c++ - 为什么 std::scan_is 在 vi​​sual studio 编译器中发出运行时错误？

Question

示例 here 在 Visual Studio 2013 中发出内存访问冲突的运行时错误。

#include <locale>
#include <iostream>
#include <iterator>

int main()
{
    auto& f = std::use_facet<std::ctype<char>>(std::locale(""));

    // skip until the first letter
    char s1[] = "      \t\t\n  Test";
    const char* p1 = f.scan_is(std::ctype_base::alpha, std::begin(s1), std::end(s1));
    std::cout << "'" << p1 << "'\n";

    // skip until the first letter
    char s2[] = "123456789abcd";
    const char* p2 = f.scan_is(std::ctype_base::alpha, std::begin(s2), std::end(s2));
    std::cout << "'" << p2 << "'\n";
}

这是为什么呢？编译器的错误实现？

Answer 1

auto& f = std::use_facet<std::ctype<char>>(std::locale("")); 行导致错误。引用 f 是空对象的别名。似乎此实现适用于 gcc C+11 及更高版本的编译器，但不适用于 Microsoft 编译器。因此，我测试过的 Visual Studio 2013 和 2015 中的正确实现是:

#include "stdafx.h"
#include <locale>
#include <iostream>
#include <iterator>

int main()
{
    std::locale loc(std::locale(""));
    //auto& f = std::use_facet<std::ctype<char>>(std::locale(""));
    auto& f = std::use_facet<std::ctype<char>>(loc);
    // skip until the first letter
    char s1[] = "      \t\t\n  Test";
    const char* p1 = f.scan_is(std::ctype_base::alpha, std::begin(s1), std::end(s1));
    std::cout << "'" << p1 << "'\n";

    // skip until the first letter
    char s2[] = "123456789abcd";
    const char* p2 = f.scan_is(std::ctype_base::alpha, std::begin(s2), std::end(s2));
    std::cout << "'" << p2 << "'\n";
}