问题描述
当创建 MyClass 的新实例作为函数的参数时：
When creating a new instance of a MyClass as an argument to a function like so:
class MyClass
{
MyClass(int a);
};
myFunction(MyClass(42));
标准是否让任何受赠方了解析构函数的时间？
，我可以假设它将在调用 myFunction（）之前的下一个语句之前被调用？
does the standard make any grantees on the timing of the destructor?
Specifically, can I assume that the it is going to be called before the next statement after the call to myFunction() ?
推荐答案
临时对象在它们所属的完整表达式的末尾被销毁。
Temporary objects are destroyed at the end of the full expression they're part of.
完整表达式是不是某个其他表达式的子表达式的表达式。通常这意味着它以; （或）结束 if ， while ， switch 等）。在你的例子中，它是函数调用的结束。
A full expression is an expression that isn't a sub-expression of some other expression. Usually this means it ends at the ; (or ) for if, while, switch etc.) denoting the end of the statement. In your example, it's the end of the function call.
请注意，您可以通过将临时表的生命周期绑定到 const 引用来延长它们的生命周期。这样做会将其生命周期延长到引用的生命周期：
Note that you can extend the lifetime of temporaries by binding them to a const reference. Doing so extends their lifetime to the reference's lifetime:
MyClass getMyClass();
{
const MyClass& r = getMyClass(); // full expression ends here
...
} // object returned by getMyClass() is destroyed here
如果你不打算改变返回的对象，这是一个不错的技巧来保存一个复制构造函数调用（与 MyClass obj = getMyClass（） code>），以防未应用返回值优化。不幸的是，它不是很知名。 （我想，C ++ 11的移动语义将会使它更少的有用，虽然。）
If you don't plan to change the returned object, then this is a nice trick to save a copy constructor call (compared to MyClass obj = getMyClass();), in case return value optimization was not being applied. Unfortunately it isn't very well known. (I suppose C++11's move semantics will render it less useful, though.)
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