CONCATENATION OF TWO STRINGS USING POINTER

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Hi,

CONCATENATION OF TWO STRINGS USING POINTER, it is not working. pls help me..........

C#

#include<iostream>
using namespace std;

int main()
{
    char *str1 = "krishna";
    char *str2 = "Prasad";
    char *str3;
    int count = 0;
    int i,j;
    for(i=0;*str1!=NULL;i++,*str1++)
    {
        count++;
    }
    for(j= 0; *str2!=NULL;j++,*str2++)
    {
        count++;
    }
    //str3 =  new char[sizeof(str1)+sizeof(str2)];
    str3 = (char*) malloc(13);


    for(i=0;str1[i]!=NULL;i++)
    {
        str3[i] = str1[i];
    }

    for(j= 0; str2[j]!=NULL;j++,i++)
    {
        str3[i] = str2[j];

    }
    cout<<endl;
    cout<<"String 3 is = ";

    while(*str3)
    {
        cout<<*str3++;
    }
    return 0;
}

Posted 24-Jul-15 8:43am

Member 11417637

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Comments

Dave Kreskowiak 24-Jul-15 15:03pm

First, WHY ARE YOU YELLING AT PEOPLE?

Second, before anyone is going to touch this, you might want to define exactly what you mean by "it is not working". That's not a problem description.

What DOES happen when you run this?

[no name] 24-Jul-15 15:09pm

In addition to what Dave said, running your code under your debugger should tell you what you code is doing and why. Learn how to debug your code.

CPallini 24-Jul-15 15:52pm

Why are you using C-like strings in C++?

3 solutions

Solution 3

Sorry, but when I see code like that I can't resist :).
C language is basic enough to don't be worth to reinvent the wheel.
Use standard routines and code it as it should be:

C++

int main()
{
    char *str1 = "krishna";
    char *str2 = "Prasad";
    char *str3;

    str3 = malloc(strlen(str1) + strlen(str2) + 1);  //Allocate space for the sum of the lenghts of the 2 strings plus the final null.
    strcpy(str3, str1);    //Copy first string in the result
    strcat(str3, str2):    //Append the second string.

    cout << str3;     //print whole new string!

    return 0;
}

Posted 25-Jul-15 3:45am

Frankie-C

v3

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bling · Accepted Answer · 2015-07-24T09:41:00

When using malloc or new to allocate an array of characters for a nul terminated string, you need to include one extra character for the terminating nul.

Replace this:

C

str3 = (char*) malloc(13);

with this:

C

str3 = (char*) malloc(count + 1);

or this:

C

str3 = new char[count + 1];

Don't forget to terminate the new string with a nul character.

Replace this:

C

for(j= 0; str2[j]!=NULL;j++,i++)
{
    str3[i] = str2[j];
}

with this:

C

for(j=0; str2[j]!=0; j++,i++)
{
    str3[i] = str2[j];
}
str3[i] = 0;

Now that the new string is nul terminated, you can replace this:

C

while(*str3)
{
    cout<<*str3++;
}

with this:

C

cout << str3;

The following code tells me you tried something but it did not work.

C

//str3 =  new char[sizeof(str1)+sizeof(str2)];

The reason is the sizeof operator does not give you the length of a string but the size of the string pointer.

CPallini · Accepted Answer · 2015-07-24T10:18:00

As suggested you should properly handle string terminators. Moreover, you should handle a possibly failing call to malloc and properly release dynamically allocated memory.

C

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

 int my_strlen(const char * s)
 {
   int l = 0;
   while ( s[l] ) ++l;
   return l;
 }

 int main()
 {
   const char *s1 = "krishna";
   const char *s2 = "Prasad";

   int l1 = my_strlen(s1);
   int l2 = my_strlen(s2);

   char * s3 = (char *) malloc(l1 + l2 + 1); // make room for the string terminator

   if ( ! s3 ) return -1; // malloc failure

   int n1, n2, n3;
   n3 = 0;
   for (n1=0; n1 < l1; ++n1, ++n3)
     s3[n3] = s1[n1];

   for (n2=0; n2 < l2; ++n2, ++n3)
     s3[n3] = s2[n2];

   s3[n3] = '\0'; // append the string terminator

   printf("string 3 is %s\n", s3);

   free(s3); // release allocated memory

   return 0;
}

CONCATENATION OF TWO STRINGS USING POINTER

3 solutions

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