抛出一个通过引用传递的临时参数

Question

inline void my_assert( bool cond, const std::exception &e = my_assert_failed() ) 
{ 
    if ( !cond ) 
      throw e;
}

标准确保：

A temporary bound to a reference parameter in a function call (5.2.2) persists until the completion of the full expression containing the call.

对于抛出的临时对象：

The temporary persists as long as there is a handler being executed for that exception.

我可以推断传递给 my_assert 的临时文件会一直存在到 catch 块完成吗？

Answer 1

来自 N4296（最终 C++14 之后的初稿）[15.1p3]：

Throwing an exception copy-initializes (8.5, 12.8) a temporary object, called the exception object. The temporary is an lvalue and is used to initialize the variable declared in the matching handler (15.3).

所以你不能假设你的临时"survives the throw"。如果抛出，类型为 std::exception 的异常对象的复制构造函数将以 e 作为参数被调用。当控制离开包含对 my_assert 的调用的完整表达式时， e 绑定到的临时对象将被销毁（在正常 return 之后或作为堆栈展开的一部分，因为你有条件地抛出异常）。

根据[12.8p31.2]：

在某些情况下可以省略异常对象的复制构造，但这不是其中之一

— in a throw-expression (5.17), when the operand is the name of a non-volatile automatic object (other than a function or catch-clause parameter) whose scope does not extend beyond the end of the innermost enclosing try-block (if there is one), the copy/move operation from the operand to the exception object (15.1) can be omitted by constructing the automatic object directly into the exception object

（强调我的）

抛出一个通过引用传递的临时参数

Throw a temporary argument passed by reference

c++

exception-handling