CODEWITHSUNDEEP
cpointers
sbhatla
sbhatla

Reputation: 1088

Unable to print a pointer after using memcpy

I'm trying to get my own version of memcpy to copy a character array from one pointer to another. However, this gives an error. It seems like there is an issue with printf.

Secondly, the official memcpy function returns the destination pointer. Is that really required? If I modify *dest, it should be reflected in *d anyway. So what is the need to return anything?

#include <stdio.h>

void memcpy2(void *dest, const void *src, size_t n)
{
    char *dp = dest;
    const char *sp = src;
    while (n--)
        *dp++ = *sp++;
}

int main(void) {
    char *c = "Hello";
    char *d=NULL;
    memcpy2(d,c,3);
    printf( "%c", *d);
    return 0;
}

Upvotes: 0

Views: 1502

Answers (1)

Iharob Al Asimi
Iharob Al Asimi

Reputation: 53006

You dereference a NULL pointer that's undefined behavior, you should allocate space and point to it with d in order for that to work, do this, instead of

d = NULL

write

d = malloc(3);

and don't forget

free(d);

after the printf().

If you want to automagically make d a valid pointer then write memcpy2() this way

void memcpy2(void **dest, const void *src, size_t n)
{
    *dest = NULL;
    if ((src == NULL) || (n == 0))
        return;
    *dest = malloc(n);
    if (*dest == NULL)
        return;

    char *dp = *dest;
    const char *sp = src;
    while (n--)
        *dp++ = *sp++;
}

and then your main would be

int main(void) {
    char *c = "Hello";
    char *d = NULL;

    memcpy2(&d, c, 3);
       /*   ^ pass the address of the pointer, so you can change where it points to */

    if (d != NULL) /* prevent NULL dereference. */
    {
        printf("%c", *d);
        free(d);
    }
    return 0;
}

Upvotes: 5

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