Yes, semantically it will be evaluated on every loop. In some cases, compilers may be able to remove the condition from the loop automatically - but not always. In particular:
void foo(const struct rect *r) {
for (int i = 0; i < r->width * r->height; i++) {
quux();
}
}
The compiler will not be able to move the multiplication out in this case, as for all it knows quux()
modifies r
.
In general, usually only local variables are eligible for lifting expressions out of a loop (assuming you never take their address!). While under some conditions structure members may be eligible as well, there are so many things that may cause the compiler to assume everything in memory has changed - writing to just about any pointer, or calling virtually any function, for example. So if you're using any non-locals there, it's best to assume the optimization won't occur.
That said, in general, I'd only recommend proactively moving potentially expensive code out of the condition if it either:
- Doesn't hurt readability to do so
- Obviously will take a very long time (eg, network accesses)
- Or shows up as a hotspot on profiling.