Q:

C++ Program To Print A Reverse Number Using Loop

0

Write A C++ Program To Print A Reverse Order Of Any Number Using Loop

Logic :- 

Very Simple Logic Just Divide a number by 10 and then proceed for next statement 

Sum=0
sum=sum*10+X

Note :- Always remember Do Not try to print the reverse Digit always try to modified 
Like you can also solve the problem

while(n!=0)
{
rev=n%10;
System.out.print("Reversed Number = "+reverse);
n=n/10
}
but in this way number is printing actually not reversing .

All Answers

need an explanation for this answer? contact us directly to get an explanation for this answer

#include<iostream>
using namespace std;
int main()
{

 int n,x,sum=0;

 cout<<"Enter The Number To Be Reverse: \n";
 cin>>n;

 while(n>0)
 {
  x=n%10;
  sum=sum*10+x;
  n=n/10;
 }
 cout<<"\nThe Reverse Number is "<<sum;
}

 

Output:

Enter The Number To Be Reverse: 

12345

The Reverse Number is 54321

need an explanation for this answer? contact us directly to get an explanation for this answer

total answers (1)

This question belongs to these collections

Similar questions


need a help?


find thousands of online teachers now