Reference initialization in C++

Question

Greetings, everyone!

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

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


        

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.