CODEWITHSUNDEEP
c
user3039950
user3039950

Reputation: 1

Strcpy Output in C

    #include <stdio.h>
#include <string.h>
int main(int argc, char* argv[])
{
    char *a = "abcde";
    char *b = "vwxyz";
    char s[10];

    strcpy(s,a+3);
    printf("%s\n",s);

    strcpy(s+2,b);
    printf("%s\n",s);
    return 0;
}

I am having a small issue figuring out why the second strcpy(s+2,b) outputs devwxyz.

I understand the first part because it points at a[3] and counts from then to the null character which is just 'de'.

The output is:

de

devwxyz

Basically, I don't know how to find "s+2" and I'm not sure why it is 'de' in the beginning of the final output. 'devwxyz'

Hope someone can help, thanks guys.

EDIT:

I attempted to figure it out with this piece of code and it seems the indexing didn't work for me..

#include <stdio.h>
#include <string.h>
int main(int argc, char* argv[])
{
    char *a = "abcde";
    char *b = "vwxyz";
    char s[10];

    strcpy(s,a+0);
    printf("%s\n",s);

    strcpy(s+1,b);
    printf("%s\n",s);
    return 0;
}

Hope someone can explain, because the first strcpy results in abcde. And with s[1] that would be upto the letter 'b' so wouldn't it be abvwxyz? The correct result is avwxyz though.

Upvotes: 0

Views: 395

Answers (2)

Jonathan Wood
Jonathan Wood

Reputation: 67195

Your code has two calls to strcpy(). The first copies "de" just as you described.

The second one copies "vwxyz" to the address of the third character in s (s + 2). Since s is just two characters long, it effectively appends "vwxyz" to "de".

Upvotes: 3

tabstop
tabstop

Reputation: 1761

You find s+2 the same way you found a+3. In this case, the third character of s is the \0 that got copied over the first time (and is replaced by this strcpy).

Upvotes: 0

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