Question

I have a question about how compiler-set symbols, in particular CPU feature flags (like SSE, AES, AVX) are actually set. For instance, if I call gcc with -mavx, is the __AVX__ symbol set regardless of whether the system the code is about to be built on actually supports AVX instructions, or does it check before?

I’m asking because I need to build a particular code path depending on CPU capabilities and would like to automate it so that the correct path is determined upon compilation based on the build system, instead of manually enabling desired features. But since the only CPU I have supports basically every feature, I cannot test my above assumption (first world problems, I know)

There is going to be a lot of code so simply keeping everything and branching at runtime is unacceptable – and it is assumed that my library will be built before being used on a given system anyway.

I mean, at worst I can force this behavior by wrapping the gcc arguments in a cpuid-aware script, but if gcc does it automatically it would be preferable. So does anyone know whether it does?

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I am mostly interested in gcc’s take on this but I am also curious to know how other C compilers behave.

Answer 1

If you pass the -mavx flag, __AVX__ will always be set for the resulting compilation (and the resulting code may not run on non-AVX machines).

If you pass the -march=native flag, gcc will enable the instruction sets supported by the build machine, so __AVX__ will only be set if the build machine supports it.