c – 只接受信件

概述这应该只接受字母,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)) { 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：

567Error!Error!678fghError!567fgh678Error!fhg687its characters!Press any key to continue . . .

由于字符串中的数字,fhg687应该输出错误.

接受的输出应仅包含字母,例如ghggjh.

解决方法 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字符,并检查该字符是否在English-letter ranges in ascii(较低的下限)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;        }     } }

总结

以上是内存溢出为你收集整理的c – 只接受信件全部内容，希望文章能够帮你解决c – 只接受信件所遇到的程序开发问题。

如果觉得内存溢出网站内容还不错，欢迎将内存溢出网站推荐给程序员好友。