生成std :: isalpha的值为true的范围 [英] generate range for which std::isalpha evaluates to true

问题描述

这可能是一些小问题，但我找不到合适的答案。我想要的是生成一个范围（让我们说一个 std :: string ）包含所有可能的 char s std :: isalpha 计算为 true 。

This may be a trivial question for some, but I cannot find an appropriate answer. What I'd like is to generate a range (let's say a std::string) that contains all possible chars for which std::isalpha evaluates to true.

例如，对于默认语言环境，字符串应为A ... Za ... z。

For example, for the default locale, the string should be "A...Za...z". However, if the locale is french, for example, then accented letters should also belong to the string.

PS：我有一个来自DieterLücking的解决方案，http://stackoverflow.com/a/25125871/3093378
它似乎适用于除了我的所有平台（ OS X 10.9.4 g ++ 4.9 ， clang ++ （clang-503.0.40）），其中尝试访问<$ p

<$ p $ <$ p $ <$ <$ p $

PS: I got a solution from Dieter Lücking, http://stackoverflow.com/a/25125871/3093378 It seems to work on all platforms except mine (OS X 10.9.4 g++4.9, clang++ LLVM version 5.1 (clang-503.0.40) ), where it just segfaults when trying to access table[i] in the line

if(table[i] & ctype::alpha)

我想知道是否有其他人可以在Mac或任何其他平台上重现此错误。

I wonder if anyone else can reproduce the error on a Mac or any other platform.

推荐答案

生成一组字符，按字母顺序，您可以直接使用ctype-table：

Instead of classifying characters by generating a set of characters, being alphabetical, you might utilize the ctype-table directly:

#include <iostream>
#include <locale>

int main() {
    typedef std::ctype<char> ctype;
    std::locale locale;
    const ctype& facet = std::use_facet<ctype>(locale);
    const ctype::mask* table = facet.table();

    // You might skip this and work with the table, only.
    std::string result;
    for(unsigned i = 0; i < facet.table_size; ++i) {
        if(table[i] & ctype::alpha)
            result += char(i);
    }
    std::cout << result << '\n';
    return 0;
}

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