用于检查输入是否为leap年的C ++程序 [英] C++ program for checking whether the input is a leap year

问题描述

我是Tony，是C ++编程的新手.我想问一个有关创建程序来检查leap年的问题.

I am Tony and I am new to c++ programming. I would like to ask a question related to creating a program to check for leap year.

在以下代码中，我尝试创建一个布尔函数来检查输入是否为a年.如果输入为负，我将提示再见！"并立即停止该程序.如果输入为正，那么我将使用自己构建的布尔函数检查是否为a年，直到输入为负数，然后退出程序.

In the following codes, I try to create a bool function to check whether the input is a leap year. If the input is negative, I will cout "Bye!" and stop the program immediately. If the input is positive, then I will check whether it is a leap year using the bool function I built until the input is a negative number then I will exit the program.

但是，我无法找到自己所犯的错误以及当前的状况，当我输入一个正值时，不会产生任何结果.如果有空，请帮忙.非常感谢您. :)

However, I am not able to find what mistakes I have made and the current situation is, when I input a positive value, there is no result generated. Please help if you are available. Much thanks to you. : )

#include <iostream>
#include <cmath>
#include <string>
#include <iomanip>

using namespace std;
bool leap_year(int year);
int main()
{
    int year; 
    while (cout << "Enter a year (or negative number to quit): ")
    {
        cin >> year;
        if (leap_year(year) == false && year <0 ) 
        {
            cout << "Bye!" << endl;
        }
        break;
        if (leap_year(year) == false && year >0 )
        {
            cout << "The year is not a leap year." << endl;
        }
        if (leap_year(year) == true && year >0 ) 
        {
            cout << "The year is a leap year." << endl;
        }
        return 0;
    }    
}
  bool leap_year(int year)
{
    bool is_leap_year = false;
    if (year % 4 == 0)
    {
       is_leap_year = true;
    }
    if (year % 100 == 0)
    {
        is_leap_year = false;
    }
    if (year % 400 == 0)
    {
        is_leap_year = true;
    }
    return is_leap_year;
}

推荐答案

首先，您(应该)需要一个while(true)循环而不是一个while(std::ostream)循环.

First of all, you (should) want a while(true) loop and not a while(std::ostream) loop.

所以替换

while (cout << "Enter a year (or negative number to quit): ")
{

使用

while (true)
{
    cout << "Enter a year (or negative number to quit): ";

正如@paddy指出的那样，您可以检查std :: ostream的返回类型以在打印时查找错误.但是在这个简单的程序中，我怀疑是否有必要.

As @paddy pointed out, you can check std::ostream`s return type to look for errors when printing out. But in this simple program I doubt it's necessary.

然后，在if语句之外有break，它将始终脱离程序(无论输入如何).替换

Then you have the break outside your if statement, which will always break out of the program (no matter the Input). Replace

if (leap_year(year) == false && year <0 ) 
{
    cout << "Bye!" << endl;
}
break;

使用

if (year < 0)
{
    cout << "Bye!" << endl;
    break;
}

(无需检查负输入是否为a年.您可以使用 if-else语句，因此您也可以像我一样将if(leap_year(year) == false && year < 0)替换为if (year < 0).)

(there's no need to check if the negative input is a leap year. You can achieve entering only 1 if-statement with if-else statments, therefore you can also replace if(leap_year(year) == false && year < 0) with just if (year < 0); as I did.)

将其应用于所有语句(不更改其内部逻辑)并在循环末尾删除return 0;时，您将获得所需的程序流.同样，删除using namespace std;更好(请在此处 ).您也不需要包含<iomanip>，<cmath>或<string>.完整代码:

When you apply this to all statements (not changing their internal logic) and remove the return 0; at the end of the loop, you get your desired program flow. Also removing using namespace std; is just better (read here why). You also don't Need to include <iomanip>, <cmath> nor <string>. Full Code:

#include <iostream>

bool leap_year(int year);
int main() {
    int year;
    while (true) {
        std::cout << "Enter a year (or negative number to quit): ";
        std::cin >> year;
        if (year < 0) {
            std::cout << "Bye!" << std::endl;
            break;
        }
        else if (leap_year(year)) {
            std::cout << "The year is a leap year." << std::endl;
        }
        else {
            std::cout << "The year is not a leap year." << std::endl;
        }
    }
}
bool leap_year(int year){
    bool is_leap_year = false;
    if (year % 4 == 0){
        is_leap_year = true;
    }
    if (year % 100 == 0){
        is_leap_year = false;
    }
    if (year % 400 == 0){
        is_leap_year = true;
    }
    return is_leap_year;
}

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