Segmentation Fault while trying to copy string in C function

Question

I have a function in 'C' that is supposed to implement my own strcpy program. Here's what I wrote. However, I am not able to debug the cause of Segmentation Fault.

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

char * mystrcpy(char *dest, char *src) {
    char * ptr = dest;
    while (*src != '\0'){
        *ptr = *src;
        ptr++; src++;
        //printf("Dest is \"%s\" and Source is \"%s\"\n",dest,src);
    }
    *ptr = '\0';
    return dest;
}


int main() {
    char str[] = "I rock always";
    char * dest = NULL;
    dest = mystrcpy(dest, str);
    printf("Source String %s and Destination String %s\n", str, dest);
}

Can someone explain this behavior to me ?

Please note that there is the one-liner K&R version of strcpy: while( *dest++ = *str++) {;} You might find it ugly, but it is a good way to learn to understand the precedence rules. — wildplasser
– wildplasser, Commented Jun 24, 2012 at 9:59
I guess, since the goal is to copy the entire string; it makes sense to escape this '\0' check, so that it gets copied into destination string. But then, the question is, when does it know to stop ? — Code4Fun
– Code4Fun, Commented Jun 24, 2012 at 10:29
'I am not able to debug the cause of Segmentation Fault' -- only because you clearly haven't tried. — Jim Balter
– Jim Balter, Commented Jun 24, 2012 at 10:33

rodrigo · Accepted Answer · 2012-06-24 09:45:21Z

3

You have to allocate memory for the destination string:

int main() {
    char str[] = "I rock always";
    char * dest = (char*)malloc(strlen(str) + 1);
    dest = mystrcpy(dest, str);
    printf("Source String %s and Destination String %s\n", str, dest);
}

Of course, it is good manners to free the memory eventually:

    free(dest);

answered Jun 24, 2012 at 9:45

rodrigo

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6 Comments

wildplasser Over a year ago

1) the cast is not necessary. 2) the strlen() is not necessary; sizeof will do, too.

rodrigo Over a year ago

@wildplasser - 1) True, but may prevent a warning in pedantic compilers. 2) Also true, but this is more generic: he may want to move the malloc into the mystrcpy function, and then the sizeof would not work.

wildplasser Over a year ago

1) "pedantic compilers"? You mean C++ compilers? 2) In that case he could call his function mystrdup().

Code4Fun Over a year ago

Thanks rodrigo and wildpasser for your valuable comments !! I understood the root cause.

Code4Fun Over a year ago

@wildplasser : I have a question; Why does sizeof stop working, when the pointer reaches an external function, as you mentioned above?

|

WW. · Accepted Answer · 2012-06-24 09:45:17Z

1

You never allocate any memory for *dest to point to.

It starts pointing to NULL, and then you attempt: *ptr = *src.

answered Jun 24, 2012 at 9:45

WW.

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Comments

wildplasser · Accepted Answer · 2012-06-24 10:05:31Z

0

#include <stdio.h>

    /* this the K&R version of strcpy */
char * mystrcpy(char *dest, char *src) {
    char *ptr = dest;
    while ( *ptr++ = *src++ ) {;}

    return dest;
}


int main(void) {
    char str[] = "I rock always";
    char dest[sizeof str];
    char *result;

    result = mystrcpy(dest, str);
    printf("Source String %s and Destination String %s and result %s\n", str, dest, result);
    return 0;
}

answered Jun 24, 2012 at 10:05

wildplasser

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Comments

Arjun K P · Accepted Answer · 2012-06-24 10:07:29Z

Allocate memory and try....

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

         char * mystrcpy(char *dest, char *src) 
         {
                     char * ptr = dest;
                     int index=0;
                     while (*src != '\0')
                     {
                         *(ptr+index) = *(src+index);
                           index++;
                     }
                     *(ptr+index) = '\0';
                     return dest;

          }




          int main() 
          {
                char str[] = "I rock always";
                char * dest = (char*)malloc((strlen(str)+1)*sizeof(char));
                dest = mystrcpy(dest, str);
                printf("Source String %s and Destination String %s\n", str, dest);
                free(dest);
                return 0;
          }

Moog · Accepted Answer · 2012-06-24 10:13:49Z

0

You must allocate memory for the destination string buffer.

Memory management is a very big part of C ... you need to be very careful that you allocate AND deallocate memory as it is used and becomes surplus to requirements. Failure to do so will result in segmentation faults (failing to allocate memory) and memory leaks (failing to free memory)

Other answers here give you examples of malloc so I won't repeat them here.

You should use the functions already available to you as much as possible, rather than writing your own. This avoids errors in implementation as somebody has already debugged and optimized it for you.

http://www.cs.cf.ac.uk/Dave/C/node19.html

edited Jun 24, 2012 at 10:13

answered Jun 24, 2012 at 9:45

Moog

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1 Comment

Code4Fun Over a year ago

Yeah ... absolutely true. I am just learning to code, so that I would be able to develop such robust libraries some day.

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