Thread: Is {0,} okay?

void *memset(void *s, int c, size_t n);
The memset() function fills the first n bytes of the memory area pointed to by s with the constant byte c.

Take a look at the results

Code:

int arr[10];
memset(arr, 300, 10);
memset(arr, 300, sizeof(arr));

Originally Posted by xeddiex

Why not?

Because memset works on bytes, not ints. It's fine if you want to initalise an int array to zero, but if you want an arbitrary value (like 2), then memset(array, 2, size); will fill every byte with a value 2. An int is generally larger than a byte.

quzah: you should submit that tragic code to the international obfuscated C code contest Cute misuse of the function.

Quzah,

Yes, I had thought you might reply with something like that

Unfortunately, not everybody uses little endian machines

My output from your code:

Code:

arr[0] is 33554432
arr[1] is 33554432
arr[2] is 33554432
arr[3] is 33554432
arr[4] is 33554432
arr[5] is 33554432
arr[6] is 33554432
arr[7] is 33554432
arr[8] is 33554432
arr[9] is 33554432

My more portable version:

Code:

#include<string.h>
#include<stdio.h>

int bigendian;

int main( void )
{
    int arr[10], x;
    int y = 1;
    if (*(char *)&y != 1)
        bigendian = 1;


    memset( arr, 0, sizeof arr );
    for( x = 0; x < 10; x++ )
    {
        int *offset = arr + x;

        if (bigendian)
            offset = (char *)offset + sizeof(int)-1;

        memset(offset, 2, 1 );
    }

    for( x = 0; x < 10; x++ )
    {
        printf("arr[%d] is %d\n", x, arr[x] );
    }

    return 0;
}

True, a more efficient version like your first one that retains portability:

Code:

#include<string.h>
#include<stdio.h>

int bigendian;

int main( void )
{
    int arr[10], x;
    int y = 1;
    char *offset = (char *)arr;

    if (*(char *)&y != 1)
        offset += sizeof(int)-1;


    memset( arr, 0, sizeof arr );

    for( x = 0; x < 10; x++ )
    {
        memset( (int *)offset + x, 2, 1 );
    }

    for( x = 0; x < 10; x++ )
    {
        printf("arr[%d] is %d\n", x, arr[x] );
    }

    return 0;
}

*runs away*

Originally Posted by quzah

I thought you'd reply something like that. I did it with pointers initially, which worked as you have it. I just had to reply something to prove it could be done using memset. I was going to have 2 memset calls per loop iteration. One to set "sizeof( int ) -1" bytes to zero, and one to set the other. But this is the "optimized" version, and so the first just zeros it outside.


Quzah.

Actually all you proved is that it'd work with small int values

yeah but in only one call to malloc? and not just malloc written once?