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Q66841: Does /ND Override /Aw?

Article: Q66841
Product(s): See article
Version(s): 6.00   | 6.00
Operating System(s): MS-DOS | OS/2
Keyword(s): ENDUSER | | mspl13_c
Last Modified: 18-NOV-1990

Question:

I have two separate modules to an application that I compile with
/Alfw and /ND. The documentation states that /Aw will assume SS!=DS
and DS is not loaded. However, when I examine the assembler output
(/Fc or /Fa), I see that DS is loaded for each function in the modules
compiled with /ND. Is this a problem in the compiler? Does /ND
override /Aw?

Response:

You are confusing the functionality of these two switches. The /Aw
switch is used by the compiler to determine if local variables
(typically stack based) can be based on DS, as well as SS. Take the
case where a variable needs to be copied from local data to global
data, which would normally include a couple of MOV instructions. If
the compiler can assume DS=SS, it can generate a LEA instruction on
the stack-based variable and set up ES:DI with the destination. DS
will already be set to the correct value. If DS != SS, the compiler
will have to generate more code to correctly access the stack-based
variable.

Now, for the /ND switch. In this case, the data segment for the module
is not the default segment. The compiler will do an explicit DS load
at function entry point because of just this fact, that is, there is
no way for the compiler to know that DS is properly set to the
nonstandard segment name (not DGROUP). While this behavior is indeed
the same behavior as the _loadds keyword, it has nothing to do with
the main reason for /Aw (DS!=SS).

To summarize, the /Aw switch doesn't cause the segment load, but it
also doesn't stop it. This is one case where one switch (/ND) can
override one of the effects of another switch (/Aw). However, the main
effect of the /Aw switch is not changed (don't assume DS = SS).

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