CODEWITHSUNDEEP
cstring
Jozef
Jozef

Reputation: 193

How come there is '\0' at the end of the string while nobody put it there?

I was trying to implement strcat by myself, It seems to work but I don't understand how come p has '\0' at its end? it didn't copy it from b and it shouldn't be there without putting it. Any explanation? The output is "yesMichaelJudy".

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


char* strcat1( char *s1, const char *s2 )
{
    register char  *p = s1;

    while ( *p )
          ++p;

    while (*s2){

        *p=*s2;
        ++p;
        ++s2;

    }

    if (*p=='\0') printf("yes");
    ++p;
    p='\0';

    return s1;
}


int main(){

    char* a;
    char* b;

    char* result;
    result=(char *)calloc(20,sizeof(char));
    a=(char *) calloc(20,sizeof(char));
    b=(char *) calloc(20,sizeof(char));
     strcpy (a,"Michael");
    strcpy (b,"Judy");
    result=strcat1(a,b);

    printf(result);

    getch();
    return 1;

}

Upvotes: 1

Views: 979

Answers (6)

Ernest Friedman-Hill
Ernest Friedman-Hill

Reputation: 81724

You're allocating the space larger than it needs to be and you're using calloc() which by definition clears all the characters to zero; therefore the extra characters at the end are zero.

Upvotes: 4

AnT stands with Russia
AnT stands with Russia

Reputation: 320719

p in your strcat1 points into the array passed through parameter s1. And parameter s1 corresponds to argument a in main. The content of array a in main is filled with data by strcpy (a,"Michael"). This is what put that '\0' at the end of that data. strcpy did that. (You can also say that that '\0' came from "Michael" since "Michael" also has a '\0' at the end, even if it is not specified explicitly).

Upvotes: 0

tomato
tomato

Reputation: 749

calloc zero initializes ( ie fills it with NUL or '\0') the allocated buffer.

Upvotes: 0

morningstar
morningstar

Reputation: 9162

p='\0';

That's nonsense. It does nothing useful. It will also generate at least a compiler warning if not an error. The code should be

*p='\0';

Which is what puts \0 at the end of the string.

Upvotes: 1

itsme86
itsme86

Reputation: 19526

strcpy() null-terminates the string. But even if it didn't, calloc() puts it there. calloc() zeroes out the allocated memory. As soon as you allocated the memory, a and b are just a bunch of string-terminators.

Upvotes: 0

Jerry Coffin
Jerry Coffin

Reputation: 490623

strcpy copies the NUL character at the end of a string. Even if it didn't, you're using calloc to allocate your a and b, and calloc allocates and zeros the memory. Since you're allocating more space than you use (allocating 20 bytes, using 4 for Judy and 7 for Michael) you have some zero bytes after the strings anyway.

Upvotes: 2

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