有人可以解释我这段代码吗？ 它给出了我们输入的数字的反面

Question

#include <iostream>

using namespace std;

int main()

{

    long long n,i;
    cout << "Please enter the number:- \n";
    cin >> n;
    while (n!=0) 
    {

        i=n%10;   // what is the use of % here? 
        cout << i;
        n=n/10;
    }
return (1);


}

I want to understand how this code works. Why does it do what it does? If we enter 5123 it will give out 3215. How ?

Answer 1

The very basics of C family languages , operator % gives the remainder of its left hand side argument when divided by the right hand side one. So, n % 10 is n 's last digit in decimal notation (for instance, 3 for 5123.)

Answer 2

Long story short:

operator / performs floor division : only integer part returned, without remainder,

operator % returns remainder of division operation

If needed, below is line by line explanation.

long long n,i;

defines variables n and i of type long long (64-bit unsigned integer).

cout << "Please enter the number:- \n";

prints out a prompt message hinting what is the expected input.

cin >> n;

is a command waiting for standard input, that is saved in int variable n . Please note that invalid value (non-numeric) will cause an error later.

while (n!=0)

starts a loop that would be terminated when/if n becomes equal to zero.

i=n%10;   // what is the use of % here? 

This command returns a remainder of division of value of variable n by 10 . The result of this operation is saved to variable i . Eg

5123 % 10 = 3
 512 % 10 = 2
  51 % 10 = 1
   5 % 10 = 5

Next,

cout << i;

prints the value of variable i to stdout and keeps the cursor on same line, next column. Doing so in a loop it would print out each new value of variable i in a single line, faking a reverse integer number.

And, finally,

n=n/10;

performs operation of "floor division" (division without remainder - returns integer part only) of value of variable n by 10 . Result is saved back to variable n :

5123 / 10 = 512
 512 / 10 = 51
  51 / 10 = 5
   5 / 10 = 0    // after this iteration the loop will terminate