Question

我试图运行这个程序，你在其中插入两个双值，然后你得到两个数字中哪一个更大，一个消息＆＃34;数字相等＆＃34;如果两个double的值相同，并且消息＆＃34;数字几乎相等＆＃34;如果两个数字之间的差异小于1 / 10.000.000。

我写的代码是：

    #include "stdafx.h"
    #include <iostream>
    #include <string>
    #include <vector>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    inline void keep_window_open() { char ch; cin >> ch; }

    double square(double x)
    {
        return x * x;
    }

    int main() {
        double number;
        double number2;
        while (cin >> number, cin >> number2) {
            if (number == '|' || number2 == '|') break;
            else {
                cout << number << " " << number2 << endl;
                if (number > number2) cout << number << " is bigger than " << number2 << endl;
                if (number < number2) cout << number2 << " is bigger than " << number << endl;
                double difference = number - number2;
                if (difference > - 1/10000000 && difference < 0) cout << "the numbers are almost equal" << endl;
                if (difference < 1 / 10000000 && difference > 0) cout << "the numbers are almost equal" << endl;
                if (number == number2) cout << "the numbers are equal" << endl;
            }

        }
    }

当我运行此程序并插入两个差异小于1 / 10.000.000的数字时，程序应该返回＆＃34;数字几乎相等＆＃34;，但它没有。< / p>

如果我插入＆＃34; 1＆＃34;它返回的是什么（当我运行程序并插入两个数字时）和＆＃34; 1.0000000000001＆＃34;：

    "1 1.0000000000001

    1 1

    1 is bigger than 1"

我不明白为什么程序认为我加入的两个数字是1，并返回＆＃34; 1＆＃34;和＆＃34; 1＆＃34;作为价值观。

它也应该给予回报＆＃34;数字几乎相等＆＃34;自从我写道：

     if (difference > - 1/10000000 && difference < 0) cout << "the numbers are almost equal" << endl;
                        if (difference < 1 / 10000000 && difference > 0) cout << "the numbers are almost equal" << endl;

但它没有。

如何让程序返回＆＃34;数字几乎相等＆＃34;当我插入两个相差小于1 / 10.000.000的双值时（例如＆＃34; 1＆＃34;和＆＃34; 1.0000000000001＆＃34;）？

Answer 1

当您尝试打印时，

cout会将这些双精度值四舍五入。见here

对于您的代码，整数除法（1/10000000）将始终返回一个整数，因此您需要执行双值或十进制值的除法（1.0 / 10000000.0）

这是一个工作代码，对逻辑进行了一些细微的修改

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
inline void keep_window_open() { char ch; cin >> ch; }

double square(double x)
{
    return x * x;
}

int main() {
    double number;
    double number2;
    while (cin >> number, cin >> number2) {
        if (number == '|' || number2 == '|') break;
        else {
            cout.precision(15);
            cout << number << " " << number2 << endl;
            double difference = number - number2;
            if (difference > - 1/10000000.0 && difference < 0) cout << "the numbers are almost equal" << endl;
            else if (difference < 1/10000000.0 && difference > 0) cout << "the numbers are almost equal" << endl;
            else if (number == number2) cout << "the numbers are equal" << endl;
            else if (number > number2) cout << number << " is bigger than " << number2 << endl;
            else if (number < number2) cout << number2 << " is bigger than " << number << endl;
        }

    }
}

用C ++逼近双变量

1 个答案: