输入验证以确保只有数字c ++ [英] Input Validation to make sure only number c++

问题描述

好吧，我想尽力使用指针，所以我试图写一个输入验证的用户输入，以确保任何不是一个数字被正确处理。当我使用isdigit（）不是为我工作。我仍然得到一个例外，当我输入一个字母表。有什么建议么？谢谢。请查看：

 ＃include< iostream> 
＃include< algorithm> 
＃include< string> 
＃include< cctype> 
 using namespace std; 
 
 void EnterNumbers（int *，int）; 
 
 
 int main（）
 {
 int input = 0; 
 int * myArray; 
 
 cout<< 请输入考试成绩的数字\\\
\\\
; 
 cin>>输入; 
 
 //分配数组
 myArray = new int [input]; 
 
 EnterNumbers（myArray，input）; 
 
 
 delete [] myArray; 
 return 0; 
} 
 
 void EnterNumbers（int * arr，int input）
 {
 
 for（int count = 0; count< input; count ++） 
 {
 cout<< \\\
\\\
输入成绩编号<< count + 1< \t; 
 cin>> arr [count] 
 
 if（！isdigit（arr [count]））
 {
 cout< 不是数字; 
} 
} 
} 
  

解决方案>

如果你测试 if（！（cin>> arr [count]））... instead - isdigit（arr [数字]）测试 arr [digit] 的值是否为数字的ASCII码[或可能匹配日语，中文或阿拉伯语是，作为一个阿拉伯文字体，不是一个0-9像我们的阿拉伯语一个数字]。所以，如果你输入48到57，它会说它的确定，但如果你键入6或345，它抱怨，它不是一个数字...

一次你发现了一个非数字，你还需要退出或清除输入缓冲区从垃圾。 cin.ignore（1000，'\\\
'）; 将读取到下一个换行符或1000个字符，以先到为准。如果有人输入了一百万数字，可能会烦人，但否则，应该解决问题。

您当然还需要一个循环来再次读取数字，直到输入有效的数字。

Ok, I'm trying to get good at using pointers so I'm trying to write a input validation for the user input to make sure that anything that isn't a number is handled correctly. When I use isdigit() isn't working for me. I still get an exception when I enter a alphabet. Any suggestions? Thanks. Check this out:

#include<iostream>
#include<algorithm>
#include<string>
#include<cctype>
using namespace std;

void EnterNumbers(int * , int);


int main()
{
    int input = 0;
    int *myArray;

    cout << "Please enter the number of test scores\n\n";
    cin >> input;

    //Allocate Array
    myArray = new int[input];

    EnterNumbers(myArray,input);


    delete[] myArray;
    return 0;
}

void EnterNumbers(int *arr, int input)
{

    for(int count = 0; count < input; count++)
    {
        cout << "\n\n Enter Grade Number  " << count + 1 << "\t";
        cin >> arr[count];

        if(!isdigit(arr[count]))
        {
            cout << "Not a number";
        }
    }
}

解决方案

If you test if (!(cin >> arr[count])) ... instead - isdigit(arr[digit]) tests if the value of arr[digit] is the ASCII code of a digit [or possibly matches Japanese, Chinese or Arabic (that is, as an Arabic script typeface, not that it's a 0-9 like our "Arabic" ones) digit]. So if you type in 48 to 57, it will say it's OK, but if you type 6 or 345, it's complaining that it is not a digit...

Once you have discovered a non-digit, you will also need to either exit or clean out the input buffer from "garbage". cin.ignore(1000, '\n'); will read up to the next newline or a 1000 characters, whichever happens first. Could get annoying if someone has typed in a million digits, but otherwise, should solve the problem.

You will of course also need a loop to read the number again, until a valid number is entered.

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