() contained fewer than 3 elements, since your increment() function initializes the third element of the array.
In answer to your second question
Why do I get a compile time error whey I pass &y to A's constructor ?
In your original constructor,
// Get the address of the Variable that is passed in main. x will now have &y2.
A (int* &arr, int* &x):
_x(x)
{
*_arr = arr;
}
you are dereferencing _arr before it has been initialized. One way to solve this would be to do this:
// Get the address of the Variable that is passed in main. x will now have &y2.
A (int* &arr, int* &x):
_x(x)
{
_arr = new (int*);
*_arr = arr;
}
// Destructor
~A ()
{
delete _arr;
}
As an aside, you also use new in main(). Whenever you use new, you should also use delete to avoid a memory leak. So at the bottom of your program, before the return statement, add the following:
delete obj1;
I'm teaching myself C++ and had some questions about arrays and pointers. My understanding is that arrays are really just pointers, however, arrays are address constants which cannot be changed.
If this is the case, I was wondering why in my function show2() I was able to change the address of the pointer list. Unlike variables, I thought arrays are passed by reference so I was expecting a compiler error when calling function show2() since I incremented the address of list. But the code works just fine. Can someone please explain?
Thank you!
#include<iostream>
#include<iomanip>
using namespace std;
void show1(double *list, int SIZE)
{
for(int i=0; i < SIZE; i++)
{
cout << setw(5) << *(list+i);
}
cout << endl;
return;
}
void show2(double *list, int SIZE)
{
double *ptr = list;
for(int i=0; i < SIZE; i++)
cout << setw(5) << *list++;
cout << endl;
return;
}
int main()
{
double rates[] = {6.5, 7.2, 7.5, 8.3, 8.6,
9.4, 9.6, 9.8, 10.0};
const int SIZE = sizeof(rates) / sizeof(double);
show1(rates, SIZE);
show2(rates, SIZE);
return 0;
}
My understanding is that arrays are really just pointers
Let's get that out of the way. No, arrays are not pointers. Arrays are a series of objects, all of the same type, contiguous in memory.
Arrays can be passed by reference, but that is not what is usually done. What is usually done, which is what you are doing, is passing a pointer to the first element of the array. Arrays can and will "decay" to a pointer to their first element upon demand. And that's what is happening when you pass rates to show1 and show2.
Inside show1 and show2, list starts out as a pointer to rates[0]. You're free to modify this pointer to point at any other double.
If you wanted to pass an array by reference, it would look like this:
void show3(double (&list)[9]) { ... }
Or the more versatile:
template<size_t SIZE>
void show3(double (&list)[SIZE]) { ... }
Note that what you can't do is pass an array by value (unless it is contained within another object). If you write a function which looks like it is taking an array by value, e.g.
void show4(double list[9]) { ... }
It is actually a pointer, and that number 9 is meaningless. Native arrays suck.
First, arrays are converted to a pointer to the first element when passed as the function argument. BTW, arrays are not pointers, as one example, sizeof(rates) in your code isn't the size of a pointer.
Second, arrays are passed by value since you are not using references.
So in the function show2, you are modifying a pointer, which is fine.
Arrays are not pointers. C++ has inherited "Array-Pointer Equivalence" from C which means that a well-known array variable can decay to a pointer, primarily for the purpose of offset math and for avoiding passing arrays by value:
int array[64];
int* a = array; // equivalent to a = &array[0];
Array's aren't pointers. If you use an array variable name in a pointer context, it will "decay" to a pointer - that is, lose the extended attributes available from an array object.
int array[64];
int* a = array;
std::cout << "array size = " << sizeof(array) << "\n";
std::cout << "a size = " << sizeof(a) << "\n";
std::cout << "(int*)(array) size = " << sizeof((int*)array)) << "\n";
"Array size" will be 256 (int is 4 bytes, 64 of them = 256 bytes), "a size" will be 4 or 8 bytes depending on 32/64 bits, and "(int*)(array)" size will be the same size as the pointer.
People often think that arrays are passed by value. This is not true: http://ideone.com/hAeH18
#include <iostream>
void bump(int arr[3]) {
for (size_t i = 0; i < 3; ++i)
arr[i]++;
}
int main() {
int array[] = { 1, 2, 3 };
bump(array);
for (size_t i = 0; i < 3; ++i)
std::cout << array[i] << "\n";
return 0;
}
This outputs "2, 3, 4" not "1, 2, 3".
This occurs because arrays decay to pointers when passed as function arguments. But to support the syntax for receiving arrays as arrays, C has to be able to treat pointers like arrays in some contexts:
void f1(int* a) { a[0]++; }
void f2(int* a) { (*a)++; }
void f3(int a[]) { a[0]++; }
void f4(int a[]) { (*a)++; }
void f5(int a[1]) { a[0]++; }
void f6(int a[1]) { (*a)++; }
All of these functions produce the same code.
In C, this originated from the fact that array information is lost at compile time. So this function:
void f(int array[])
has no way to tell how large the array it is receiving is. They wanted programmers to be conscious of this and be careful about how/if they passed size information - e.g. in the case of char arrays, instead of size, we have the nul terminator byte.
Unfortunately they didn't choose to make it obvious by diasllowing the representation that makes it look like you are receiving an array with size information intact :(