memcpy implementation issue

Question

I am writing a small implementation of memcpy as follows.

#include "stdio.h"
int main( )
{
    int i=5;
    int j=4;
    printf("i=%d\t",i);
    swap(&i,&j,sizeof(int));
    printf("i=%d",i);
    return 0;
}


int swap(void *vp1,void *vp2,int size)
{
    char *a=(char *)vp1;
    char *b=(char *)vp2;

    for(int i=0;i<size;i++)
    {
        *a=*b;
        a++;
        b++;
    }

    return 0;
}

The output of this code is

i=5 i=33

rather than

i=5 i=4

Can anybody explain what is wrong with the code?

Some obvious problems with your code: 1. The `memcpy()` workalike is named `swap()`, which is horribly confusing. 2. The signature doesn't match. 3. It fails to use `const` for the source pointer. 4. Needless casting from `void *` to `char *`. 5. Will break if given a negative size. 6. Pointless return value. 7. Calling function without first declaring it isn't a sign of good code hygiene. That said, it looks like it should work. — unwind, Oct 02 '12 at 14:23
Also `size` is not int, but size_t. I guess that this could cause problems on a 64-bit, big endian system since the integer might not be truncated correctly. — Krumelur, Oct 02 '12 at 16:52

score -3 · Answer 1 · edited Oct 02 '12 at 16:35

-3

#include "stdio.h"
int swap(void *vp1,void *vp2,int size)

{

     int i;

    char *a=(char *)vp1;
    char *b=(char *)vp2;

    for(i = 0; i<size;i++)
    {
     *a = *b;
     a++;
     b++;
     }
    return 0;
}

int main( )

{

    int i=5;
    int j=4;
    printf("i=%d\t",i);
    swap(&i,&j,sizeof(int));
    printf("i=%d",i);
    return 0;
}

edited Oct 02 '12 at 16:35

Bo Persson

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answered Oct 02 '12 at 15:54

Hello-World

1

1

Please explain any code solutions you give so that the asker can learn how to solve their problem. – Joseph Mansfield Oct 02 '12 at 17:38
You basically switched the declarations of `swap` and `main`, it doesn't answer the question at all. – netcoder Oct 02 '12 at 17:50

memcpy implementation issue

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