Difference between "pointer to int" and "pointer to array of ints"

Question

int main()
{
    int (*x)[5];                 //pointer to an array of integers
    int y[6] = {1,2,3,4,5,6};    //array of integers
    int *z;                      //pointer to integer

    z = y;
    for(int i=0;i<6;i++)
        printf("%d ",z[i]);

    x = y;
    for(int i=0;i<6;i++)
        printf("%d ",(*x)[i]);

    return 0;
}

Both the above printfs print numbers 1 through 6.
If both "pointer to array of integers" and "pointer to integer" can do the same thing, do they have the same internal representation?
EDIT: This code does give warnings when compiled as pointed out by the answers below, however it does print the values correctly both the time on my x86_64 machine using gcc

Answer 1

Who knows - this code exhibits undefined behavior:

printf("%d ",(*x)[i]);

Answer 2

To answer your question from the title, from the comp.lang.c FAQ: Since array references decay into pointers, if arr is an array, what's the difference between arr and &arr?

However, the code you've posted has other issues (you're assigning y, not &y to x, and y is a 6-element array, but *x is a 5-element array; both of these should generate compilation warnings).

Answer 3

Firstly, your code will not compile. The array has type int[6] (6 elements), while the pointer has type int (*)[5]. You can't make this pointer to point to that array because the types are different.

Secondly, when you initialize (assign to) such a pointer, you have to use the & on the array: x = &y, not just a plain x = y as in your code.

I assume that you simply typed the code up, instead of copy-pasting the real code.

Thirdly, about the internal representation. Generally, in practice, you should expect all data pointers to use the same internal representation. Moreover, after the above assignments (if written correctly), the pointers will have the same numerical value. The difference between int (*)[5] and int * exists only on the conceptual level, i.e. at the level of the language: the types are different. It has some consequences. For example, if you increment your z it will jump to the next member of the array, but if you increment y, it will jump over the whole array etc. So, these pointers do not really "do the same thing".

Answer 4

The short answer: There is a difference, but your example is flawed.

The long answer:

The difference is that int* points to an int type, but int (*x)[6] points to an array of 6 ints. Actually in your example,

x = y;

is undefined** behavior, you know these are of two different types, but in C you do what you want. I'll just use a pointer to an array of six ints.

Take this modified example:

int (*x)[6];                 //pointer to an array of integers
int y[6] = {1,2,3,4,5,6};    //array of integers
int *z;                      //pointer to integer
int i;

z = y;
for(i = 0;i<6;i++)
    printf("%d ",z[i]);

x = y; // should be x = &y but leave it for now!

for(i = 0;i<6;i++)
    printf("%d ",x[i]); // note: x[i] not (*x)[i]

First,

1 2 3 4 5 6

Would be printed. Then, we get to x[0]. x[0] is nothing but an array of 6 ints. An array in C is the address of the first element. So, the address of y would be printed, then the address of the next array in the next iteration. For example, on my machine:

1 2 3 4 5 6 109247792 109247816 109247840 109247864 109247888 109247912

As you can see, the difference between consecutive addresses is nothing but:

sizeof(int[6]) // 24 on my machine!

In summary, these are two different pointer types.

** I think it is undefined behavior, please feel free to correct my post if it is wrong.

Answer 5

what is mean by c? how to use it