https://gcc.gnu.org/bugzilla/show_bug.cgi?id=53383
--- Comment #22 from H.J. Lu <hjl.tools at gmail dot com> --- (In reply to Andy Lutomirski from comment #21) > $ touch foo.c > $ gcc -c -mno-sse -mpreferred-stack-boundary=3 -mincoming-stack-boundary=3 > foo.c > foo.c:1:0: error: -mincoming-stack-boundary=3 is not between 4 and 12 > $ gcc -c -mno-sse -mpreferred-stack-boundary=3 foo.c I will fix it. > > > > > This makes no sense, since they should be equivalent. > > > > > > Also, I find the docs to be unclear as to what different values of the > > > incoming and preferred stack boundaries mean. > > > > > > Finally, why is -mno-sse required in order to set a low stack boundary? > > > Couldn't gcc figure out that the existence of a stack variable (SSE, > > > alignas, __attribute__((aligned(32))), etc) should force dynamic stack > > > alignment? > > > > Since the x86-86 psABI says that stack must be 16 byte aligned, if the stack > > isn't 16-byte aligned, the code with SSE insn, which follows the psABI, > > will crash when called with 8-byte aligned stack. > > I'm confused here. I agree in principle, but I don't actually think that > gcc works this way, or, if it does, it shouldn't. > > If I compile with -mpreferred-stack-boundary=3 and create an aligned(32) > local variable, then gcc will dynamically align the stack and the variable > will have correct alignment even if the incoming stack was not 16-byte > aligned. That is correct. > Sure, the generated SSE code will be less efficient with > -mpreferred-stack-boundary=3 (because neither "and $-16,%rsp" nor the > required frame pointer is free), but it should still work, right? It works only if ALL codes are compiled with -mpreferred-stack-boundary=3.
