Modifying a array in a function in C

Question

#include <stdio.h>
#include <stdlib.h>

void array_tester(int *array);

int main(int argc,char *argv[])
{   
    int test[] = {1,1,1,1};
    int print;
    for(print = 0; print < sizeof(test) / sizeof(int); print++)
    {
        printf("%d\n",test[print] );
    }

    array_tester(test);
    for(print = 0;print < sizeof(test)/sizeof(int);print++)
    {
        printf("%d\n",test[print] );
    }

    return EXIT_SUCCESS;
}


void array_tester(int array[])
{   
    int i ;
    for(i = 0; i < sizeof(array) / sizeof(int); i++)
    {
        array[i] = i;
    } 
}

The problem is I want to modify the the array in the array_tester function but I'm unable to do so. What do I use inside sizeof()?

this · Accepted Answer · 2013-12-13 14:04:06Z

5

You have to manually pass the information about the size of the array.

void array_tester( int* array , int size ) ;

And then call it:

array_tester( test , sizeof(test)/sizeof(*test) ) ;

The reason behind this is that array passed to a function will decay to a pointer and the sizeof will return the size of pointer not the original array.

answered Dec 13, 2013 at 14:04

this

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1 Comment

kviiri Over a year ago

Well explained, +1 in particular for explaining how the replaced solution differs from the original one.

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