Question

Let’s say I have a char *example that contains 20 chars. I want to remove every char from example[5] to example[10] and then fix up the array so that example[11] comes right after example[4].

Basically shifting all the characters after the deleted region to when the deleted region started.

Any ideas?

EDIT: I think there might be a way using memcpy? But I’m not sure how to do it.

Answer 1

You can’t use memcpy() reliably because there’s overlap between the source and target; you can use memmove(). Since you know the lengths, you use:

memmove(&example[5], &example[11], (20 - 11 + 1));

Remember you need to copy the null terminator too.

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

int main(void)
{
    char array[] = "abcdefghijklmnopqrst";
    char *example = array;
    printf("%.2zu: <<%s>>\n", strlen(example), example);
    memmove(&example[5], &example[11], (20 - 11 + 1));
    printf("%.2zu: <<%s>>\n", strlen(example), example);
    return(0);
}

Compiled with a C99 compiler, that yields:

20: <<abcdefghijklmnopqrst>>
14: <<abcdelmnopqrst>>

If you have a C89 compiler (more specifically, C library), you’ll have to worry about the z in the format string, which indicates a size_t argument. It’s simplest to remove the z and cast the result of strlen() with (unsigned).