为什么std :: exception析构函数不是noexcept [英] Why is the std::exception destructor not noexcept

问题描述

C ++ 11的析构函数 < a href = http://en.cppreference.com/w/cpp/error/exception rel = nofollow noreferrer> std :: exception 基类不是 noexcept ，因此（理论上）可能会引发异常，从而对其所有派生类（包括 std :: bad_alloc 和 std :: runtime_error ）。 C ++ 98 std :: exception 的析构函数具有 throw（）异常规范，表明它是不允许抛出异常。为什么会有所不同？为什么现在允许引发异常？鉴于 std :: exception 构造器现在是 noexcept ：您可以安全地构造这样的对象，但是不能安全地销毁它：与正常情况相反行为。

The destructor of the the C++11 std::exception base class is not noexcept, and thus may (in theory) throw an exception, which consequent relaxed permission for all its derived classes (including std::bad_alloc and std::runtime_error). The destructor of the C++98 std::exception however had a throw() exception specification, indicating it was not permitted to throw exceptions. Why the difference? Why is it now permitted to throw exceptions? The permission is particularly odd given that the std::exception constructors are now noexcept: you can safely construct such an object, but you can not safely destroy it: the opposite of the normal behaviour.

让异常类的析构函数抛出异常通常是灾难性的。是什么会导致 std :: exception ::〜exception 引发异常？

Having the destructor of an exception class throw an exception is usually disastrous. What can cause std::exception::~exception to throw an exception?

推荐答案

实际上是 noexcept（true）。从C ++ 11开始，对于析构函数，没有明确指定异常的规定被认为是 noexcept（true）。

It is noexcept(true) indeed. Since C++11, for destructors with no exception specification explicitly provided the exception specification is considered to be noexcept(true).

除非没有异常说明如果明确提供，则将异常规范视为隐式声明的析构函数将使用的规范（请参见下文）。在大多数情况下，这是 noexcept（true）。因此，必须将抛​​出的析构函数显式声明为 noexcept（false）。 （自C ++ 11起）

Except that if no exception specification is explicitly provided, the exception specification is considered to be one that would be used by the implicitly-declared destructor (see below). In most cases, this is noexcept(true). Thus a throwing destructor must be explicitly declared noexcept(false). (since C++11)

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