C Linux Pro

Interrupt service routines often set variables that are tested in main line code. For example, a serial port interrupt may test each received character to see if it is an ETX character (presumably signifying the end of a message). If the character is an ETX, the ISR might set a global flag. An incorrect implementation of this might be:

int etx_rcvd = FALSE;

void main()
{
...
while (!ext_rcvd)
{
// Wait
}
...
}

interrupt void rx_isr(void)
{ ...
if (ETX == rx_char)
{
etx_rcvd = TRUE;
}
...
}

With optimization turned off, this code might work. However, any half decent optimizer will "break" the code. The problem is that the compiler has no idea that etx_rcvd can be changed within an ISR. As far as the compiler is concerned, the expression !ext_rcvd is always true, and, therefore, you can never exit the while loop. Consequently, all the code after the while loop may simply be removed by the optimizer. If you are lucky, your compiler will warn you about this. If you are unlucky (or you haven't yet learned to take compiler warnings seriously), your code will fail miserably. Naturally, the blame will be placed on a "lousy optimizer."

The solution is to declare the variable etx_rcvd to be volatile. Then all of your problems (well, some of them anyway) will disappear.