C++: Life span of temporary arguments?

Question

When creating a new instance of a MyClass as an argument to a function like so:

class MyClass {   MyClass(int a); };      myFunction(MyClass(42)); 

Does the standard make any guarantees on the timing of the destructor?

Specifically, can I assume that it is going to be called before the next statement after the call to myFunction() ?

Answer 1

Temporary objects are destroyed at the end of the full expression they're part of.

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.

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 

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.)

Answer 2

Everyone has rightly cited 12.2/3 or similar, which answers your question:

Temporary objects are destroyed as the last step in evaluating the full-expression that (lexically) contains the point where they were created.

I find it amusing that over the next page in my printing of the standard, 12.2/4 says:

There are two contexts in which temporaries are destroyed at a different point than the end of the full-expression.

Neither of them applies to your example, they both relate to the use of temporaries in initializers. But it does go to show that you have to keep your wits about you when dealing with a tricky beast like the C++ standard.