Result of calling strcpy is different than expecte

问题:

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

int main()
{
   char src[]="123456";
   strcpy(src, &src[1]);
   printf("Final copied string : %s\n", src);
}

When I use the Visual Studio 6 Compiler it gives me the expected answer "23456".

How come this program prints "23556" when compiled with gcc 4.7.2?

回答1:

strcpy(src, &src[1]); is undefined behavior:

C11 §7.24.2.3 The strcpy function

The strcpy function copies the string pointed to by s2 (including the terminating null character) into the array pointed to by s1. If copying takes place between objects that overlap, the behavior is undefined.

By the way, memcpy is similar (but not memmove). See C FAQ: What's the difference between memcpy and memmove.

回答2:

This is undefined behaviour. Use the memmove function instead. memmove is designed to allow overlapping of source and destination buffers.

memmove(src, &src[1], strlen(&src[1]) + 1) ;  // + 1 for copying the terminating zero

回答3:

From ISO/IEC 9899:TC3 (c99)

7.21.2.3 The strcpy function

Synopsis

1

#include <string.h>

char *strncpy(char * restrict s1, const char * restrict s2, size_t n);

Description

2 The strcpy function copies the string pointed to by s2 (including the terminating null character) into the array pointed to by s1. If copying takes place between objects that overlap, the behavior is undefined.

So what you are doing is simply undefined behaving ;)

You can also see the ANNEX J.2

Stating cases of undefined behavior with a note how to prevent:

The behavior is undefined in the following circumstances:

[...]

—An attempt is made to copy an object to an overlapping object by use of a library function, other than as explicitly allowed (e.g., memmove) (clause 7).