C programming strcat using pointer

Question

I am a beginner in C. I wanted to make strcat function using pointers. I made it but don't know what is wrong with it. I used gcc compiler and it gave segmentation fault output.

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

char scat(char *,char *);

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

    char *q=scat(s,t);
    while(*q!='\0') printf("the concatenated string is %c",*q);
}

char *scat(char *s,char *t)
{
    char *p=s; 
    while(*p!='\0'){
        p++;
    } 
    while(*t!='\0'){
        *p=*t;
        p++;
        t++;
    }
    return p-s-t;
}

BTW, the main function should return an int instead of void. PLease make sure that your compiler is configered to emit warnings. — hugomg
– hugomg, Commented Jan 10, 2013 at 14:03

Bart Friederichs · Accepted Answer · 2013-01-10 14:32:44Z

8

This one works:

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

char *scat(char *,char *);                 /* 1: your prototype was wrong */

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

    char *q=scat(s,t);   
    printf("cat: %s\n", q);               /* 2: you can use %s to print a string */
    free(q);
}

char *scat(char *s,char *t)
{
    char *p=malloc(strlen(s)+strlen(t)+1);    /* 3: you will have to reserve memory to hold the copy. */
    int ptr =0, temp = 0;                   /* 4 initialise some helpers */

    while(s[temp]!='\0'){                  /* 5. use the temp to "walk" over string 1 */
        p[ptr++] = s[temp++];
    }
    temp=0;
    while(t[temp]!='\0'){                   /* and string two */
        p[ptr++]=t[temp++];
    }
    return p;
}

edited Jan 10, 2013 at 14:32

answered Jan 10, 2013 at 14:05

Bart Friederichs

33.8k16 gold badges112 silver badges219 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Kundan Negi Over a year ago

Your way of explanation was amazing by pointing out mistakes in comments. Thanks buddy !!!!!!

kame Over a year ago

But this doesn't answer the question. You asked to do it with pointers instead of arrays.

GrandMarquis · Accepted Answer · 2013-01-10 13:59:54Z

3

You have to allocate new space to copy at the end of s. Otherwise, your while loo[ will go in memory you don't have access to.

You shoul learn about malloc() here.

answered Jan 10, 2013 at 13:59

GrandMarquis

1,9111 gold badge18 silver badges30 bronze badges

Comments

hmjd · Accepted Answer · 2013-01-10 14:06:00Z

3

It is undefined behaviour to modify a string literal and s, and eventually p, is pointing to a string literal:

char* s = "james";

s is passed as first argument to scat() to which the local char* p is assigned and then:

*p=*t;

which on first invocation is attempting to overwite the null character an the end of the string literal "james".

A possible solution would be to use malloc() to allocate a buffer large enough to contain the concatentation of the two input strings:

char* result = malloc(strlen(s) + strlen(p) + 1); /* + 1 for null terminator. */

and copy them into it. The caller must remember to free() the returned char*.

You may find the list of frequently asked pointer questions useful.

edited Jan 10, 2013 at 14:06

answered Jan 10, 2013 at 14:00

hmjd

122k21 gold badges216 silver badges259 bronze badges

Comments

hmatar · Accepted Answer · 2013-01-10 13:59:07Z

2

Because p goes till the end of the string and then it starts advancing to illegal memory. That is why you get segmentation fault.

answered Jan 10, 2013 at 13:59

hmatar

2,4372 gold badges19 silver badges27 bronze badges

2 Comments

Kundan Negi Over a year ago

what do you mean by advancing to illegal? pointer p was incremented to till end of string,so p-s-t was done by me to reach the first address of string and then use indirection operator * to get the characters.Thanks in advance.

hmatar Over a year ago

char *scat(char *s,char *t) { char *p=s; while(*p!='\0'){ // here means you have reached end of "string" p++; // you are advancing pointer } while(*t!='\0'){ *p=*t; p++; // here you are advancing past end of string. t++; } return p-s-t; }

anishsane · Accepted Answer · 2013-01-10 13:58:44Z

0

It's because s points to "james\0", string literal & you cannot modify constant.

