Conditional defines?

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Azu

Joined: 16 Dec 2008
Posts: 1159

Azu 19 Aug 2009, 08:03

This doesn't compile. Sad

Please tell me how to fix it..

Code:

if foo eq      1
   bar =   "blah"
else if foo eq      2
   bar =   1,2,3
else
       bar =   0
end if

foo  equ     2
db bar

19 Aug 2009, 08:03

shoorick

Joined: 25 Feb 2005
Posts: 1614
Location: Ukraine

shoorick 19 Aug 2009, 09:22

maybe place foo equ 2 before if

19 Aug 2009, 09:22

Azu

Joined: 16 Dec 2008
Posts: 1159

Azu 19 Aug 2009, 09:28

Hmm. Actually.. it does compile in the example I gave. But it compiles wrong (to 0x0 instead of 0x010203). If I move the equ to the top it doesn't compile at all. Why is this? I thought that equs were parsed before ifs (which is why I use = instead of equ to set bar).. but were that the case it wouldn't matter whether the "foo equ 2" was at the top or bottom.

Now I'm even more confused.

19 Aug 2009, 09:28

revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 20417
Location: In your JS exploiting you and your system

revolution 19 Aug 2009, 12:43

You can't define symbols later in the source. This doesn't work:

Code:

display foo
foo     equ     1

Also you can't assign bar = "blah", you will need to use bar equ "blah". This mean that you also have to use match to process the bar equ.

Code:

foo equ 2
bar equ   0
match foo,1 {bar equ "blah"}
match foo,2 {bar equ 1,2,3}
db bar

19 Aug 2009, 12:43

Azu

Joined: 16 Dec 2008
Posts: 1159

Azu 19 Aug 2009, 21:20

Thanks, but it doesn't work.

Code:

foo equ 2 
bar equ   0 
match foo,2 {bar equ 1,2,3} 
match foo,1 {bar equ "blah"} 
db bar

Outputs "blah". Sad

Code:

bar equ   0 
match foo,2 {bar equ 1,2,3} 
match foo,1 {bar equ "blah"} 
db bar

Shouldn't even compile.. but outputs "blah". Sad

Please help.. the method you posted has exactly the same problems as the one I originally posted..

19 Aug 2009, 21:20

LocoDelAssembly
Your code has a bug

Joined: 06 May 2005
Posts: 4624
Location: Argentina

LocoDelAssembly 19 Aug 2009, 21:47

Code:

foo equ 2
bar equ 'INVALID'
match =2,foo {bar equ '1','2','3'}
match =1,foo {bar equ "blah"}
display bar

19 Aug 2009, 21:47

Azu

Joined: 16 Dec 2008
Posts: 1159

Azu 19 Aug 2009, 21:50

Thank you

it works.

19 Aug 2009, 21:50

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