即使输入是另一种数据类型，C++ cin.fail() 也会执行并移至下一行

2022-01-19 00:00:00 validation switch-statement c++ cin

我正在使用 get.fail() 来检查输入中是否有任何字符，如果有，我想给用户一个重新输入的机会.但是，无论哪种情况，只要输入前面有一个整数，该程序似乎仍然接受用户输入.说 w1 和 1w，程序会告诉用户它只接受整数，而后者接受输入并移动到下一行，这会导致另一个问题.

I am using get.fail() to check if there's any char in the input, and if there is I would like to give the user a chance to re-enter. However, the program seems to still accept the user input whenever there's an integer in front of the input no matter the case. Say w1 and 1w, the program will tell the user that the it only accepts integers while the latter one accepts the input and moves over to the next line which then causes another problem.

    void userChoice(int input){

        switch(input)
        {
            case 1:
                insert();
                break;
            case 2:
                display();  
                break;
            case 3:
                update_data();
                break;
            case 4:
                delete_position();
                break;
            case 5:
                cout<<"Thank you for using the program
";
                exit(0);
                break;
            case 6:
                tellSize();
                break;
            default:
                cout<<"Not an option
";
                cin>>input;

                while(cin.fail())
                {
                    cin.clear();
                    cin.ignore(INT_MAX, '
'); 
                    cin>>input;
                    break; 
                }
                userChoice(input);
        }

    }

参考上面的代码，假设我给出了输入 1w.该程序仍将执行案例 1，就好像没有任何问题一样，然后 w 以某种方式传递给 insert() 函数，这不是我想要的.我希望程序让用户重新输入输入，无论是 1w 还是 w1，总之我不希望程序移到下一个如果整数输入中有一个字符，则为行.

Referring to the code above, say I give the input 1w. The program will still execute case 1 as if there's nothing wrong, and then w is somehow passed into the insert() function which is not what I want. I would like the program to let the user re-enter the input no matter if it's 1w or w1, in short I do not want the program to move over to the next line if there's a char in an integer input.

tl;博士:为什么当cin为1w时下面的代码仍然执行，不应该打印仅输入数字"，因为那里有一个字符?

tl;dr: Why does the code below still execute when the cin is 1w, shouldn't it print "Enter number only" since there's a character in there?

这是我制作的一个快速程序，为了重现我面临的错误，我首先输入 1h 这是我面临的第一个错误，为什么程序仍然在执行时有输入中的字符 h?然后在第二个输入中，我输入 2w 并且程序打印出 2，程序不应该循环 while 循环，因为有一个 char w 在输入中?

Here's a quick program I made, to reproduce the error I am facing, I first enter 1h and here's the first bug I'm facing, why is the program still executing when there's a char h in the input? Afterwards in the second input, I enter 2w and the program prints out 2, shouldn't the program loop the while loop since there's a char w in the input?

#include<iostream>
#include<iomanip>
#include<limits>
using namespace std;

void option_1()
{
    int amount_input;

    cout<<"Enter the amount of cake"<<endl;
    cin>>amount_input;
    while(cin.fail())
    {
        cin.clear();
        cin.ignore(numeric_limits<streamsize>::max(), '
'); 
        cout<<"Enter number only
";
        cin>>amount_input;
    }
        cout<<amount_input;
}

void options(int input)
{
    bool fail;

    switch(input)
    {
        case 1:
            option_1();
            break;


        default:
            cout<<"Not an option"<<endl;
            cin>>input;
            while(cin.fail())
            {
                cin.clear();
                cin.ignore(numeric_limits<streamsize>::max(), '
'); 
                cin>>input;
                break; 
            }
            options(input);

    }
}

void menu(){
    int user_input;

    cout<<"Enter 1 to print something
";
    cin>>user_input;
    options(user_input);
}

int main(){
    menu();
}

推荐答案

从您的评论来看，您似乎只希望整数输入，并且不希望允许在整数之后输入其他字符，例如 1w，即使 1 将被转换为 int 而 w 未被读取以被 删除.在调用 .clear() 之后忽略(...).(如上所述，您现在对 .clear() 和 .ignore(...) 的使用是正确的.

From your comments, it appears you want integer input exclusively and do not want to allow input of additional character after the integer like 1w, even though 1 would be converted to an int while leaving w unread to be removed by .ignore(...) after your call to .clear(). (as mentioned above, your use of .clear() and .ignore(...) are now correct.

如果这是您的意图，您需要一种方法来检查用户输入的整数之后是否还有其他内容.如果实际上什么都不存在，要以非阻塞方式执行此操作，您有几个选择.(例如，您可以在下一个字符处 .peek() - 但您只能得到一个，或者您可以使用 line-oriented 输入方法) line-面向 方法允许您将用户输入的整行读入 string，然后提取整数值并检查输入行中是否包含其他任何内容.

