Re: ARM compiler bug?

Yes I have verified that.You will notice that I print out the value of z and
it is 0x7FFFFFFF.

Your right that x is unsigned (and is really a very positive number), but
the conditions use by the ldr (ldrgt and ldrle) are signed greater than and
signed less than or equal. It is going to treat the previous conditions
flags as if they where signed values and not unsigned values.


"anonymous" <anonymous@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:B26A44B6-C293-4DF6-89F8-D88DDC27F2BF@xxxxxxxxxxxxxxxx
> "static unsigned int x = 0x80000000; // Very negative"
>
> This is an unsigned number and is +2147483648(dec).
>
> Have you verified that z is actually 0x7FFFFFFF ?
>
> "Danny" wrote:
>
> > Hi All,
> >
> > The following sample code gives me unexpected results.
> >
> > void FindBug()
> > {
> > static unsigned int x = 0x80000000; // Very negative
> > static unsigned int y = 1; // Positive
> > int z = x - y; // z = 0x7FFFFFFF
> > if (z > 0) {
> > OALMSG(TRUE, (L"z (0x%X) > 0\r\n", z));
> > } else {
> > OALMSG(TRUE, (L"z (0x%X) <= 0?\r\n", z));
> > }
> > }
> >
> > Looking at this code seems that I should get "z (0x7FFFFFFF) > 0",
however,
> > I actually get "z (0x7FFFFFFF) <= 0?".
> > I have attached the generated assembler file below. I have embedded some
> > comments in the assembler.
> >
> > 00000 |FindBug| PROC
> >
> > ; 37 : {
> >
> > 00000 |$L14233|
> > 00000 e52de004 str lr, [sp, #-4]!
> > 00004 |$M14231|
> >
> > ; 38 : static unsigned int x = 0x80000000; // Very negative
> > ; 39 : static unsigned int y = 1; // Positive
> > ; 40 : int z = x - y; // z = 0x7FFFFFFF
> >
> > 00004 e59f3024 ldr r3, [pc, #0x24]
> > 00008 e5932004 ldr r2, [r3, #4]
> > 0000c e5933000 ldr r3, [r3]
> > 00010 e0521003 subs r1, r2, r3
> >
> > ; 41 : if (z > 0) {
> > ; 42 : OALMSG(TRUE, (L"z (0x%X) > 0\r\n", z));
> >
> > -> At this point z = 0x7FFFFFFF (positive number).
> >
> > 00014 c59f0010 ldrgt r0, [pc, #0x10]
> >
> > -> We expect the ldrgt to execute but it does not (ldrle does).
> > -> The condition for gt are Z = 0, N==V.
> > -> Z = 0 We pass the first condition.
> > -> N = 0 (bit 31) and V = 1 (overflow from negative to positive) We fail
the
> > second condition.
> > -> Therefore, N != V and the ldrgt (>) does not execute, even though
that is
> > what we expect.
> >
> > -> Looks like a compiler bug.
> >
> > ; 43 : } else {
> > ; 44 : OALMSG(TRUE, (L"z (0x%X) <= 0?\r\n", z));
> >
> > 00018 d59f0008 ldrle r0, [pc, #8]
> > 0001c eb000000 bl NKDbgPrintfW
> >
> > ; 45 : }
> > ; 46 : }
> >
> > 00020 e49de004 ldr lr, [sp], #4
> > 00024 e12fff1e bx lr
> > 00028 |$L14236|
> > 00028 00000000 DCD
> >
|??_C@_1CC@EMHAJMPE@?$AAz?$AA?5?$AA?$CI?$AA0?$AAx?$AA?$CF?$AAX?$AA?$CJ?$AA?5
> > ?$AA?$DM?$AA?$DN?$AA?5?$AA0?$AA?$DP?$AA?$AN?$AA?6?$AA?$AA@|
> > 0002c 00000000 DCD
> >
|??_C@_1BO@FIKAIAKN@?$AAz?$AA?5?$AA?$CI?$AA0?$AAx?$AA?$CF?$AAX?$AA?$CJ?$AA?5
> > ?$AA?$DO?$AA?5?$AA0?$AA?$AN?$AA?6?$AA?$AA@|
> > 00030 00000000 DCD |?y@?1??FindBug@@9@9|
> > 00034 |$M14232|
> >
> > ENDP ; |FindBug|
> >
> > Danny
> >
> >
> >


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