Question

程序应读取2个整数并根据键盘引入的符号计算总和或乘积。如果您在任何给定的设备上按q，则必须退出。

#include "stdafx.h"
#include <iostream>
#include<conio.h>

using namespace std;

int main()
{

char k, l ='h',c;     
int a,b,s, p;             

aici: while (l != 'q')
{

    cin >> a;


    if (_kbhit() == 0) //getting out of the loop
    {
        c = a;
        if (c == 'q')
        {
            l = 'q';
            goto aici;
        }
    }


    cin >> b;

    if (_kbhit() == 0)
    {
        c = b;
        if (c == 'q')
        {
            l = 'q';
            goto aici;
        }
    }


    k = _getch();

    if (_kbhit() == 0)
    {
        c = k;
        if (c == 'q')
        {
            l = 'q';
            goto aici;
        }
    }


    if (k == '+')
    {

        s =(int)(a + b);
        cout << s;
    }
    if (k == '*')
    {
        p = (int)(a*b);
        cout << p;
    }
}
return 0;
}

它希望a和b都是整数，因此输入＆＃39; q＆＃39;弄得一团糟。是否有可能使程序工作而不将a和b声明为chars？

Answer 1

你不需要在cin中使用goto和kbhit（）。一个简单的方法是：

#include <iostream>
#include <string>

using namespace std;

int main()
{
    int a,b;
    string A, B
    char k;

    while(1)
    {
        cin >> A;
        if(A == "q") break;
        cin >> B;
        if(B == "q") break;

        a = atoi(A.c_str());
        b = atoi(B.c_str());

        cin >> k;
        if(k == 'q') break;

        if (k == '+')
            cout << (a + b);
        if (k == '*')
            cout << (a*b);

    }
}

Answer 2

您无法获得想要走这条路的位置。标准输入流读取将阻止，阻止您查找＆＃39; q＆＃39;并退出。

而是查看＆＃39; q＆＃39;的所有输入。因为它有它并在你有完整的消息后将它转换为你需要的值。类似的东西：

while (int input = _getch()) != 'q') // if read character not q
{
    accumulate input into tokens
    if enough complete tokens
       convert numeric tokens into number with std::stoi or similar
       perform operation
       print output 
}

你可以按照

的方式做点什么

std::stringstream accumulator;
while (int input = _getch()) != 'q') // if read character not q
{
    accumulator << std::static_cast<char>(input);
    if (got enough complete tokens)// this be the hard part
    {
        int a;
        int b;
        char op;
        if (accumulator >> a >> b >> op)
        { // read correct data
            perform operation op on a and b
            print output 
        }
        accumulator.clear(); // clear any error conditions
        accumulator.str(std::string()); // empty the accumulator
    }
}

std::stringstream documentation.

检查是否按下了字符

2 个答案: