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Index > Compiler Internals > "db (-1) shl 1" and "db $ff shl 1"

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edemko



Joined: 18 Jul 2009
Posts: 549
edemko
Code:
db (-1) shl 1 ; = $ffffffff'fffffffe ; allows
db $FF shl 1  ; = $00000001'fffffffe ; does not allow
    
Post 19 May 2010, 10:15
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baldr



Joined: 19 Mar 2008
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baldr
edemko,

-1*2 == -2, fits in (signed) byte;
255*2 == 510, doesn't fit in byte.
Post 19 May 2010, 10:27
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edemko



Joined: 18 Jul 2009
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edemko
isn't $FF same -1

edit #2
$FF would have to be internally byte_into_qword extended by fasm as DB was used
Post 19 May 2010, 10:32
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edemko



Joined: 18 Jul 2009
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edemko
decision: use hex and binary
Post 19 May 2010, 10:39
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baldr



Joined: 19 Mar 2008
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baldr
edemko wrote:
$FF would have to be internally byte_into_qword extended by fasm as DB was used
Numeric expressions are evaluated in the same way regardless of intended usage of their results. Doesn't "byte should be sign-extended into qword because «define byte» is used" sound contradictory? $FF has no signs of sign. Wink
Post 19 May 2010, 11:05
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edemko



Joined: 18 Jul 2009
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edemko
thanks
due my logics db -1 would be impossible
Post 19 May 2010, 11:30
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Tomasz Grysztar



Joined: 16 Jun 2003
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Tomasz Grysztar
edemko wrote:
isn't $FF same -1

No, $FF is 255, a positive number, it is NOT the same as -1. If you want negative number, you have to use the "-" sign, only then fasm is able understand what you are doing and adjust calculations accordingly (since in assembly you don't distinguish "unsigned" and "signed" data types, assembler needs to take them both into account.
Post 19 May 2010, 11:41
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
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revolution
But also note that this assembles okay:
Code:
db 0xffffffffffffffff    
since internally fasm stores -1 as 0xffffffffffffffff, thus it cannot distinguish between the two inputs.
Post 20 May 2010, 09:41
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Tomasz Grysztar



Joined: 16 Jun 2003
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Location: Kraków, Poland
Tomasz Grysztar
revolution wrote:
But also note that this assembles okay:
Code:
db 0xffffffffffffffff    
since internally fasm stores -1 as 0xffffffffffffffff, thus it cannot distinguish between the two inputs.
Yes, this on the other hand is a kind of "bug", that comes from the limitations of fasm's architecture. Something that is planned to be better in fasm 2.
Post 20 May 2010, 09:52
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baldr



Joined: 19 Mar 2008
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baldr
Tomasz Grysztar,

It's not a bug as I see it. Argument for db is an expression, if it evaluates to something (signed 64-bit) that fits in byte (value>>7 is -1, 0 or 1) then it's OK to accept it.
Post 20 May 2010, 10:29
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revolution
When all else fails, read the source


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revolution
baldr wrote:
It's not a bug as I see it. Argument for db is an expression, if it evaluates to something (signed 64-bit) that fits in byte (value>>7 is -1, 0 or 1) then it's OK to accept it.
It depends upon one's interpretation of "bug". This:
Code:
db 1 shl 999    
Gives no error.
Post 20 May 2010, 10:48
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baldr



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baldr
revolution,

Do you think it should? May I ask why, then?
Post 20 May 2010, 11:07
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revolution
When all else fails, read the source


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revolution
baldr wrote:
Do you think it should? May I ask why, then?
Overflow.

How can 2^999 be stored in one byte?
Post 20 May 2010, 11:18
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baldr



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baldr
revolution,

Exactly. For now, we're stuck with [-2**63, 2**63-1] range, thus 1 shl 999 equals to 0 and everything is fine. Wink

Without rules to define certain expression as being of certain type (s/ubyte and such), there is no chance to detect overflow condition. It's common issue for HLLs, they regularly define rules to attribute and coerce values.
Post 20 May 2010, 12:22
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Tomasz Grysztar



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Tomasz Grysztar
Actually, fasm can easily detect such overflow condition. It ignores it just for convenience.
Post 20 May 2010, 12:24
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revolution
When all else fails, read the source


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Location: In your JS exploiting you and your system
revolution
baldr wrote:
Exactly. For now, we're stuck with [-2**63, 2**63-1] range, thus 1 shl 999 equals to 0 and everything is fine. Wink

Without rules to define certain expression as being of certain type (s/ubyte and such), there is no chance to detect overflow condition. It's common issue for HLLs, they regularly define rules to attribute and coerce values.
Indeed. Like I said, it depends upon one's interpretation of "bug". If you are happy to say that 2^999==0 then fine, it is not a bug under that definition.
Post 20 May 2010, 12:26
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baldr



Joined: 19 Mar 2008
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baldr
revolution wrote:
If you are happy to say that 2^999==0 then fine, it is not a bug under that definition.
I'm not "happy", just "understanding the limitations". While (1 shl X) shr X seems to be equal to 1 for any X, in real world it's not.
Post 20 May 2010, 19:19
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edemko



Joined: 18 Jul 2009
Posts: 549
edemko
continuing numbers dispute: dt $20HexDigitsMayBeCool
Code:
macro tf name*, exponent*, significand*{
  label name tbyte at $
  dq significand
  dw exponent
}

tf f1,$3fff-0000,$8000000000000000 ;2^0*1
tf f2,$3fff-0064,$8000000000000000 ;2^-64*1

entry $

fld     tbyte[f2]
fld     tbyte[f1]
fadd    st0,st1   ;st0+st1=st0, here is why it sAcks!
    


%LAST_REMAINDER as a global variable, could it be good? More over with read/write access.
Post 24 May 2010, 11:33
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