Why does strcpy omit the first character in source string when using destination string of size 1?

Question

In C the strcpy function is used to copy a source into a destination string.

But when I use a destination char array of size 1 the strcpy correctly copies the source into the destination. But it also changes the source char array. I want to understand how this works in C.

I have done some research on how to correctly use strcpy in a program but all of them uses destination size more than 1. I did the program using destination size equal to 1. That's where the problem is.

char a[] = "String ABC";
char b[1];

strcpy(b, a);
int i;
// printf("%c\n", *(&(a[0])-1));

printf("%s\n",a);
printf("%s\n",b);

I expect the output to be

String ABC
String ABC

but the output I get is

tring ABC
String ABC

Answer 1

The problem is that you are copying to 1 byte string a longer string resulting in undefined behaviour.

If you run this program:

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

int main(int argc, char *argv[])
{
    char a[] = "String ABC";
    char b[1];
    printf("%p\n", &a);
    printf("%p\n", &b);

    strcpy(b, a);
    int i;
    printf("%c\n", *(&(a[0])-1));
    printf("%c\n", a[0]);
    printf("%s\n",a);
    printf("%s\n",b);
    printf("%p\n", &a);
    printf("%p\n", &b);
}

you see b and a have contiguous addresses and b is stored in a memory address before a. Most likely strcpy copies the string to b but since b is not allocated to store such a long string, it overwrites the next contiguous memory cell which seems to be a.

Let me indicate with || a memory cell storing a char. Suppose -b- is the cell storing one char long string. Before copy you have

|-b-|---a memory allocation--|
|-b-|S|t|r|i|n|g| |A|B|C|D|\n|

Now a is copied into b: the second cell is the one of a which now contain t

  |--a memory allocation-|
|S|t|r|i|n|g| |A|B|C|D|\n|

This is what I suppose it is happening. But remember that copying a longer string into a shorter one will result in undefined behaviour.