On 29/10/2019 09:28, Luc Van Oostenryck wrote:
> On Tue, Oct 29, 2019 at 02:32:15AM +0000, Ramsay Jones wrote:
>>> + static const struct arch {
>>> + const char *name;
>>> + int mach;
>>> + int bits;
>>> + } archs[] = {
>>> + { "aarch64", MACH_ARM64, 64 },
>>> + { "arm64", MACH_ARM64, 64 },
>>> + { "arm", MACH_ARM, 32 },
>>> + { "i386", MACH_I386, 32 },
>>> + { "m68k", MACH_M68K, 32 },
>>> + { "mips", MACH_MIPS64 },
>>> + { "powerpc", MACH_PPC64 },
>>> + { "ppc", MACH_PPC64 },
>>> + { "riscv", MACH_RISCV64 },
>>
>> I would rather these were explicitly set to 0.
>
> Hmm ... To me, the difference pop out better like so.
> The absence of a value is supposed to mean "there is not
> a known size for this, it needs to be calculated/guessed".
> I find that an explicit 0 conveys this les well.
OK, authors choice! ;-)
>
>>> + { "s390", MACH_S390X, 64 },
>>> + { "s390x", MACH_S390X, 64 },
>>> + { "sparc", MACH_MIPS64 },
>>
>> Er, I suppose this should be MACH_SPARC64, right? (also 0 init).
>
> Ooops, yes, thank you. Bad copy-paste, bad.
>
>>> + if (bits == 0) {
>>> + // guess the size of the architecture
>>> + if (!strcmp(suf, "")) {
>>> + if (arch_m64 == ARCH_LP32)
>>> + bits = 32;
>>> + else
>>> + bits = 64;
>>
>> So, this is a 50-50 bet. ;-)
>
> No, not really.
> The -m32/-m64 flags are still taken in account, before the
> --arch and after it too. If no -m32/-m64 is given at all
> then the used size is the one of the native arch (because
> arch_m64 is initialized so).
heh, yes, I just meant that (if -m32/-m64 has _not_ been given)
then you have a 50/50 chance that you are cross-compiling to a
system that has the same 'bit-ness' as your current platform.
(well, actually, I suppose both are likely to be 64-bit these
days - so, maybe not 50/50! ;-) ).
> I think it's quite natural to
> use but it lacks some error checking (e.g.: --arch=arm -m64
> will purposely *not* select arm64, aka aarach64, but no
> warnings will be issued).
Thanks!
ATB,
Ramsay Jones
