Re: Is this a bug?

[Songtao Chen <songtao@cisco.com>]

Hi Neil,

Thanks for the explanation.

My original problem was this,

#include<stdio.h>

int main(void)
{
    int r, i;

    i = 0;
    i = ++i % 2;
    printf("test 1 : i = %d\n", i);

    i = 0;
    i = i++ % 2;
    printf("test 2 : i = %d\n", i);

    return 0;
}

Test 2 gives 0. I was expecting 1. Should it not?

Here is the disassembled version, maybe you can tell
what's going on. I don't know much about assembly code.

    i = 0;
    1083c:       c0 27 bf e8     clr  [ %fp + -24 ]
    i = ++i % 2;
    10840:       d0 07 bf e8     ld  [ %fp + -24 ], %o0
    10844:       92 02 20 01     add  %o0, 1, %o1
    10848:       90 10 00 09     mov  %o1, %o0
    1084c:       d0 27 bf e8     st  %o0, [ %fp + -24 ]
    10850:       93 3a 20 1f     sra  %o0, 0x1f, %o1
    10854:       95 32 60 1f     srl  %o1, 0x1f, %o2
    10858:       92 02 00 0a     add  %o0, %o2, %o1
    1085c:       95 3a 60 01     sra  %o1, 1, %o2
    10860:       92 10 00 0a     mov  %o2, %o1
    10864:       95 2a 60 01     sll  %o1, 1, %o2
    10868:       90 22 00 0a     sub  %o0, %o2, %o0
    1086c:       d0 27 bf e8     st  %o0, [ %fp + -24 ]
    printf("test 1 : i = %d\n", i);
    10870:       13 00 00 42     sethi  %hi(0x10800), %o1
    10874:       90 12 61 78     or  %o1, 0x178, %o0     ! 10978 
<_lib_version+0x8>
    10878:       d2 07 bf e8     ld  [ %fp + -24 ], %o1
    1087c:       40 00 40 6d     call  20a30 <_PROCEDURE_LINKAGE_TABLE_+0x6c>
    10880:       01 00 00 00     nop

    i = 0;
    10884:       c0 27 bf e8     clr  [ %fp + -24 ]
    i = i++ % 2;
    10888:       d0 07 bf e8     ld  [ %fp + -24 ], %o0
    1088c:       92 02 20 01     add  %o0, 1, %o1
    10890:       d2 27 bf e8     st  %o1, [ %fp + -24 ]
    10894:       93 3a 20 1f     sra  %o0, 0x1f, %o1
    10898:       95 32 60 1f     srl  %o1, 0x1f, %o2
    1089c:       92 02 00 0a     add  %o0, %o2, %o1
    108a0:       95 3a 60 01     sra  %o1, 1, %o2
    108a4:       92 10 00 0a     mov  %o2, %o1
    108a8:       95 2a 60 01     sll  %o1, 1, %o2
    108ac:       90 22 00 0a     sub  %o0, %o2, %o0
    108b0:       d0 27 bf e8     st  %o0, [ %fp + -24 ]
    printf("test 2 : i = %d\n", i);
    108b4:       13 00 00 42     sethi  %hi(0x10800), %o1
    108b8:       90 12 61 90     or  %o1, 0x190, %o0     ! 10990 
<_lib_version+0x20>
    108bc:       d2 07 bf e8     ld  [ %fp + -24 ], %o1
    108c0:       40 00 40 5c     call  20a30 <_PROCEDURE_LINKAGE_TABLE_+0x6c>
    108c4:       01 00 00 00     nop

    return 0;


Thanks for your help.

Songtao





At 09:45 PM 1/2/2003 +0000, Neil Booth wrote:
>Songtao Chen wrote:-
>
> > Hi Neil,
> >
> > Thanks for replying.
> >
> > Can you tell me why the parentheses are irrelevant,
> > any doc to support that?
>
>For the same reason that (2) is the same as 2, and not 3.  Like I said,
>post-increment returns the value before incrementing.
>
>Neil.
>
> >
> > Songtao
> >
> > At 09:23 PM 1/2/2003 +0000, Neil Booth wrote:
> > >Songtao Chen wrote:-
> > >
> > >> Hi there,
> > >>
> > >> It appears to be a bug to me.
> > >
> > >No, i++ returns the value before increment.  Parentheses are irrelevant.
> > >
> > >Neil.
> >