只接受信件

Question

这应该只接受字母,但它还不正确:

#include <iostream>
#include <string>
#include <sstream>
using namespace std;

int main()
{
    std::string line;
    double d;

    while (std::getline(std::cin, line))
    {
        std::stringstream ss(line);
        if (ss >> d == false && line != "") //false because can convert to double
        {
            std::cout << "its characters!" << std::endl;
            break;
        }
        std::cout << "Error!" << std::endl;
    }
    return 0; 
}

这是输出:

567
Error!

Error!
678fgh
Error!
567fgh678
Error!
fhg687
its characters!
Press any key to continue . . .

fhg687 应该输出错误,因为字符串中的数字.

接受的输出应仅包含字母,例如ghggjh.

Answer 1

使用std::all_of适当的谓词,你可以更好地使用字符串.在你的情况下,该谓词将是std::isalpha.(标题<algorithm>和<cctype>要求)

if (std::all_of(begin(line), end(line), std::isalpha))
{
    std::cout << "its characters!" << std::endl;
    break;
}
std::cout << "Error!" << std::endl;

Answer 2

更新:显示更全面的解决方案.

最简单的方法可能是迭代输入中的每个char并检查该char是否在ascii(upper + lower)的英文字母范围内:

char c;

while (std::getline(std::cin, line))
{
    // Iterate through the string one letter at a time.
    for (int i = 0; i < line.length(); i++) {

        c = line.at(i);         // Get a char from string

        // if it's NOT within these bounds, then it's not a character
        if (! ( ( c >= 'a' && c <= 'z' ) || ( c >= 'A' && c <= 'Z' ) ) ) {

             std::cout << "Error!" << std::endl;

             // you can probably just return here as soon as you
             // find a non-letter char, but it's up to you to
             // decide how you want to handle it exactly
             return 1;
        }
     }
 }