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> Main > Understanding $ with EQU and = |
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badc0de02 07 Sep 2014, 14:54
weird...
hmm i think there is a bug somewhere im using this to simulate the situation because you only show a part of your code: Code: org 7c00h E820Buffer: times 24 db 0 E820BufferLength db ($ - E820Buffer) out put in hex: Code: 00000000000000000000000000000000000000000000000018 output in decimal: 24 no problem quite the same with that other way i think you have a bug in your code. or it is a bug in flatassembler |
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07 Sep 2014, 14:54 |
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smiddy 07 Sep 2014, 15:36
Well, I wonder if the 'equ' is part of the issue. Unfortunately I don't have time to relook at it presently. It must be within the usage of locale boundary difference, perhaps? However that doesn't make sense either, since they essentially are saying the very same thing.
I won't get an opportunity to re-look at it for a few days unfortunately.  |
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07 Sep 2014, 15:36 |
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smiddy 07 Sep 2014, 15:39
badc0de02 wrote: ...i think you have a bug in your code. I'll post the rest of the code snippets later. Smiddy |
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07 Sep 2014, 15:39 |
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l_inc 07 Sep 2014, 15:52
smiddy
Quote: Well, I wonder if the 'equ' is part of the issue It is. equ is a preprocessor directive and $ is an assembly time constant/variable. This means, that the preprocessing time constant E820BufferLength is expanded long before $ is recognized. Consider the following code fragment: Code: E820Buffer: times 24 db 0 E820BufferLength equ ($ - E820Buffer) rb 0x100 dd E820BufferLength Now consider what the assembler will see after the preprocessing stage: Code: E820Buffer: times 24 db 0 rb 0x100 dd ($ - E820Buffer) Would you expect the last dword to contain 24? Or maybe something else? _________________ Faith is a superposition of knowledge and fallacy |
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07 Sep 2014, 15:52 |
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typedef 07 Sep 2014, 19:39
E820Buffer: times 24 db 0
E820BufferLength = ($ - E820Buffer) |
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07 Sep 2014, 19:39 |
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smiddy 07 Sep 2014, 22:41
l_inc wrote: smiddy rb would reserve bytes elsewhere, especially if you declare a data (this is the wrong verbiage) a data segment, somewhere else. So I don't really know. |
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07 Sep 2014, 22:41 |
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smiddy 07 Sep 2014, 22:41
typedef wrote: E820Buffer: times 24 db 0 |
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07 Sep 2014, 22:41 |
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l_inc 07 Sep 2014, 22:58
smiddy
Quote: rb would reserve bytes elsewhere Oh, I didn't expect your answer to be that much wrong. There is no such thing as "data segment" for fasm (1). fasm does not differentiate data and code at all. rb reserves bytes at the very same location where it appears in the listing. In my example rb 0x100 is effectively the same as times 0x100 db 0 . So if you are for some reason more familiar with db you can consider my example as follows: Code: E820Buffer: times 24 db 0 times 0x100 db 0 dd ($ - E820Buffer) (1) Don't misunderstand me here. fasm does support data and code specification for corresponding executable file formats, but this specification does not make any semantical difference for the compiler: you can put both wherever you want and everything will be put into the resulting file in the same order it appears in the listing. Hence the name "flat assembler". _________________ Faith is a superposition of knowledge and fallacy |
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07 Sep 2014, 22:58 |
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smiddy 08 Sep 2014, 02:40
When I started using "flat assembler" was before rx (where x is any of the data type declarations) and never went to any length of study or use them. I did know they were 'reserved' however, but expected them not to be "inline" or "at the very locale" as were they exist in the code, that does make sense though. Does that just mean they are not initialized to any value, and they use what ever is there in those memory locales?
So the question is, when E820BufferLength is recognized first, and then it goes back and assigns the EQU, why doesn't it assign the correct value? Is your point that length is aligned to some boundary value? If so, how does one turn that off boundary alignments? Is there a difference between '=' and 'EQU'? I assume '=' works, I need to go back and try it. Unfortunately that will be a while. |
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08 Sep 2014, 02:40 |
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neville 08 Sep 2014, 04:06
smiddy wrote: Is there a difference between '=' and 'EQU'? I assume '=' works, I need to go back and try it. Unfortunately that will be a while. So this topic should really be renamed to "Understanding EQU and ="  _________________ FAMOS - the first memory operating system |
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08 Sep 2014, 04:06 |
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smiddy 08 Sep 2014, 11:02
Maybe neville, but if I do this:
Code: E820Buffer: times 24 db 0 E820BufferLength equ 24 Code: E820Buffer: times 24 db 0 E820BufferLength equ ($ - E820Buffer) |
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08 Sep 2014, 11:02 |
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l_inc 08 Sep 2014, 12:02
smiddy
Quote: Does that just mean they are not initialized to any value Yes and no. The memory you get from rx is de facto set to zero (same as dx ?). It is however marked as uninitialized. Practically this means that this memory will not become a part of the compiled executable in some cases such as at the end of a file and at the end of a section. A bit different case are macros, that make use of the "uninitialized" semantics. E.g., procedure macros interpret uninitialized data definition directives in a locals block as stack space allocation, in which case you in fact will get whatever it was in the stack before. Quote: So the question is, when E820BufferLength is recognized first, and then it goes back and assigns the EQU, why doesn't it assign the correct value? equ is recognized first and defines E820BufferLength. The $ is however not recognized yet. Then E820BufferLength is recognized, as it was defined by equ. It's correct value is not the result of the calculation but the whole sequence of tokens ($ - E820Buffer). This sequence of tokens is put at the location of E820BufferLength and is then evaluated at that location. As the value of $ is location dependent, you will get different values of the expression ($ - E820Buffer) depending on where you use E820BufferLength and completely independent on where you define E820BufferLength with the equ directive. Quote: Is your point that length is aligned to some boundary value? No. Absolutely not. Quote: Is there a difference between '=' and 'EQU'? neville has already pointed out that difference: equ defines a symbol for further symbolic substitution, and the equality sign induces actual calculation of an expression following it. But it is also important to remember that equ and everything defined by it is processed at the preprocessing stage. $ is meaningless at this early stage and cannot be evaluated. The equality sign is processed during the assembly stages, when $ already can be evaluated as current address. _________________ Faith is a superposition of knowledge and fallacy |
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08 Sep 2014, 12:02 |
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revolution 08 Sep 2014, 12:37
Perhaps you can see the difference here:
Code: E820Buffer: times 0x24 db 0 E820BufferLength equ ($ - E820Buffer) E820BufferLength2 = ($ - E820Buffer) dd ($ - E820Buffer) ;<---- places 0x00000024 dd ($ - E820Buffer) ;<---- places 0x00000028 dd ($ - E820Buffer) ;<---- places 0x0000002c dd E820BufferLength ;<---- places 0x00000030 dd E820BufferLength2;<---- places 0x00000024 |
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08 Sep 2014, 12:37 |
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smiddy 08 Sep 2014, 17:00
Thanks, now I get it (I think): with EQU and $, the $ defines 'where' it is being used (ala E820BufferLength used well into my code, get evaluated there), not the location in the ASM file. I will use = instead.
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08 Sep 2014, 17:00 |
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typedef 08 Sep 2014, 17:13
$.....Current address/offset
EQUATE(s).....same as #define C/C++ =.....Evaluate expression then assign result to |
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08 Sep 2014, 17:13 |
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