How to pass &arr to a function in c-CodePudding

I simply want to pass an &arr in C language.

For example:

#include <stdio.h>

void test( PARAMETER??? )
{
    return;
}

int main()
{
    int arr[] = {1,2,3,4,5,6,7,8};
    test(&arr);
    return 0;
}

How should I declare the parameter?

As it is of type int (*)[8]

I simply want to pass &arr in C language. I know I can pass arr and length argument, but how can I via this?

CodePudding user response：

As it is of type int (*)[8]

If you want to pass &arr, then that is exactly the type you need to declare the parameter as, eg:

void test(int (*param)[8])
{
    // use param as needed...
}

CodePudding user response：

you can pass array in function with array size. but you need to use sizeof() to calculate the size of array

#include <stdio.h>

void test( int arr[], size_t size_of_array )
{
    return 0;
}

int main()
{
    int arr[] = {1,2,3,4,5,6,7,8};
    size_t size_of_array = sizeof(arr)/sizeof(arr[0]);  /* calculate of array size */
    test(arr, size_of_array );
    return 0;
}

CodePudding user response：

You can pass it directly (not an address) to a function like

void test(int prm[restrict 8]);

It will not allow to pass int arr[7], and will allow to pass int arr[9], but you will not know the array size, all what you know, it has the size at least 8.

CodePudding user response：

You can't pass an array as an argument.

You can pass a pointer to its first element:

#include <stdio.h>

void test( size_t n, int *p ) {       // `p` is a pointer an `int`.
   printf( "%zu\n", sizeof(p) );      // `sizeof( int* )`, 8 for me.
   printf( "%zu\n", sizeof(*p) );     // `sizeof( int )`, 4 for me.
   printf( "%d\n", p[0] );            // 1
   printf( "%d\n", p[1] );            // 2
}

int main( void ) {
   int arr[] = {1,2,3,4,5,6,7,8,9,10,11};
   test( sizeof(arr)/sizeof(*arr), arr );
}

Same thing in disguise:

#include <stdio.h>

void test( size_t n, int p[n] ) {     // `p` is a pointer an `int`.
   printf( "%zu\n", sizeof(p) );      // `sizeof( int* )`, 8 for me.
   printf( "%zu\n", sizeof(*p) );     // `sizeof( int )`, 4 for me.
   printf( "%d\n", p[0] );            // 1
   printf( "%d\n", p[1] );            // 2
}

int main( void ) {
   int arr[] = {1,2,3,4,5,6,7,8,9,10,11};
   test( sizeof(arr)/sizeof(*arr), arr );
}

Finally, you could pass a pointer to the array:

#include <stdio.h>

void test( size_t n, int (*p)[n] ) {  // `p` is a pointer to an `int[11]`
   printf( "%zu\n", sizeof(p) );      // `sizeof( int* )`, 8 for me.
   printf( "%zu\n", sizeof(*p) );     // `sizeof( int[11] )`, 44 for me.
   printf( "%d\n", (*p)[0] );         // 1
   printf( "%d\n", (*p)[1] );         // 2
}

int main( void ) {
   int arr[] = {1,2,3,4,5,6,7,8,9,10,11};
   test( sizeof(arr)/sizeof(*arr), &arr );
}

In any case, you will need the pass the size of the array if you need it, because it's not stored anywhere in the array itself.