如何验证数字输入C ++

I'd like to know how to limit an input value to signed decimals using std::cin.

我想知道如何使用std :: cin将输入值限制为带符号的小数。

7 个解决方案

#1

If the backing variable of the cin is a number, and the string provided is not a number, the return value is false, so you need a loop:

如果cin的后备变量是一个数字,并且提供的字符串不是数字,则返回值为false,因此您需要一个循环:

int someVal;

while(!(cin >> someVal)) {
   cin.reset();
   cout << "Invalid value, try again.";
}

#2

double i;

//Reading the value
cin >> i;

//Numeric input validation
if(!cin.eof())
{
    peeked = cin.peek();
    if(peeked == 10 && cin.good())
    {
             //Good!
             count << "i is a decimal";
        }
        else
        {
             count << "i is not a decimal";
         cin.clear();
         cin >> discard;
        }
}

This also gives an error message with the input -1a2.0 avoiding the assignation of just -1 to i.

这也给出了输入-1a2.0的错误消息,避免了将-1分配给i。

#3

cin's >> operator works by reading one character at a time until it hits whitespace. That will slurp the whole string -1a2.0, which is obviously not a number so the operation fails. It looks like you actually have three fields there, -1, a, and 2.0. If you separate the data by whitespace, cin will be able to read each one without problem. Just remember to read a char for the second field.

cin的>>运算符通过一次读取一个字符直到它碰到空白来工作。这将淹没整个字符串-1a2.0,这显然不是一个数字,因此操作失败。看起来你实际上有三个字段,-1,a和2.0。如果按空格分隔数据,cin将能够毫无问题地读取每个数据。只记得读第二个字段的字符。

#4

Combining the techniques from the top answer here and this website, I get

结合这里的最佳答案和本网站的技术,我得到了

input.h

#include <ios>  // Provides ios_base::failure
#include <iostream>  // Provides cin

template <typename T>
T getValidatedInput()
{
    // Get input of type T
    T result;
    cin >> result;

    // Check if the failbit has been set, meaning the beginning of the input
    // was not type T. Also make sure the result is the only thing in the input
    // stream, otherwise things like 2b would be a valid int.
    if (cin.fail() || cin.get() != '\n')
    {
        // Set the error state flag back to goodbit. If you need to get the input
        // again (e.g. this is in a while loop), this is essential. Otherwise, the
        // failbit will stay set.
        cin.clear();

        // Clear the input stream using and empty while loop.
        while (cin.get() != '\n')
            ;

        // Throw an exception. Allows the caller to handle it any way you see fit
        // (exit, ask for input again, etc.)
        throw ios_base::failure("Invalid input.");
    }

    return result;
}

Usage

inputtest.cpp

#include <cstdlib>  // Provides EXIT_SUCCESS
#include <iostream>  // Provides cout, cerr, endl

#include "input.h"  // Provides getValidatedInput<T>()

int main()
{
    using namespace std;

    int input;

    while (true)
    {
        cout << "Enter an integer: ";

        try
        {
            input = getValidatedInput<int>();
        }
        catch (exception e)
        {
            cerr << e.what() << endl;
            continue;
        }

        break;
    }

    cout << "You entered: " << input << endl;

    return EXIT_SUCCESS;
}

Sample run

Enter an integer: a
Invalid input.
Enter an integer: 2b
Invalid input.
Enter an integer: 3
You entered: 3.

输入整数:a无效输入。输入整数:2b输入无效。输入一个整数:3您输入了:3。

#5

I'm not trying to be rude. I just wanted to share a solution I provided which I believe is more robust and allows for better input validation.

我不是想要粗鲁。我只想分享我提供的解决方案,我相信它更强大,并允许更好的输入验证。

Please refer to: My Solution to Input Validation

请参阅:我的输入验证解决方案

#6

I tried many techniques for reading integer input from the user using the >> operator, but in a way or another all my experiments have failed.

我尝试了许多使用>>运算符从用户读取整数输入的技术,但是在某种程度上我的所有实验都失败了。

Now I think that getline() function (not the method with the same name on std::istream) and the strtol() function from the include cstdlib is the only predictable consistent solution for this problem. I would appreciate if someone proved me wrong. Here is something like the one I use:

现在我认为getline()函数(不是std :: istream上具有相同名称的方法)和include cstdlib中的strtol()函数是此问题唯一可预测的一致解决方案。如果有人证明我错了,我将不胜感激。这是我使用的那个:

#include <iostream>
#include <cstdlib>

// @arg prompt The question to ask. Will be used again on failure.
int GetInt(const char* prompt = "? ")
{
    using namespace std; // *1
    while(true)
    {
        cout << prompt;
        string s;
        getline(cin,s);
        char *endp = 0;
        int ret = strtol(s.c_str(),&endp,10);
        if(endp!=s.c_str() && !*endp)
            return ret;
    }
}

*1: Placing using namespace whatever; to the global scope may lead to broken "unity builds" (google!) on larger projects, so should be avoided. Practice to not use that way, even on smaller projects!

* 1:使用命名空间放置任何东西;在全球范围内可能导致在大型项目上破坏“统一构建”(google!),因此应该避免。练习不使用这种方式,即使是在较小的项目上!

Reading integers from files is a very different matter. Raúl Roa's approach can be good for that if properly worked out. I also suggest that wrong input files should not be tolerated, but it really depends on the application.

从文件中读取整数是一个非常不同的问题。 RaúlRoa的方法如果得到适当的解决方案可能会有好处。我还建议不应该容忍错误的输入文件,但它实际上取决于应用程序。

Be warned that using >> and getline() in the same program on cin will lead to some problems. Use one of them only, or google to know how to handle the issue (not too hard).

请注意,在cin上的相同程序中使用>>和getline()会导致一些问题。只使用其中一个,或谷歌知道如何处理问题(不是太难)。

#7

Something like:

double a;
cin >> a;

Should read your signed "decimal" fine.

应该阅读您签署的“十进制”罚款。

You'll need a loop and some code to make sure it handles invalid input in a sensible way.

你需要一个循环和一些代码来确保它以合理的方式处理无效输入。

Good luck!

#1