Question

我试图制作一个用户只能输入0到1的cin。如果用户没有输入这些数字，那么他应该收到错误，说“请输入0到1的范围。”

但它不起作用。

我做错了什么？

   int alphaval = -1;
    do
    {
        std::cout << "Enter Alpha between [0, 1]:  ";
        while (!(std::cin >> alphaval)) // while the input is invalid
        {
            std::cin.clear(); // clear the fail bit
            std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n'); // ignore the invalid entry
            std::cout << "Invalid Entry!  Please Enter a valid value:  ";
        }
    }
    while (0 > alphaval || 1 < alphaval);

    Alpha = alphaval;

Answer 1

试试这个：

int alphaval;
cout << "Enter a number between 0 and 1: ";
cin >> alphaval;
while (alphaval < 0 || alphaval > 1)
{
        cout << "Invalid entry! Please enter a valid value: ";
        cin >> alphaval;
}

Answer 2

如果你想捕获空行我会使用std::getline，然后解析字符串以查看输入是否有效。

这样的事情：

#include <iostream>
#include <sstream>
#include <string>

int main()
{
    int alphaval = -1;
    for(;;)
    {
        std::cout << "Enter Alpha between [0, 1]:  ";

        std::string line;
        std::getline(std::cin, line);
        if(!line.empty())
        {
            std::stringstream s(line);
            //If an int was parsed, the stream is now empty, and it fits the range break out of the loop.
            if(s >> alphaval && s.eof() && (alphaval >= 0 && alphaval <= 1))
            {
                break;
            }
        }
        std::cout << "Invalid Entry!\n";
    }
    std::cout << "Alpha = " << alphaval << "\n";

    return 0;
}

如果您想要一个不同的错误提示，那么我会将初始提示放在循环之外，并将内部提示更改为您喜欢的内容。

Answer 3

C ++第一周，从Peggy Fisher's Learning C++ on Lynda.com开始。这就是我提出的。喜欢收到反馈。

int GetIntFromRange(int lower, int upper){
    //variable that we'll assign input to
    int input; 
    //clear any previous inputs so that we don't take anything from previous lines
    cin.clear(); 
    cin.ignore(numeric_limits<streamsize>::max(), '\n');

    //First error catch. If it's not an integer, don't even let it get to bounds control
    while(!(cin>>input)) {
        cout << "Wrong Input Type. Please try again.\n";
        cin.clear();
        cin.ignore(numeric_limits<streamsize>::max(), '\n');
   }

    //Bounds control
    while(input < lower || input > upper) {
        cout << "Out of Range. Re-enter option: ";
        cin.ignore(numeric_limits<streamsize>::max(), '\n');

        //Second error catch. If out of range integer was entered, and then a non-integer this second one shall catch it
        while(!(cin>>input)) {
            cout << "Wrong Input Type. Please try again.\n";
            cin.clear();
            cin.ignore(numeric_limits<streamsize>::max(), '\n');
        }
    }

    //return the cin input
    return input;
}

由于练习是为了命令汉堡包，这就是我要求的金额：

int main(){
    amount=GetIntFromRange(0,20);
}

Answer 4

我在寻找一种类似的方法，发现那不错对照所有不是0或1的东西进行检查。

size_t input;

 cout << "Enter Alpha between [0, 1]: ";

do
{
    cin.clear();
    cin.ignore(numeric_limits<streamsize>::max(), '\n'); 
    cout << " "Please enter within the range of 0 to 1."

} while((!(cin >> input)) || cin.bad() || input <= 0 || input >= 1);

CIN在一定范围内

4 个答案: