What is the difference when array is declared as array[n] or as pointer array* according to example below? I guess that for example both 'a' and 'c' point at the first element of array, but they behave different.
#include <iostream>
int main() {
int a[3] = {1};
int b[5];
std::cout << *a << std::endl; //prints 1 - ok
//a = b; //error during compilation
int* c = new int[3];
c[0] = 2;
int* d = new int[5];
std::cout << *c << std::endl; //prints 2 - ok
c = d; //works ok!
return 0;
}
Long story short - they are essentially the same, but ever so slightly different.
From what I've gathered from http://c-faq.com/aryptr/aryptr2.html , whilst they can both act as a pointer to the front of an array, when you declare an array as
int a[3];
you are essentially binding the size of '3' to your variable a, along with the fact it's an array. Hence, when you try to assign b, of size 5 to a, you get a compilation error.
In contrast, when you write
int * a;
You are merely saying 'this is a pointer that may point to an array', with no promise on the size.
Subtle, isn't it?
The difference between the following two lines:
int g[10];
and
int* h = new int[10];
is that the second is dynamically-allocated, whereas the first is statically-allocated.
For most purposes, they are identical, but where in memory they end up living is different.
Related
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 am having trouble with a function that takes a pointer to a fixed array. I have a simple pointer and the compiler will not allow me to static_cast it to the pointer-to-array type.
Here's some example code:
int main()
{
typedef int (*Arr3)[3];
int a[3] = {1,2,3};
int* p = &a[0];
Arr3 b = static_cast<Arr3>(p);
}
Error message:
prog.cpp:11:10: error: static_cast from 'int *' to 'Arr3' (aka 'int (*)[3]') is not allowed
Arr3 b = static_cast<Arr3>(p);
^~~~~~~~~~~~~~~~~~~~
1 error generated.
I am pretty sure I could use a reinterpret_cast, but is that really my only option here? Or am I missing something?
You can't exactly static_cast here, but there is a trick that’s more type-safe than reinterpret_cast:
#include <iostream>
using std::cout;
typedef int (*arr3p)[3];
typedef int arr3[3];
inline arr3& to_arr3(arr3 p)
{
return *(arr3p)(p);
}
inline arr3p to_arr3p(arr3 p)
{
return (arr3p)(p);
}
int main() {
arr3 a = {1, 2, 3};
int *p = &a[0];
arr3p foo = &to_arr3(p);
cout << (*foo)[0] << ", ";
arr3p bar = to_arr3p(p);
cout << (*bar)[1] << ", ";
arr3& baz = to_arr3(p);
cout << baz[2] << std::endl;
return 0;
}
I believe your cast is incorrect
Arr3 b = static_cast<Arr3>(p);
should be
Arr3 b = static_cast<Arr3>(&a);
In fact, the cast is unnecessary. Simply
Arr3 b = &a;
will do.
One-dimension array is compatible with pointer, but that's not true for multi-dimension array. For example,
int *p = (int*)1;
int (*s)[3] = (int (*)[3])2;
printf("%d,%d\n", sizeof(int*), sizeof(int (*)[3]));
printf("%d,%d\n", ++p,++s);
The output is:
8,8
5,14
They are both pointer type, so their size is 8 bytes int 64-bit machine.++s, this array pointer will advance 3 elements, but ++p, p will only advance 1 element. When we declare pointer array, we must specify size in every dimension except 1th dimension. The compiler needs to known that information to compute pointer arithmetic operation.
C++11 code:
int a[3];
auto b = a; // b is of type int*
auto c = &a; // c is of type int(*)[1]
C code:
int a[3];
int *b = a;
int (*c)[3] = &a;
The values of b and c are the same.
What is the difference between b and c? Why are they not the same type?
UPDATE: I changed the array size from 1 to 3.
The sizeof operator should behave differently, for one, especially if you change the declaration of a to a different number of integers, such as int a[7]:
int main()
{
int a[7];
auto b = a;
auto c = &a;
std::cout << sizeof(*b) << std::endl; // outputs sizeof(int)
std::cout << sizeof(*c) << std::endl; // outputs sizeof(int[7])
return 0;
}
For me, this prints:
4
28
That's because the two pointers are very different types. One is a pointer to integer, and the other is a pointer to an array of 7 integers.
The second one really does have pointer-to-array type. If you dereference it, sure, it'll decay to a pointer in most cases, but it's not actually a pointer to pointer to int. The first one is pointer-to-int because the decay happened at the assignment.
Other places it would show up is if you really did have two variables of pointer-to-array type, and tried to assign one to the other:
int main()
{
int a[7];
int b[9];
auto aa = &a;
auto bb = &b;
aa = bb;
return 0;
}
This earns me the error message:
xx.cpp: In function ‘int main()’:
xx.cpp:14:8: error: cannot convert ‘int (*)[9]’ to ‘int (*)[7]’ in assignment
aa = bb;
This example, however, works, because dereferencing bb allows it to decay to pointer-to-int:
int main()
{
int a;
int b[9];
auto aa = &a;
auto bb = &b;
aa = *bb;
return 0;
}
Note that the decay doesn't happen on the left side of an assignment. This doesn't work:
int main()
{
int a[7];
int b[9];
auto aa = &a;
auto bb = &b;
*aa = *bb;
return 0;
}
It earns you this:
xx2.cpp: In function ‘int main()’:
xx2.cpp:14:9: error: incompatible types in assignment of ‘int [9]’ to ‘int [7]’
*aa = *bb;
The identity of any object in C++ is determined by the pair of its type and its address.
