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.
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)
.
spoiler:
void cut_the_string(char *str, size_t cutpos, size_t cutlen)
{
size_t len;
len = strlen(str);
if (cutpos >= len) return;
if (cutpos + cutlen > len) cutlen = len-cutpos;
memmove( str+cutpos, str+cutpos+cutlen, 1+len - cutpos - cutlen);
return;
}
Any ideas?
I would prepare a parallel array, copying all the characters from the source array but ignoring those characters in the range specified. Optionally, one can release the memory used by source array if it is not on stack.
memmove(&example[5], &example[11], strlen(example) - 11 + 1);
Note that the behavior of memcpy
is undefined when the memory blocks overlap, so memmove
is more appropriate here.