What does while(!volatile); do?

Question

I saw this in a part that never gets called in a coworker's code:

volatile unsigned char vol_flag = 0;
// ...
while(!vol_flag);

vol_flag is declared in the header file, but is never changed. Am I correct that this will lead to the program hanging in an infinite loop? Is there a way out of it?

Answer 1

Usually a code like this indicates that vol_flag is expected to be changed externally at some point. Here, externally may mean a different thread, an interrupt handler, a piece of hardware (in case of memory mapped IO) etc. This loop effectively waits for the external event which changes the flag.

The volatile keyword is a way for the programmer to express the fact that it is not safe to assume what is apparent from the code: namely that the flag is not changed in the loop. Thus, it prevents the compiler from making optimizations which could compromise the intentions behind the code. Instead, the compiler is forced to make a memory reference to fetch the value of the flag.

Note that (unlike in Java) volatile in C/C++ does not establish happens-before relationship and does not guarantee any ordering or visibility of memory references across volatile access. Moreover, it does not ensure atomicity of variable references. Thus, it is not a tool for communication between threads. See this for details.