Thread: Simple Program, null terminator

Hey, i'm trying to learn C and am stuck trying to answer a simple question. ( Copying sequence of chars )

My Problem is that my program is printing out a random symbol afterwards , when trying to copy a sequence of chars into a new buffer.

Code:

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




int tokenCopy(char* dest, const char* src, int destSize);


int main()
{
    char buff[3];
    int n = tokenCopy(buff, "This is a string", 3);
    printf("%d '%s'\n", n, buff);
    
    return EXIT_SUCCESS;
}
int tokenCopy(char* dest, const char* src, int destSize){
    
    int num_chars = 0;
    
    
    for(int i = 0; i < destSize; i++){
        
        if(src[i] == '\0' || src[i] == ' '  ){
        break;
        }
        
        
        dest[i] = src[i];
        num_chars+= 1;
        
        
        
    }
    
    return num_chars;
}

Could someone tell me what i need to change? I think its something to do with null terminator?

Correct you are!

If the dest size is 3, then you need to copy just two chars, leave the for loop, and add

Code:

dest[i]='\0';

Just after the for loop in tokenCopy(). NOW you have a string. Without that, you don't have a string - (unless C adds the end of string marker char for you). You have just a bunch of chars - not a string - and they won't print correctly using the %s, in most cases.

Originally Posted by Adak

Correct you are!

If the dest size is 3, then you need to copy just two chars, leave the for loop, and add

Code:

dest[i]='\0';

Just after the for loop in tokenCopy(). NOW you have a string. Without that, you don't have a string - (unless C adds the end of string marker char for you). You have just a bunch of chars - not a string - and they won't print correctly using the %s, in most cases.

thanks!