1

Please run it on your machine using gcc and tell if it also gives you a segmentation fault output. I don't think there is any problem with the program. I am a beginner in C. So Help!!

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

char *scat(char *,char *);

void main()
{
    char *s="james";
    char *t="bond";

    char *w=scat(s,t);
    printf("the con: %s\n", w);
    free(w);
}

char *scat(char *s,char *t)
{ 
    char *p=malloc(strlen(s)+strlen(t)+1); 
    int temp=0 , ptr=0;

    while(s[temp]!='\0'){
        p[ptr++]=s[temp++];
    } 
    temp=0;
    while(t[temp]='\0'){
        p[ptr++]=t[temp++];
    }
    return p;
}
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4
  • 3
    If you get a segmentation fault, so the program has a problem... Commented Jan 10, 2013 at 19:41
  • You should probably try formatting your code better. Commented Jan 10, 2013 at 19:41
  • 2
    First mistake: "I don't think there is any problem with the program" Commented Jan 10, 2013 at 19:41
  • memcpy() is your friend. Commented Jan 10, 2013 at 19:42

4 Answers 4

Reset to default
1

The second loop has no effect:

while(t[temp]='\0') { // <<== Assignment!!!
    p[ptr++]=t[temp++];
}

This should be a != not =, or better yet, you can drop zero altogether:

while(t[temp]) { // Zero-checking is implicit in C
    p[ptr++] = t[temp++];
}

Since you are not writing to s or t, it's probably a good idea to declare them both const. This would have caught the assignment in the while loop above.

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11 Comments

It should probably be !=. Good catch!
@wildplasser Thanks for catching the == vs. != :)
And since string literals are passed in, that is probably where it crashes.
@DanielFischer Right, so it's probably a good idea to declare both params const.
have run it on your gcc compiler and seen it?
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1

You don't 0-terminate the string in scat. You could add:

p[ptr] = 0;

Right before returning.

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2 Comments

please run it own your gcc compiler and tell me?
Naw man, it could wake Cthulhu because of the undefined behavior. WHOOPS!
0
  1. Please run it on your machine using gcc and tell if it also gives you a segmentation fault output.
  2. I don't think there is any problem with the program.

I don't follow your reasoning. If the program segfaults on your system then it is very likely that the program is at fault.

FYI this is what GCC(*) outputs. It doesn't even compile.

NOTE: It's g++ actually, but it shouldn't make too much of a difference.

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7 Comments

was that output with -Wall, mr strawhat?
i copied this same program some a person. his program works and mine not. i rechecked both the programs 10 times but still the problem persist
@Mike it's compiled with -Wall -Werror -Wextra -pedantic-errors
@KundanNegi - Google search or look at the man for gcc, there are a lot of extra flags that will give you more details and help you find errors/warnings with your code
+1 for -Wall & friends, -1 for compiling C code as C++ :) so you're even ;)
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0

There is no need to emulate the functions that exist in string.h:

#include <string.h>

char *scat(char *s,char *t)
{ 
    char *p;
    size_t len1, len2;
    len1 = strlen(s);
    len2 = strlen(t);

    p=malloc(len1+len2+1);
    if (!p) barf();
    memcpy(p,s, len1);
    memcpy(p+len1, t, len2+1);

    return p;
}
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4 Comments

ya i know my friend but it is an exercise i want to try.
talking about functions that exist in string.h: strcat()
@KundanNegi reimplementing functions that are part of the C standard is generally a bad idea. Try exercising the use of said functions with a more complex example instead and you'll learn more useful aspects of the language.
Well, if you really want to avoid library functions, you could roll your own strlen(), too.

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