C++ Program to Copy One String to Another

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In this article, you will learn and get code on copy one string to another in C++. The program is created with following approaches:

Copy String without using strcpy() Function
Using Pointer
Using user-defined Function
Using library function, strcpy()

Copy String without using strcpy() Function

This program copies the string (entered by user at run-time) without using any library function like strcpy() Let's have a look at the program first. Its explanation is given later on:

The question is, write a program in C++ that copies one string to another. Here is its answer:

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    char strOrig[100], strCopy[100], i=0;
    cout<<"Enter the string: ";
    gets(strOrig);
    while(strOrig[i]!='\0')
    {
        strCopy[i] = strOrig[i];
        i++;
    }
    strCopy[i] = '\0';
    cout<<"\nEntered String: "<<strOrig;
    cout<<"\nCopied String: "<<strCopy;
    cout<<endl;
    return 0;
}

This program was build and run under Code::Blocks IDE. Here is its sample run:

Now enter the string say codescracker and press ENTER key to copy it to another string. Here is the final snapshot of its sample run:

The dry run of above program with user input, codescracker goes like:

Initial value, i=0
When user enters the string say codescracker, it gets stored in a variable, strOrig[] in following way:
- strOrig[0]=c
- strOrig[1]=o
- strOrig[2]=d
- strOrig[3]=e
- strOrig[4]=s
- strOrig[5]=c
- strOrig[6]=r
- strOrig[7]=a
- strOrig[8]=c
- strOrig[9]=k
- strOrig[10]=e
- strOrig[11]=r
That is, the first character (c) of the string (codescracker) gets stored at very first index (0) of the strOrig[]. And the second character (o) gets stored at second index (1), and so on.
Because the maximum size of strOrig[] is 100, and the entered string is only of 12 character long, therefore at 12^th index (index after the last character's index), a null terminated character (\0) gets automatically initialized. That indicates the end of string, or no any character left in strOrig from this (12^th) index.
Now the condition, strOrig[i]!='\0' or strOrig[0]!='\0' or c!='\0' evaluates to be true, therefore program flow goes inside the loop and the 0^th index's character of strOrig[] gets initialized to strCopy[] at same index
That is, strOrig[i] or strOrig[0] or c gets initialized to strCopy[i] or strCopy[0]
Now the value of i gets incremented. So i=1
Program flow goes back and evaluates the condition of while loop again.
Because a null terminated character (\0) is available at 12^th index, therefore both the statements (availabe in the body of while loop) gets executed 12 times
Therefore all the characters gets copied to strCopy[] variable one by one
And when the execution of while loop gets ended, don't forgot to initialize a null terminated character at the index after last character's index
Now just print the value of both the variables say strOrig[] and strCopy[]

Copy String using Pointer

This program copies the string using pointer. The question is, write a program in C++ to copy the content of one string to another using pointer. Here is its answer:

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    char strOrig[100], strCopy[100];
    char *origPtr, *copPtr;
    cout<<"Enter the string: ";
    gets(strOrig);
    origPtr = &strOrig[0];
    copPtr = &strCopy[0];
    while(*origPtr)
    {
        *copPtr = *origPtr;
        origPtr++;
        copPtr++;
    }
    *copPtr = '\0';
    cout<<"\nEntered String: "<<strOrig;
    cout<<"\nCopied String: "<<strCopy;
    cout<<endl;
    return 0;
}

Here is its sample run with user input, Hello codescracker

The statement,

origPtr = &strOrig[0];

initializes the initial address of strOrig[] variable. That is its 0^th index's address. And the statement,

copPtr = &strCopy[0];

also initializes the 0^th index's address of strCopy[]

Note - If a pointer type variable holds address of any variable that contains some values, then anything happened with the variable (pointer) makes changes to original. Because it holds its address, not value. So any operation perform through pointer type variable, directly effects the value at that address.

The * is called as value at operator. And & is called as address of operator.

The statement, origPtr++; means, origPtr now holds next index's address.

So using pointer also, all the character of original string gets copied (in one-by-one manner) to strCopy[]

Copy String using User-defined Function

This program also copies one string to another, but using a user-defined function named cpystr()

#include<iostream>
#include<stdio.h>
using namespace std;
void cpystr(char *, char *);
int main()
{
    char strOrig[100], strCopy[100];
    cout<<"Enter the string: ";
    gets(strOrig);
    cpystr(strOrig, strCopy);
    cout<<"\nEntered String: "<<strOrig;
    cout<<"\nCopied String: "<<strCopy;
    cout<<endl;
    return 0;
}
void cpystr(char *origPtr, char *copPtr)
{
    while(*origPtr)
    {
        *copPtr = *origPtr;
        origPtr++;
        copPtr++;
    }
    *copPtr = '\0';
}

This program produces the same output as of previous program.

Using strcpy() Function

This is the last program that uses a library function of C++, named strcpy() to copy string. It takes two string as its argument. The value of second argument gets copied to first.

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
    char strOrig[100], strCopy[100];
    cout<<"Enter the string: ";
    gets(strOrig);
    strcpy(strCopy, strOrig);
    cout<<"\nEntered String: "<<strOrig;
    cout<<"\nCopied String: "<<strCopy;
    cout<<endl;
    return 0;
}

Same Program in Other Languages

C++ Online Test

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