There are two distinct objects with the same address in your example: The array itself, and the first element of the array. The first has type int[1], the second has type int. Two distinct objects can have the same address if one is a subobject of the other, as is the case for array elements, class members, and class base subobjects.
Your example would be clearer if you wrote:
int a[5];
int (*ptr_to_array)[5] = &a;
int * ptr_to_array_element = &a[0];
But you have taken advantage of the fact that the id-expression a for the array decays to a pointer to the array's first element, so a has the same value as &a[0] in your context.
Consider this example:
#include<stdio.h>
int main()
{
int myArray[10][10][10][10]; //A 4 Dimentional array;
//THESE WILL ALL PRINT THE SAME VALUE
printf("%d, %d, %d, %d, %d\n",
myArray,
myArray[0],
myArray[0][0],
myArray[0][0][0],
&myArray[0][0][0][0]
);
//NOW SEE WHAT VALUES YOU GET AFTER ADDING 1 TO EACH OF THESE POINTERS
printf("%d, %d, %d, %d, %d\n",
myArray+1,
myArray[0]+1,
myArray[0][0]+1,
myArray[0][0][0]+1,
&myArray[0][0][0][0]+1
);
}
You will find that all the 5 values printed in first case are all equal. Because they point to the same initial location.
But just when you increment them by 1 you see that different pointers now jump (point) to different locations. This is because myArray[0][0][0] + 1 will jump by 10 integer values that is 40 bytes, while myArray[0][0] + 1 will jump by 100 integer values i.e by 400 bytes. Similarly myArray[0] + 1 jumps by 1000 integer values or 4000 bytes.
So the values depend on what level of pointer you are referring to.
But now, if I use pointers to refer all of them:
#include<stdio.h>
int main()
{
int myArray[10][10][10][10]; //A 4 Dimentional array;
int * ptr1 = myArray[10][10][10];
int ** ptr2 = myArray[10][10];
int *** ptr3 = myArray[10];
int **** ptr4 = myArray;
//THESE WILL ALL PRINT THE SAME VALUE
printf("%u, %u, %u, %u\n", ptr1, ptr2, ptr3, ptr4);
//THESE ALSO PRINT SAME VALUES!!
printf("%d, %d, %d, %d\n",ptr1+1,ptr2+1,ptr3+1,ptr4+1);
}
So you see, different levels of pointer variables do not behave the way the array variable does.
I want to do something like this below:
int main() {
int a[10];
int *d = generateArrayOfSize(10) // This generates an array of size 10 on the heap
a = d;
print(a); // Prints the first 10 elements of array.
}
However above code gives compilation error (incompatible types in assignment of ‘int*’ to ‘int [10]’).
What can I do to make the above code to work?
Arrays are non-assignable and non-copyable, so you'd have to copy each element by hand (in a loop), or using std::copy.
If you're using C++, then use C++ arrays rather than C style arrays and pointers. Here's an example
#include <array>
#include <iostream>
template<size_t N>
std::array<int, N> generateArrayOfSize(void)
{
std::array<int, N> a;
for (int n=0; n<N; ++n)
a[n] = n;
return a;
}
template<size_t N>
void print(std::array<int, N> const &a)
{
for (auto num : a)
std::cout << num << " ";
}
int main() {
std::array<int, 10> a;
std::array<int, 10> d = generateArrayOfSize<10>();
a = d;
print(a); // Prints the first 10 elements of array.
}
which outputs 0 1 2 3 4 5 6 7 8 9
Arrays are not pointers.
You can't do :
int a[10];
int *d;
a = d;
Change it to :
int *a;
int *d;
a = d;
Main differences between arrays and pointers in C programming :
Pointer | Array
-------------------------------------------|-------------------------------------------
A pointer is a place in memory that keeps | An array is a single, pre allocated chunk
address of another place inside | of contiguous elements (all of the same
| type), fixed in size and location.
-------------------------------------------|-------------------------------------------
A pointer can point to a dynamically | They are static in nature. Once memory is
allocated memory. In this case, the memory | allocated , it cannot be resized or freed
allocation can be resized or freed later. | dynamically.
-------------------------------------------|-------------------------------------------
You have a quite good explanation here : https://stackoverflow.com/a/7725410/1394283
An array is not a pointer (although a name of an array often decays to a pointer to its first element).
To make the above code to work, you can declare a as a pointer: int *a;. The print function takes an int* (or a decayed array) anyway.
If you really want to have two arrays and copy contents from one array to another, you should copy the data in a loop.
