Thread: finding size of empty char array

Originally Posted by prog-bman

This will do:

Code:

sizeof(array) / sizeof(array[0])

No it won't. Read the OP.

Now I pass the array to a function:

function(myarray);

How can I find out inside the function the size allocated to the array?

Code:

#include<stdio.h>

void foo( int array[] )
{
    printf( "sizeof( array ) is %d\n", sizeof( array ) );
    printf( "sizeof( array[ 0 ] ) is %d\n", sizeof( array[ 0 ] ) );
    printf( "sizeof( array ) / sizeof( array[ 0 ] ) is %d\n",
        sizeof( array ) / sizeof( array[ 0 ] ) );
}

void bar( char array[] )
{
    printf( "sizeof( array ) is %d\n", sizeof( array ) );
    printf( "sizeof( array[ 0 ] ) is %d\n", sizeof( array[ 0 ] ) );
    printf( "sizeof( array ) / sizeof( array[ 0 ] ) is %d\n",
        sizeof( array ) / sizeof( array[ 0 ] ) );
}

int main( void )
{
    int array[ 10 ];
    char barray[ 10 ];

    foo( array );
    bar( barray );

    return 0;
}

/*
sizeof( array ) is 4
sizeof( array[ 0 ] ) is 4
sizeof( array ) / sizeof( array[ 0 ] ) is 1
sizeof( array ) is 4
sizeof( array[ 0 ] ) is 1
sizeof( array ) / sizeof( array[ 0 ] ) is 4
*/

Arrays degrade to pointers to their first element when passed to functions. No cigar for you.


Quzah.

>>The size of the array must be stored somewhere

No it doesn't - arrays do not have built in bounds checking

>>The size of array n is undetermined before compile time

So you are using a dynamic memory allocation i guess - how
does that know/calculate the size of the array?

Just pass it as a parameter. I suppose your function could read
the array until the string termination character - but does your
program guarantee that the char array passed is an actual
string?