April 21, 2021

C String Library and String Copy Function – strcpy()



C Programming: C String Library and strcpy() Function in C Programming.
Topics discussed:
1) Introduction to C string library.
2) String Copy function (strcpy).
3) The prototype of strcpy() function.
4) Example use of strcpy() function.
5) The prototype of strncpy() function.
6) Example use of strncpy() function.

C Programming Lectures:

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Comments
  • This is the correct code for strncpy
    #include <stdio.h>
    #include <string.h>
    int main()
    {
    char str1[6] = "Hello";
    char str2[5] = "";
    strncpy(str2, str1, sizeof(str2) – 1);
    printf("%s", str2);
    return 0;
    }
    (OR)
    #include <stdio.h>
    #include <string.h>
    int main()
    {
    char str1[6] = "Hello";
    char str2[5];
    strncpy(str2, str1, sizeof(str2));
    str2[sizeof(str2)-1]='';
    printf("%s", str2);
    return 0;
    }

    Output for both the programs : Hell

  • at 6:32 the output is "Hell" but on my computer it's showing "HellHello" why is that????
    and when source>= destination so that means at 6:32 strncpy() din put a null character in str2[4] then why your output is "Hell", without showing an undefined behaviour ? tho in my computer it showed "HellHello" with same code

  • What do you mean by prototype here. I saw in some previous videos too but din quite understand what is prototype. We're using only strcpy(destination,source) but the prototype consist more. What is prototype and what it is used for???

  • For str1[6] and str2[4]
    We have to add str2[4]='' after the strncpy line.Otherwise it will print HellHello. (5:25)

    #include<stdio.h>
    #include<string.h>
    int main()
    {
    char str1[6]="Hello";
    char str2[4];
    strncpy(str2,str1,sizeof(str2));
    str2[4]='';
    printf("%s",str2);
    return 0;
    }

  • The correct code to print "Hell" for str2 is (5:25) :

    #include <stdio.h>

    #include <string.h>

    int main(){

    char str1[6] = "Hello";

    char str2[5];

    strncpy(str2, str1, sizeof(str2)-1);

    printf("%s" , str2);

    return 0;

    }

    Please check if I am correct.
    (the problem was about the string terminator of str2 as I mentioned in my previous comment).

  • In the case of strncpy : (5:25)
    If str2 is "Hell". As str2 is of size 4. then what about the string terminator?
    shouldn't it be "Hel" only?

  • The prototype is having arguments as char pointers. But, the problem explanation has char arrays. Why because we cannot change or modify char pointers? But, we can copy, right?

  • But sir why not use a for loop like this

    char str1[6] = "Hello";
    char str2[6];

    for(int i = 0; i < sizeof(str1); ++i)
    str2[i] = str1[i];

    But anyways fantastic lessons:)

  • char c=48;

    int i, mask=01;

    for(i=1; i<=5; i++)

    {

    printf("%c", c|mask);

    mask = mask<<1;

    }

    anyone explain me output of this..plz help

  • @neso Academy, what if size or str1 and str2 where both 5, adding the nul byte at the end wouldn’t overwrite ‘o’??

  • #include<stdio.h>

    #include<string.h>

    int main()

    {

    char str[6]="hello";

    char str1[0];

    strcpy( str1,str);

    printf("%s",str1);

    }
    but this code is exuted in my computer please help………….
    output is hello

  • Fantastic vedio sir it is really helpful to understand full concept we are really thankful to you sir 👍🙏🙏🙏

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