使用 BOOST_STRONG_TYPEDEF 区分 arg 类型但导致 seg 错误

Question

我有一个方法需要多个相同枚举类型的变量。为了让编译器检测我是否传递了错误的参数，我使用了BOOST_STRONG_TYPEDEF。但是，当我创建一个实例并在 IF 语句中进行比较时，我遇到了一个段错误。

Boost 版本是 1.74

enum class Testable
{
    UNDEFINED,
    A,
    B
};

BOOST_STRONG_TYPEDEF(Testable, SomeType)

int main()
{  
    SomeType abc{Testable::UNDEFINED};
    std::cout << "START" << std::endl;
    
    if(abc == Testable::UNDEFINED)  // Seg faults here
    {
        volatile int j = 0;
    }
    
    std::cout << "FINISH" << std::endl;
}

GDB 回溯表明这是堆栈溢出/递归调用：

#1    0x00007ffff74c5d9d in boost::operators_impl::operator== (y=@0x7fffffcc9e44:
#2    0x00007ffff74c5d9d in boost::operators_impl::operator== (y=@0x7fffffcc9e44:
#3    0x00007ffff74c5d9d in boost::operators_impl::operator== (y=@0x7fffffcc9e44:
#4    0x00007ffff74c5d9d in boost::operators_impl::operator== (y=@0x7fffffcc9e44:
#5    0x00007ffff74c5d9d in boost::operators_impl::operator== (y=@0x7fffffcc9e44:
#6    0x00007ffff74c5d9d in boost::operators_impl::operator== (y=@0x7fffffcc9e44:

BOOST_STRONG_TYPEDEF 的文档不多。是不是我用错了？

Boost 版本是 1.74。我正在使用 Clang。

Answer 1

消毒剂说

==3044==ERROR: AddressSanitizer: stack-overflow on address 0x7ffcc58b3ff8 (pc 0x56310c340e84 bp 0x7ffcc58b4000 sp 0x7ffcc58b3ff
0 T0)
    #0 0x56310c340e84 in boost::operators_impl::operator==(Testable const&, SomeType const&) /home/sehe/custom/boost_1_75_0/boo

问题在于 Boost 的 STRONG_TYPEDEF 使派生类型与基类型完全排序：

struct SomeType
    : boost::totally_ordered1<SomeType, boost::totally_ordered2<SomeType, Testable>>
{
    Testable t;
    explicit SomeType(const Testable& t_) noexcept((boost::has_nothrow_copy_constructor<Testable>::value)) : t(t_) {}
    SomeType() noexcept( (boost::has_nothrow_default_constructor<Testable>::value)) : t() {}
    SomeType(const SomeType& t_) noexcept( (boost::has_nothrow_copy_constructor<Testable>::value)) : t(t_.t) {}
    SomeType& operator=(const SomeType& rhs) noexcept((boost::has_nothrow_assign<Testable>::value)) {
        t = rhs.t;
        return *this;
    }
    SomeType& operator=(const Testable& rhs) noexcept((boost::has_nothrow_assign<Testable>::value)) {
        t = rhs;
        return *this;
    }
    operator const Testable&() const { return t; }
    operator Testable&() { return t; }
    bool operator==(const SomeType& rhs) const { return t == rhs.t; }
    bool operator<(const SomeType& rhs) const { return t < rhs.t; }
};

如果您删除隐式转换的来源：

struct SomeType
    : boost::totally_ordered1<SomeType
      /*, boost::totally_ordered2<SomeType, Testable>*/>
{
     // ...

它 JustWorks(TM)。我认为您也应该使用转换运算符explicit 并始终进行强制转换：

Live On Coliru

#include <boost/serialization/strong_typedef.hpp>
#include <iostream>

enum class Testable { UNDEFINED, A, B };
       
struct SomeType
    : boost::totally_ordered1<SomeType
      /*, boost::totally_ordered2<SomeType, Testable>*/>
{
    Testable t;
    explicit SomeType(const Testable& t_) noexcept((boost::has_nothrow_copy_constructor<Testable>::value)) : t(t_) {}
    SomeType() noexcept( (boost::has_nothrow_default_constructor<Testable>::value)) : t() {}
    SomeType(const SomeType& t_) noexcept( (boost::has_nothrow_copy_constructor<Testable>::value)) : t(t_.t) {}
    SomeType& operator=(const SomeType& rhs) noexcept((boost::has_nothrow_assign<Testable>::value)) {
        t = rhs.t;
        return *this;
    }
    SomeType& operator=(const Testable& rhs) noexcept((boost::has_nothrow_assign<Testable>::value)) {
        t = rhs;
        return *this;
    }
    explicit operator const Testable&() const { return t; }
    explicit operator Testable&() { return t; }
    bool operator==(const SomeType& rhs) const { return t == rhs.t; }
    bool operator<(const SomeType& rhs) const { return t < rhs.t; }
};

int main() {
    SomeType abc{ Testable::UNDEFINED };
    std::cout << "START" << std::endl;

    if (abc == SomeType{Testable::UNDEFINED}) {
        volatile int j = 0;
    }

    std::cout << "FINISH" << std::endl;
}