2009/7/20 Gonzalo Tornaria <torna...@math.utexas.edu>
>
> Really funny... What would you expect the following program to output?
>
> #include <stdio.h>
> #include <string.h>
>
> int main (void)
> {
> int i = 1 , j = -1;
> char c = 1, d = 1;
>
> memset(&c, 2, i + j);
> memset(&d, 2, i + j);
>
> printf("%d %d\n", c, d);
>
> return 0;
> }
>
> On the T2 with gcc (tried a few different versions) it outputs "1 2".
> Dunno how to run the sun compiler.
>
> The asm looks reasonable; excerpt:
>
> mov 1, %g1
> st %g1, [%fp-8]
> mov -1, %g1
> st %g1, [%fp-4]
> mov 1, %g1
> stb %g1, [%fp-9]
> mov 1, %g1
> stb %g1, [%fp-10]
>
> ld [%fp-8], %g2
> ld [%fp-4], %g1
> add %g2, %g1, %g1
> add %fp, -9, %g2
> mov %g2, %o0
> mov 2, %o1
> mov %g1, %o2
> call memset, 0
> nop
>
> ld [%fp-8], %g2
> ld [%fp-4], %g1
> add %g2, %g1, %g1
> add %fp, -10, %g2
> mov %g2, %o0
> mov 2, %o1
> mov %g1, %o2
> call memset, 0
> nop
>
> Notation:
>
> i -> [%fp-8]
> j -> [%fp-4]
> c -> [%fp-9]
> d -> [%fp-10]
>
>
> Maybe you can try compiling to assembler on the T2 and compiling this
> assembler on your blade to see if it gives some clue?
>
> Gonzalo
>
I would have expected 1 1, as you have twice tried to modify zero bytes
(i+j=0) That's what I get on my Blade 2000. But on 't2' I get 1 2.
Sticking another printf I get 1 1 after the first memset call, then 1 2
after the second memset call.
How to I force gcc to output the assember like this?
Since t2 is on a maintence contract, I will call Sun today and log a bug
report. I'll do that in the evening here as the phone call will be cheaper -
I need to phone from the UK to the USA.
Dave
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