Inconsistent labels?

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Rahsennor

Joined: 07 Jul 2007
Posts: 61

Rahsennor 04 Dec 2010, 02:02

This may or may not be a bug in fasm. If it isn't, feel free to ignore me. Very Happy

This code:

Code:

x equ edi+1

if x eqtype edi
 display 'register',13,10
end if

if x eqtype 1
 display 'number',13,10
end if

if x eqtype edi+1
 display 'both',13,10
end if

Does this:

Code:

flat assembler  version 1.68  (16384 kilobytes memory)
both
1 passes, 0 bytes.

But this code:

Code:

label x at edi+1

if x eqtype edi
 display 'register',13,10
end if

if x eqtype 1
 display 'number',13,10
end if

if x eqtype edi+1
 display 'both',13,10
end if

Does this:

Code:

flat assembler  version 1.68  (16384 kilobytes memory)
number
1 passes, 0 bytes.

If you don't get what I mean, try this:

Code:

if x eqtype 1
 display 'number',13,10
 db x
end if

if x eqtype edi+1
 display 'both',13,10
 db x-edi
end if

If x is a label, it picks the first statement and whinges about the resulting "db edi+1". If x is an equate:

Code:

flat assembler  version 1.68  (16384 kilobytes memory)
both
temp.asm [10]:
 db x-edi
error: invalid argument.

Which is the same result as for "db edi+1-edi"; no register algebra. However, the same db with x as a label does work: x-edi = 1, assembly succeeds.

I see two things: first, eqtype behaves differently depending on whether its arguments are labels; and second, db permits or forbids register algebra depending on whether or not its argument contains a label. Are these behaviours intentional or not? Question

04 Dec 2010, 02:02

JohnFound

Joined: 16 Jun 2003
Posts: 3499
Location: Bulgaria

JohnFound 04 Dec 2010, 07:16

in your first example, "equ" works in preprocessing stage and the assembler never can see the "label" X (it is not label but symbolic constant) it can see only the string "edi+1" replaced in the text of the source.
On the other hand, all labels are numbers once defined, so in your second example the assembler will see the label X, not the text "edi+1".

04 Dec 2010, 07:16

Rahsennor

Joined: 07 Jul 2007
Posts: 61

Rahsennor 04 Dec 2010, 10:10

The preprocessor is not involved. I used equ simply to illustrate. Substitute like so:

Code:

label x at edi+1

if x eqtype edi+1 ; fails!
 display 'with label',13,10
end if

if edi+1 eqtype edi+1 ; same address, but now it works
 display 'without label',13,10
end if

And the same for this:

Code:

label x at edi+1

db x-edi ; works
db edi+1-edi ; same value: fails

The problem is that the address is the same, but the result is different. This does not, to me, make any sense. Can anyone enlighten me? Confused

EDIT: To elaborate, say I want to make a macro that takes an address as an argument. That macro must decide if it can save the address directly with dd or if it must use an alternative encoding. How do I do it?

04 Dec 2010, 10:10

Tomasz Grysztar

Joined: 16 Jun 2003
Posts: 8367
Location: Kraków, Poland

Tomasz Grysztar 04 Dec 2010, 10:44

This is related to how parser sees the source. Unless in a special context (after the AT operator, or inside the square brackets), "edi+1" is seen as two separate expressions. "edi" - register, and "+1" - a numerical expression. That's why this works:

Code:

if eax 4 eqtype edi+1
 display 'register and expression'
end if

and this:

Code:

push edi+1

generates two push instructions. It is related to the tokenization (as in lexical analysis) of source that is done by preprocessor. Strings like "edi+1" or "edi +1" or "edi+ 1" are all tokenized with exactly the same result - three consecutive tokens, "edi", "+" and "1". And then the parser is greedy to see the register operand when it can, so in these three tokens it sees two operands - register and then numerical expression.
So the fact that this:

Code:

db edi+1-edi

fails is the side-effect of this parsing process, since DB directive sees the register operand, which is not allowed there, and thus fails.

As I wrote earlier, AT and PTR operators (and the square brackets, which are the same thing as PTR) provide a special context where "numerical" expressions starting with registers are allowed. Thus:

Code:

if at eax 4 eqtype at edi+1
 display 'Not anymore!'
end if

label x at edi+1

if at x eqtype at edi+1
 display 'And this one is OK'
end if

Last edited by Tomasz Grysztar on 04 Dec 2010, 10:46; edited 1 time in total

04 Dec 2010, 10:44

baldr

Joined: 19 Mar 2008
Posts: 1651

baldr 04 Dec 2010, 10:45

Rahsennor wrote:

To elaborate, say I want to make a macro that takes an address as an argument. That macro must decide if it can save the address directly with dd or if it must use an alternative encoding. How do I do it?

Win32AX.Inc header uses mov instruction in pushd macro to distinguish kind of address that was given.

04 Dec 2010, 10:45

Rahsennor

Joined: 07 Jul 2007
Posts: 61

Rahsennor 04 Dec 2010, 11:18

Merci beaucoup! It makes sense now.

"if at edi+1 eqtype at 1" still passes, making it useless for my purpose. It was a silly purpose anyway. Laughing

@baldr: that is some hack. Nice.

04 Dec 2010, 11:18

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