Difference between strncpy and memcpy?

Question

How can I access s[7] in s?

I didn't observe any difference between strncpy and memcpy. If I want to print the output s, along with s[7] (like qwertyA), what are the changes I have to made in the following code:

#include <stdio.h> #include <stdlib.h>  int main() {     char s[10] = "qwerty", str[10], str1[10];     s[7] = 'A';     printf("%s\n", s);     strncpy(str, s, 8);     printf("%s\n", str);     memcpy(str1, s, 8);     printf("%s\n", str1);     return 0; } 

Output:

qwerty qwerty qwerty 

Answer 1

Others have pointed out your null-termination problems. You need to understand null-termination before you understand the difference between memcpy and strncpy.

The main difference is that memcpy will copy all N characters you ask for, while strncpy will copy up to the first null terminator inclusive, or N characters, whichever is fewer.

In the event that it copies less than N characters, it will pad the rest out with null characters.

Answer 2

You are getting the output querty because the index 7 is incorrect (arrays are indexed from 0, not 1). There is a null-terminator at index 6 to signal the end of the string, and whatever comes after it will have no effect.

Two things you need to fix:

1. Change the 7 in s[7] to 6
2. Add a null-terminator at s[7]

The result will be:

char s[10] = "qwerty"; s[6] = 'A'; s[7] = 0; 

Original not working and fixed working.

As for the question of strncpy versus memcpy, the difference is that strncpy adds a null-terminator for you. BUT, only if the source string has one before n. So strncpy is what you want to use here, but be very careful of the big BUT.