本文介绍了只接受字母的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
这应该只接受字母,但它不正确:
This should accept only letters, but it is not yet correct:
#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: / p>
Here is the output:
567
Error!
Error!
678fgh
Error!
567fgh678
Error!
fhg687
its characters!
Press any key to continue . . .
fhg687
接受的输出只能包含字母,例如 ghggjh
。
Accepted output should contain letters only, such as ghggjh
.
推荐答案
更新:显示更完整的解决方案。
Updated: to show a fuller solution.
方法可能是遍历输入中的每个字符,并检查该字符是否位于 ascii中的英文字母范围(upper + lower):
The simplest approach would probably be to iterate through each char in the input and check whether that char is within English-letter ranges in ascii (upper + lower):
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;
}
}
}
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