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cstrcpy

Use the STRCPY C

I code:

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

void main()
{
    char string[]="stackflow";
    strcpy(&string[0],&string[1]);
    puts(string);
    getch();
}

The result is "tackflow". I don't understand about it. Who are you can explain? Thank in advance.

Solution

  • The call strcpy(&string[0],&string[1]); means: copy the NULL-terminated string starting at the address of string[1] to the address of string[0]. This means you are copying the string starting at offset 1 (tackflow) to the address of offset 0. Or put it another way: you are moving the string one character to the left, overwriting the s.

    strcpy is used to copy an array of bytes that are terminated by a NULL byte (a C string) from one address (the second parameter) to another address (the first parameter). Usually, it is used to copy a string from one memory location to a totally different memory location:

    char string[] = "stackflow";
char copied_string[10]; // length of "stackflow" + NULL byte

strcpy(copied_string, string);
puts(copied_string);

    As @PascalCuoq correctly points out, your call is undefined behavior, which means anything may happen. The standard says:

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

    So you are lucky you got a "sane" output at all.