If this is your intent, you need a way to check if there is anything else following the integer input by the user. To do that in a non-blocking way if nothing actually exists, you have a couple of options. (e.g. you can .peek() at the next character -- but you only get one, or you can use a line-oriented input approach) The line-oriented approach allows you to read the entire line of user input into a string and then extract the integer value and check whether there is anything else contained in the line of input.

最直接的方法是创建一个 std::basic_stringstream从使用 getline() 读取的数据行.这种方法消耗了整行用户输入，并为您提供了从该行中提取您可能想要的任何信息所需的所有工具.它还以一种不会影响任何后续用户输入的方式执行此操作.

The most straight forward way is to create a std::basic_stringstream from the line of data read with getline(). This approach, consumes the entire line of user input and provides you with all the tools you need to extract whatever information you may want from the line. It also does this in a way that does not effect any of your subsequent user inputs.

虽然我建议您将 void menu() 和 void options(int input) 函数结合起来，这样您只需一个函数来处理菜单的输入处理 -- 除了可能重复几行代码之外，将其分成两部分并没有什么问题.以下只是关于如何处理您的 menu() 函数以仅允许整数输入的建议.您可以根据代码的其余部分对其进行调整.

While I would recommend you combine your void menu() and void options(int input) functions so you simply have one function to handle input processing for your menu -- there is nothing wrong with breaking it in two other than the possibility of a few lines of code being duplicated. The following is just a suggestion on how to handle your menu() function to only allow integer input. You can adapt it to the remainder of your code.

您将需要一些额外的包括:

You will need a couple of additional includes:

#include <sstream>
#include <string>

我还会#define第一个和最后一个可接受的菜单条目，这样您的代码中就有这些常量可用，并且可以在添加到菜单时轻松更改，例如

I would also #define the first and last acceptable menu entries so you have those constants available in your code in a place that can be easily changed as you add to your menu, e.g.

#define MENUFIRST 1     /* first valid entry */
#define MENULAST  1     /* last valid entry */

(注意:只允许输入 1 作为有效的菜单项)

(note: that will allow only 1 be entered as a valid menu entry)

要使用上述方法限制您的 menu() 功能，您可以这样做:

To limit you menu() function using the approach outlined above, you could do:

void menu(){

    int user_input = 0;
    string line, unwanted;

    for (;;) {  /* loop continually until valid input received */
        cout << "
Enter 1 to print something: ";
        if (!getline (cin, line)) { /* read an entire line at a time */
            cerr << "(user canceled or unrecoverable stream error)
";
            return;
        }
        stringstream ss (line);         /* create a stringstream from line */
        if (!(ss >> user_input)) {      /* test if valid integer read */
            /* test eof() or bad() */
            if (ss.eof() || ss.bad())   /* if not, did user cancel or err */
                cerr << "(empty-input or unreconverable error)
";
            else if (ss.fail())         /* if failbit - wasn't an int */
                cerr << "error: invalid integer input.
";
        }
        else if (ss >> unwanted) {      /* are there unwanted chars? */
            cerr << "error: additional characters following user_input.
";
            user_input = 0;             /* reset user_input zero */
        }       /* was int outside MENUFIRST-to-MENULAST? */
        else if (user_input < MENUFIRST || MENULAST < user_input)
            cerr << "error: integer not a valid menu selection.
";
        else    /* valid input!, break read loop */
            break;
    }

    options(user_input);
}

鉴于上述讨论，评论应该是不言自明的，但如果您有任何问题，请告诉我.将该函数与您的其余代码一起使用(并注释未使用的 //bool fail;)，您可以测试它是否满足您的要求，例如

The comments should be self-explanatory given the discussion above, but let me know if you have questions. Using the function with the rest of your code (and commenting the unused // bool fail;), you can test whether it meets your requirements, e.g.

使用/输出示例

$ ./bin/menu_int

Enter 1 to print something: w1
error: invalid integer input.

Enter 1 to print something: $#%&^#&$ (cat steps on keyboard) !#$%%^%*()
error: invalid integer input.

Enter 1 to print something: 1w
error: additional characters following user_input.

Enter 1 to print something: -1
error: integer not a valid menu selection.

Enter 1 to print something: 24
error: integer not a valid menu selection.

Enter 1 to print something: 1
Enter the amount of cake
3
3

另外请注意，您的 menu() 功能现在将正确捕获由用户按 Ctrl+d(或 Ctrl+z on windows)取消输入并优雅退出.

Also note, your menu() funciton will now properly trap a manual EOF generated by the user pressing Ctrl+d (or Ctrl+z on windows) to cancel input and exit gracefully.

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