Change char *s="james"; to char s[50]="james";.

answered Jan 10, 2013 at 13:58

anishsane

21.1k5 gold badges46 silver badges77 bronze badges

Comments

Ahmed Kato · Accepted Answer · 2013-01-10 14:01:17Z

0

You need to understand the basics of pointers.

a char * is not a string or array of characters, it's the address of the beginning of the data.

you can't do a char * - char* !!

This is a good tutorial to start with

you will have to use malloc

answered Jan 10, 2013 at 14:01

Ahmed Kato

1,7076 gold badges29 silver badges54 bronze badges

2 Comments

Kundan Negi Over a year ago

But as you said in your 2nd line that its points to addresses of the beginning of data, so that address must be in integer. So why can't i subtrat an int from int

Ahmed Kato Over a year ago

you can and it will give you a new address but it is useless.

Mecki · Accepted Answer · 2013-01-10 14:09:09Z

0

You get a segmentation fault because you move the pointer to the end of s and then just start writing the data of p to the memory directly following s. What makes you believe there is writable memory available after s? Any attempt to write data to non-writable memory results in a segmentation fault and it looks like the memory following s is not writable (which is to expect, since "string constants" are usually stored in read-only memory).

answered Jan 10, 2013 at 14:09

Mecki

135k36 gold badges270 silver badges278 bronze badges

Comments

Nobilis · Accepted Answer · 2013-01-10 14:09:30Z

0

Several things look out of order.

First keep in mind that when you want to return a pointer to something created within a function it needs to have been malloc'ed somewhere. Much easier if you pass the destination as an argument to the function. If you follow the former approach, don't forget to free() it when you're done with it.

Also, the function scat has to return a pointer in the declaration i.e. char *scat, not char scat.

Finally you don't need that loop to print the string, printf("%s", string); will take care of printing the string for you (provided it's terminated).

answered Jan 10, 2013 at 14:09

Nobilis

7,4982 gold badges36 silver badges72 bronze badges

Comments

nvseenu · Accepted Answer · 2013-01-10 14:36:59Z

At first, your code will be in infinte loop because of the below line. you were supposed to use curely braces by including "p++; t++ " statements.

while(*t!='\0')
 *p=*t;

though you do like this, you are trying to alter the content of the string literal. which will result in undefined behavior like segmentation fault.

A sequence of characters enclosed with in double quotes are called as string literal. it is also called as "string". String is fixed in size. once you created, you can't extend its size and alter the contents. Doing so will lead to undefined behavior.

To solve this problem , you need to allocate a new character array whose size is sum of the length of two strings passed. then append the two strings into the new array. finally return the address of the new array.

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

char* scat(char *,char *);
void append(char *t , char *s);

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

    char *n = scat(s,t);        
    printf("the concatenated string is %s",n);

    return 0;
}

char* scat(char *s,char *t)
{
    int len = strlen(s) + strlen(t);
    char *tmp = (char *)malloc(sizeof(char)* len);

    append(tmp,s);
    append(tmp,t);

    return tmp;
} 


void append(char *t , char *s)
{   
     //move pointer t to end of the string it points. 
    while(*t != '\0'){
        t++;
    }

    while( *s != '\0' ){
        *t = *s;
        t++;
        s++;    
    }       
}

I think you have to change the code to: tmp[0] = '\0'; append(tmp,s); tmp[strlen(s)] = '\0'; append(tmp,t); tmp[len] = '\0';

Collectives™ on Stack Overflow

C programming strcat using pointer

9 Answers 9

2 Comments

Comments

Comments

2 Comments

Comments

2 Comments

Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

9 Answers 9

2 Comments

Comments

Comments

2 Comments

Comments

2 Comments

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related