How is strcpy implemented?

Question

I have a question about using strcpy. I know the ANSI C standard says : source and destination must not overlap, otherwise the behaviour is unpredictable. I show you a piece of code that works as I expect if it is compiled using an old gnu C compiler under Linux.

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

char S[80],*P;

int main() {
    strcpy(S,"abcdefghi\r\njklmnopqr\r\nstuvwxyz\r\n");
    for (P=S; P=strchr(P,'\r'); P++) strcpy(P,P+1);
    printf("%s\n",S);
    return 0;
}

This sequence removes every \r (carriage return) from the input string. I know (from Kernigham and Ritchie) that a very simple implementation for strcpy is the following

while (*t++=*s++) ;

Now I compiled my program using gcc (Gentoo 4.5.4 p1.0, pie-0.4.7) 4.5.4 and it prints this:

abcdefghi
jklmnpqr          <-- missing 'o'
stuvwxxyz         <-- doubled 'x'

I suppose this compiler (in fact its library) uses a very sophisticated sequence for strcpy, and I don't understand the reason.

Answer 1

You were warned not to do that. The reason is that a byte-for-byte copy is actually quite slow and requires a lot of looping to get through a string. The compiler can optimize this easily (for instance, by copying an int-sized chunk at a time, or using some platform-specific parallellization.)

But if the strings overlap, then those optimizations make assumptions about your data that are no longer valid. As a result, they give you unspecified results. It is likely your older GCC simply didn't do any such optimizations.

Since the documentation for strcpy() says not to use overlapping strings, don't.