Why strcpy(buffer +i, buffer + i +1) doesn't work in the same way as using a temporay array?

Question

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

int main() {
    char buffer[200] = "This is a string";
    char buffer1[200];
    strcpy(buffer1, buffer);
    char vowels[] = "aeiouAEIOU";
    int i;

    for (i = 0; i < strlen(buffer); i++) {
        if (strchr(vowels, buffer[i]) != NULL) {
            char tmp[1000];
            strcpy(tmp, buffer + i + 1);
            strcpy(buffer + i, tmp);
            
            strcpy(buffer1 + i, buffer1 + i + 1);
            break;
        }
    }

    printf("buffer=%s\n", buffer); // prints "Ths is a string", which is OK
    printf("buffer1=%s\n", buffer1); // prints "Ths is asstring", which is not OK.
    return 0;
}

I'm trying to remove a character from a string.
In the code below I used to 2 ways to achieve this, with the variables buffer and buffer1. But, one of them, buffer1, prints a weird result.

What makes buffer and buffer1 contain different values?

Answer 1

Do not overlap

char *strcpy(char * restrict s1, const char * restrict s2); has restrict which informs the caller to not provide pointers to overlapping data.

OP's code used overlapping data and so has undefined behavior (UB).
Don't do that.

---

Could use memmove() which handles overlapping data.

size_t sz = strlen(buffer1 + i + 1) + 1;
memmove(buffer1 + i, buffer1 + i + 1, sz);

Answer 2

Using strcpy for overlapping arrays invokes undefined behavior. In such a case you should use memmove.

From the C Standard (7.23.2.3 The strcpy function)

2. ... If copying takes place between objects that overlap, the behavior is undefined.

Here is a demonstration program

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

int main( void )
{
    char buffer[200] = "This is a string";
    char vowels[] = "aeiouAEIOU";

    size_t n = strlen( buffer );

    size_t pos = strcspn( buffer, vowels );

    memmove( buffer + pos, buffer + pos + 1, n - pos );

    puts( buffer );
}

The program output is

Ths is a string