对于 gcc 编译器，gcc 在不使用任何标志与 -O2 进行编译时针对哪个 x86-64 指令集？

Question

For x86-64 there are lots of instruction sets that speed up code execution. Here is a list from gcc wiki https://gcc.gnu.org/wiki/FunctionMultiVersioning :

• MMX
• SSE
• SSE2
• SSE3
• SSSE3
• SSE4.1
• SSE4.2
• POPCNT
• AVX
• AVX2

For gcc compiler, what x86-64 instruction set does gcc target when you compile without any flags versus -O2 ?

To keep things simple lets just say the question is about gcc version 12 (most recent major). But I would like to know what gcc command switches/options i need to do to so that i can see what my version of gcc version does.

I assume that gcc chooses something that is "portable" so that would mean probably something slow. But this is just my assumption... I would like to know does that mean like SSE4.2 or none?

Answer 1

If you don't pass a command-line -march option, then you get whatever was selected when gcc was compiled. The default is -march=x86-64 but it could have been overridden by whoever compiled your gcc (eg your binary package distributor). See https://gcc.gnu.org/install/configure.html and note the --with-arch option.

You can compile with -v -Q to see what option is in use. Look for the options passed line.

With -march=x86-64 you get "least common denominator" code that will run on every known x86-64 CPU, all the way back to the AMD K8. This includes SSE2, which was part of the original AMD64 spec, but not SSE3 or anything later. popcnt would not be included either.

The -march option is orthogonal to optimization options like -O2 and the -f... flags (eg -funroll-loops ). You always get code compatible with whatever is selected by -march , no matter what optimization options are in use. However -m flags (like -mavx ) can permit the use of other CPU features beyond what -march implies, in which case your code is only guaranteed to run on CPUs with those features.