我是C的新手.我正在使用命令行编写一个简单的计算器.命令行应具有以下格式：
programname firstNumber运算符secondNumber
这是我到目前为止所得到的：

#include <iostream>
#include <fstream>
using namespace std;

int main(int argc, char* argv[])
{
    if (argc != 3)
    {
        cerr << "Usage: " << argv[0] << endl;
        exit(0);
    }
    else
    {
        int firstNumber = atoi(argv[1]);
        char theOperator = atoi(argv[2]);
        int secondNumber = atoi(argv[3]);
        switch (theOperator)
        {
        case'+':
            {
                cout << "The answer is " << firstNumber + secondNumber << endl;
                break;
            }
        case '-':
            {
                cout << "The answer is " << firstNumber - secondNumber << endl;
                break;
            }
        case '*':
            {
                cout << "The answer is " << firstNumber * secondNumber << endl;
                break;
            }
        case '/':
            {
                if (secondNumber == 0)
                {
                    cout << "Can not devide by a ZERO" << endl;
                    break;
                }
                else
                {
                    cout << "The answer is " << firstNumber / secondNumber << endl;
                    break;
                }
            }
        }
    }
}

该程序无法运行.当我运行它时,它会显示一个适当的用法消息并结束程序.谁能帮帮我吗？

解决方法:

其他人已经给你答案,但你可以很容易地自己想出这个.只需在您知道代码进入的位置打印argc：

int main(int argc, char* argv[])
{
    if (argc != 3)
    {
        cout << "argc is: " << argc << endl; // Debug output that you delete later
        cerr << "Usage: " << argv[0] << endl;
        exit(0);
    }
    else

然后回过头来看看argc.当你发现argc实际上是4并且你想知道argc里面是什么时你应该写一些代码来打印它,这样你就可以搞清楚……就像这样：

int main(int argc, char* argv[])
{
    cout << "argc is: " << argc << endl; // Debug output that you delete later
    for (int i = 0; i < argc; ++i)
    {
        // Print out all of the arguments since it's not working as you expect...
        cout << "argv[" << i << "] = " << argv[i] << endl;
    }

    if (argc != 3)
    {
        cerr << "Usage: " << argv[0] << endl;
        exit(0);
    }
    else

而你很快就会弄清楚出了什么问题……

请学习如何做到这一点,因为它会保存你的,但将来你不必在这里等待答案.

此外,您的代码中还有另一个错误.

为什么你要将字符从字符串转换为int？

else
{
    int firstNumber = atoi(argv[1]);
    char theOperator = atoi(argv[2]); // <<< WTF? Why?
    int secondNumber = atoi(argv[3]);
    switch (theOperator)

你可能想摆脱那里的atoi部分,然后去：

char theOperator = argv[2][0]; // First character of the string

假设第二个参数将始终只有一个字母…您可能希望强制执行/检查.请参阅strlen()和std::string并注意argv [2]的类型是char *(指向char的指针).

我还建议您阅读从SO Howto-Ask Help Page链接的How to debug small programs.这可能会有所帮助.不,我认为你的问题不好.调试小程序是您将来需要的一项技能,如果您打算编程,那么现在就可以从中受益.

欢迎编程和C