This will print in this way when you assign a string reference to a pointer you have to use *ptr to print the value of a pointer otherwise in your case print(d) that is like cout< in c++ it will only print the location of the d[0].
int ary[5]={1,2,3,4,5};
int *d;
d=ary;
for(int i=0;i<5;i++)
cout<<*(d+i);
Because array names are non-modifiable. So you can't do
a = d;
Declare it as a pointer like this:
int *a;
Little rusty with my C++ but try something like this.
int main() {
int *a;
int *d = generateArrayOfSize(10) // This generates an array of size 10 on the heap
a = d;
print(a); // Prints the first 10 elements of array.
}
In C, it was always true when Thing X[10]; was declared, X was the constant address of the first element(i.e. &X[0]). So you could then say:
Thing *Y = X; // Equivalent to (Thing *Y = &X[0];)
But in C++, the compiler "remembers" that the Thing array X has 10 elements, and some C++ imposed type checking rules break. Imagine we add Thing Z[20]; to the discussion.
Thing *Y = X; and Thing *Y = Z; if both allowed, would imply that a single variable could be set to Thing Arrays of length 10 and 20, which are very different (ahem) "things", as a quick look at a 2D array will reveal. This sort of justifies why the C language assumed equivalent of X and &X[0] is broken in C++.
Well, at least for some versions of C++. So best not to assume it, and use
Thing *Y = &x[0]; and Thing *Y = &Z[0] instead;
This approach has two advantages. It does what is wanted, and it actually compiles. :-)
I have a question.
I created a correctly initialized integer pointer
int * p
and a correctly initialized integer array
int * array1 = new int[]
Which of the following is legal code?
p = array1;
array1 = p;
Or are both correct?
is this possible as well
p[0] since, to my pointer arithmetic knowledge, it doesn't add anything.
All in c++
If the question is trying to get at pointers versus arrays, they are not always compatible. This is hidden in the presented code because the array is immediately converted to a pointer.
int* array1 = new int[5]; // Legal, initialising pointer with heap allocated array
int array2[5] = {0}; // Declaring array directly on the stack and initalising with zeros
int *p = 0; // Declaring pointer and initialising to numm
p = array2; // Legal, assigning array to pointer
p = array1; // Legal, assigning pointer to pointer
array1 = p; // Legal, assigning pointer to pointer
array2 = p; // ILLEGAL, assigning pointer to array
Here array2 has an array type and cannot be used to store a pointer. Actually, the array cannot be reassigned at all, as it is not an l-value.
int array3[5] = {0}; // Declaring array directly on the stack and initalising with zeroes
array3 = array2; // ILLEGAL, array not an l-value
The array has a fixed address and reassiging it would be similar to trying to write:
int i = 0;
&i = p;
[Hopefully, trying to reassign the location of a variable is obvious nonsense.]
Both are legal.
The first, "p = array1", will cause your correctly initialized integer pointer to be leaked and to point p to the first occurrence of the array that array1 points to.
The second, "array1 = p", will cause the correctly initialized integer array to to be leaked and to point array1 to the single int that p points to.
So I suspect I'm missing something. Could you perhaps post complete code?
If both p and array1 are declared as "int *", as you imply here, then either assignment is legal, by definition.
both are leagal.
passing adresses to eachother. dont knw what u want to do
According to the explanation provided in this site. The first assigned in is correct but the second assignment cannot be considered valid. Please look under the title 'Pointers and arrays'.
You can assign your array variable to pointer variable. So p = array1 is correct.
You can refer this code.
#include <iostream>
using namespace std;
int main ()
{
int numbers[5];
int * p;
p = numbers; *p = 10;
p++; *p = 20;
p = &numbers[2]; *p = 30;
p = numbers + 3; *p = 40;
p = numbers; *(p+4) = 50;
for (int n=0; n<5; n++)
cout << numbers[n] << ", ";
for (int n=0; n<5; n++)
cout << p[n] << ", ";
return 0;
}
If both are pointers then no need to declare pointer as array like int *array1 = new int[].
Both variables are of type pointer to int. So both assignments are legal. The fact that one of the variables is an array doesn't matter in C. Arrays and pointers are the same after initialization.
"I created a correctly initialized integer pointer int * p and a correctly initialized integer array int * array1 = new int[]"
First of all, keeping the syntactical mistakes aside, the terminology is wrong.
int *p; - Here you are just declaring a pointer to hold an integer variable's address. You are not initializing it. Declaration is different from initialization and initialization is different from assignment.
int *array1 = new int[]; - Keeping the error aside, this is not an initialized integer array. array1 is a pointer pointing to an array of integers.
And passing nothing to [] is incorrect. A value needs to be passed that decides number of memory locations to be allocated for holding integer values.
Answering the questions.
Which of the following is legal code?
p = array1;
If, array1 is properly initialized it's correct. p points to the first integer in the array1. Even if it is not properly initialized also, it is correct. In this case, both array and p points to garbage values.
int* array1 = new int[5]; // Notice **5** being passed.
array1 = p;
Infact this operation is useless. Both p and array1 were both pointing to the same location earlier. But this is legally correct.
is this possible as well p[0] ?
Yes, it's correct. p can dereference to 0 to 4 as it's index values. If you just want to p[0], in that case -
int* array1 = new int ; // Notice nothing being passed.