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.

Answer 3

Strncpy will copy up to NULL even you specified the number of bytes to copy , but memcpy will copy up to specified number of bytes .

printf statement will print up to NULL , so you will try to print a single charater , it will show ,

printf("\t%c %c %c\t",s[7],str[7],str1[7]);

OUTPUT

  7              7

Answer 4

To make "qwertyA" you need to set s[6] = 'A' and s[7]='\\0'

Strings are indexed from 0, so s[0] == 'q' , and they need to be null terminated.

Answer 5

When you have:

char s[10] = "qwerty";

this is what that array contains:

s[0]  'q'
s[1]  'w'
s[2]  'e'
s[3]  'r'
s[4]  't'
s[5]  'y'
s[6]   0
s[7]   0
s[8]   0
s[9]   0

If you want to add an 'A' to the end of your string, that's at index 6, since array indexes start at 0

 s[6] = 'A';

Note that when you initialize an array this way, the remaining space is set to 0 (a nul terminator), While not needed in this case, generally be aware that you need to make your strings nul terminated. eg

char s[10];
strcpy(s,"qwerty");
s[6] = 'A';
s[7] = 0;

In the above example "qwerty" , including its nul terminator is copied to s . s[6] overwrites that nul terminator. Since the rest of s is not initialized we need to add a nul terminator ourselves with s[7] = 0;

Answer 6

As explained by Philip Potter, 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. Execute the below code and check the difference, you might find it useful. Thanks.

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

int main()
{
    char s[10] = "qwer\0ty", 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);
    for(int i = 0; i<8; i++)
    {
        printf("%d=%c,",i,str[i]);
    }
    printf("\n");
    for(int i = 0; i<8; i++)
    {
        printf("%d=%c,",i,str1[i]);
    }
    return 0;
}

Output:

qwer
qwer
qwer
0=q,1=w,2=e,3=r,4=,5=,6=,7=,
0=q,1=w,2=e,3=r,4=,5=t,6=y,7=A,