Reverse number in cpp is another easy and simple program which mostly asked in exams.
In this tutorial, we are going to learn how to reverse a number programmatically
Reverse Number Logic
We separate digits in number by modulus operation which gives us a digit in a number.
Rem = number % 10
After separating number we store that digit in another variable with considering its unit places which we perform as
Reverse = Reverse × 10 + Remember
We repeated step 1 and 2 till number ends.
Reverse Number Example
Number = 342
Reverse number = 243
C++ Program to Reverse a Number
| 12345678910111213141516171819202122232425262728 | #include <iostream>using namespace std;int main(){ int temporary ,n,rev=0,x;cout<<“Enter a number\n”;cin>>n;temporary =n;while(n>0){ x=n%10;rev=rev*10+x; n=n/10; }cout<<“The reverse of “<<temporary<<“is “<<rev; return 0;} |
Output :
Explanation :-
1. Here we did initialization for
temporary ->To store the entered value(i.e ‘n’) as you will come to know at the end of the program
n->To store number given by a user.
rev->To store the reverse of a number.It is initialized to zero
x->To store n%10.
2. First of all, we got a number ‘n’ from a user and then stored it in a temporary variable called as ‘temporary ‘ for restoring the value.(remember this point).
3. Now the main logic comes here
let the number ‘n’ be 321 and as 321>0, while loop gets executed
then x=321%10—>which is 1.
rev=0*10+1——–>1
n=321/10———>32
The rev for the first loop execution is rev=1.
4. Now the number ‘n’ has become ’32’ and n>0, while loop executes for the 2nd time.
then x=32%10—>which is 2.
rev=1*10+2——–>12
n=32/10———>3
The rev when loop executed the second time is rev=12.
5. Now the number ‘n’ has become ‘3’ and n>0, while loop executes for the 3rd time
then x=3%10—>which is 3.
rev=12*10+3——–>123
n=3/10———>0
The rev when loop executed the third time is rev=123.
6. Now as the number of variable ‘n’ is 0 which is not n>0 then the loop terminates.Then the final reverse is ‘123’.
7. So now I hope you understood why the temporary variable is used.It is because the value in ‘n’ becomes 0 at the end of the program so for restoring this value to print at the end we used ‘temporary ‘(as from the 2nd point).
8. Finally, it prints the value in ‘rev’.