Question

Greetings, everyone!

Examining my own code, I came up to this interesting line:

const CString &refStr = ( CheckCondition() ) ? _T("foo") : _T("bar");

Now I am completely at a loss, and cannot understand why it is legal. As far as I understand, const reference must be initialized, either with r-value or l-value. Uninitialized references cannot exist. But ()? operator executes a CheckCondition() function before it assigns value to the reference. I can see now, that while CheckCondition() is executed, refStr exists, but still not initialized. What will happen if CheckCondition() will throw an exception, or pass control with a goto statement? Will it leave the reference uninitialized or am I missing something?

Answer 1

Simpler example: const int x = foo();

This constant too has to be initialized, and for that foo() needs to be called. That happens in the order necessary: x comes into existance only when foo returns.

To answer your additional questions: If foo() would throw, the exception will be caught by a catch() somewhere. The try{} block for that catch() surrounded const int x = foo(); obviously. Hence const int x is out of scope already, and it is irrelevant that it never got a value. And if there’s no catch for the exception, your program (including const int x) is gone.

C++ doesn’t have random goto‘s. They can jump within foo() but that doesn’t matter; foo() still has to return.