CODEWITHSUNDEEP
c
redsoxlost
redsoxlost

Reputation: 1235

Segmentation Fault - String Concatenation in C Error

I am fairly new to C programming and still trying to understand all the nooks and crannies of C. I am writing a program to concatenate two strings. But I am getting an error which I don't understand. Here is the output.

Asfakul
The Length of the String name is 7
The Length of the String fullname is 7
L
Segmentation fault (core dumped) 

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

int  main(int argc, char const *argv[])
{
  char *name="Asfakul";
  char *surname="Laskar";
  char *fullname;
  int i=0;
  //Allocate Memory of 100 char 
  fullname=(char*)malloc(100*sizeof(char));
  fullname=name;
  while(*name !='\0')
  {
    i++;
    name++;
  }

  // Allocate Memory for FullName

  //fullname=(char*)malloc(100*sizeof(char));
  //Coppied the spurce String
 // fullname=name; // Here this assignement will not work as Pointer name now point to NULL character of String Name.
  puts(fullname);
  printf("The Length of the String name is %d\n",i );
  printf("The Length of the String fullname is %d\n",strlen(fullname) );

  while(*surname !='\0')
  {
    printf("%c\n",*(fullname+i+1));
    *(fullname+i+2)=*(surname);
    printf("%c\n",*(surname));
    i++;
    surname++;

  }
  puts(fullname);



  return 0;
}

Please help me understand what I am doing wrong.

Upvotes: 0

Views: 396

Answers (3)

&#212;rel
&#212;rel

Reputation: 7622

You can past char by char

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

int  main(int argc, char const *argv[])
{
  char *name="Asfakul";
  char *surname="Laskar";
  char *fullname;
  int i=0;
  //Allocate Memory of 100 char 
  fullname=malloc(100*sizeof(char));
  while(*name !='\0')
  {
    fullname[i++] = *name++;
  }

  fullname[i] = '\0';

  puts(fullname);
  printf("The Length of the String name is %d\n",i );
  printf("The Length of the String fullname is %zd\n",strlen(fullname) );
  fullname[i++] = ' ';
  while(*surname !='\0')
  {
    fullname[i++]= *surname++;    
  }
  fullname[i] = '\0';

  puts(fullname);

  free(fullname);
  return 0;
}

Upvotes: 0

Kotshi
Kotshi

Reputation: 338

*(fullname+i+2)=*(surname);

Here you are trying to cat the surname at the end of the name:

  char *name="Asfakul";

This is read-only space.

You should alloc enough space for both strings and copy them inside the allocated space.

Upvotes: 0

Bathsheba
Bathsheba

Reputation: 234715

fullname = name; assigns the pointer name to fullname. You subsequently modify the data at name. That's not allowed since name points to a read-only string literal.

You're also discarding the malloc pointer, leaving you no way to free the allocated memory! This will not end well.

You should take a deep copy of name instead: consider using a strncpy.

If you were to use const char* for the string literals then compilation should fail, so protecting yourself from these kind of things.

Upvotes: 6

Related Questions