正十进制整数的递归函数数字,以相反的顺序c

我有一项工作要写一个递归函数,该函数以相反的顺序写一个正整数的数字.我的问题是该功能无法正确显示反向.我知道我应该在显示数字时使用％或10,而在递归调用中以及在基本情况应该为<时使用10的/. 10.这是我的代码.

#include <iostream>
using namespace std;

int reverse(int,int);

int main()
{
    int number;
    int n;

    cout << " Enter number to reverse." << endl;
    cin >> number;
    cout << reverse(number % 10,0);

    return 0;
}//end main

int reverse(int number,int n)
{

    if(n < 10)
    {
        return n;
    }
    else
    {
        return reverse(number/10,n);
    }
}// end reverse

解决方法:

我认为这是您的职能应为：

void reverse(int number){
    if(number == 0) //base/basic case i.e if number is zero the problem is already solved, nothing to do, so simply return
        return;
    else{
        cout << number % 10; // print that last digit, e.g 103%10 == 3 
        reverse(number/10); //solve the same problem but with smaller number, i.e make the problem smaller by dividing it by 10,  initially we had 103, now 10 
    }
}

标签：digits,c,variable-assignment,reverse
来源： https://codeday.me/bug/20191011/1891469.html