On Thu, 28 Jul 2011, Jari Häkkinen wrote:
Are you sure about that? I just tried it with a little example and at least gcc compiles both variants to the exact same assembly code. Tried it with and without -O2.That would freak me out. Doesn't "++j" mean "increment j, then test" whereas "j++" means "test j, then increment"?No, for a for loop for ( [1]; [2]; [3] ) where [3] is ++j will increment j before use. However, in an if-statement the complete statement [3] is evaluated before the test [2] is done. If the compiler is smart it will produce the fastest binary code regardless ++j or j++. However, if the [3] is more complicated like a hypothetical i = ++j + k the compiler will most probably generate different binary code (compared to i = ++j + k).
Right, but j++ will increment _after_ it's used, correct? So how could ++j vs j++ generate the same assembly code and be correct?
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