pointer to an array in c

The below program gives me an error in the intialisation (*a)[5]=((*)[])&v;. When I don't type at that line then I'm still getting the error.

int main()
{
    int v[10];
    int **p;
    int (*a)[5];
    (*a)[5]=((*)[])&v;
    printf("%d\n",*a);
    printf("%d\n",sizeof(v开发者_JS百科));
    return 0;
}

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&v is a reference to the int array v, so it is of type int ** (essentially). int (*a)[5] means you're declaring a pointer to an array of 5 ints. BUT, in the next line, when you write, (*a)[5], it means you are dereferencing a to get an int pointer, and then dereferncing again to access the element stored at [5]. but arrays in C are zero indexed, so you can only access [0] to [4], and [5] is not allocated.

You want a = &v. This will make a a pointer to the int pointer v. There are other issues like *a is an int pointer so "%d" won't fly, you haven't given any values to a or v, ((*)[]) is just unnecessary, etc.

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When you mean error, you probably refer to run-time error. int (*a)[5] is a pointer to an array of 5 ints, which is never initialized. However on the next line you try to dereference it. You should initialize 'a' before using it, you may be trying to initialize it with v but that's not an array of 5 ints. If you were to redeclare v as int v[5] then a = &v should work.

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Assignment to variable a should only have a on the LHS.

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I think it is simpler than you are expecting

#include <stdio.h>

int main()
{
    int v[10];
    int **p;
    int *a[5];

    v[0] = 1234;
    a[0] = v;

    printf("%d\n", *a[0]);
    printf("%d\n", sizeof(v));

    return 0;
}

*a[5] is an array of int * pointers. v[10] is effectively also a pointer to an int.

So *a and a[0] are effectively the same thing.

ETA: Perhaps printf("%d\n", a[0][0]); is better