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Posted Aug 13, 2012 16:50 UTC (Mon) by nybble41 (subscriber, #55106)
In reply to: ACCESS_ONCE() by etienne
Parent article: ACCESS_ONCE()

Fortunately, in your example avar is always initialized to 10. :)

Assuming "loglevel = 4" was replaced with "loglevel == 4", I would expect the compiler to generate a warning in this case, since it can't _prove_ that avar was initialized before the printf() call. I would also hope that the compiler would take advantage of the fact that avar is either 10 or undefined to simply set it to 10 regardless of loglevel, for code size and performance reasons if nothing else.

In cases like this, IMHO, it would be better to use a common flag rather than testing loglevel multiple times:

void fct (void) {
int avar;
bool use_avar;
use_avar = (loglevel == 4);
if (use_avar) avar = 10;
if (use_avar) printf ("%d\n", avar);

I'm not sure whether the compiler can follow this any better than before, so there may still be a warning, but at least the coupling is visible to anyone reading the code, and doesn't depend on the implementation of external functions.

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