Translating C to PB

[Julian van der Vaart]

What would be the best alternative for the following C/C++ statement:

const double f = ...

Thanks.

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[José Roca]

MACRO f = ...

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[Julian van der Vaart]

Thanks, it did work. Until it got more complex.

I'm now starting to get nesting errors for all the macro's. The code I am translating from C++ to PB is a very complex formule.

Are there any other alternatives to 'const double'?

I did try' %f = ... ' Which does work for some, but not for things like '%f = sqr(...)'

Thanks.

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[Steven Pringels 3]

If in C, const is used, I wonder how one can still use
f = sqrt(....);

with it.

So, for const, there is only one possibility %F in PB.

Maybe you should give us the whole picture ?


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I'd rather be hated for who I am, then being loved for who I am not.

[Julian van der Vaart]

To be more specific, it's not C but C++ (does that matter??).

Here is an example from the, way to big to post here, code:

    const double ee=f*(2-f);  
    const double e=sqrt(ee);  
    const double eea = ee/(1.0-ee);

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[Simon Morgan]

This works (macros can be nested):

MACRO f=1.5
MACRO ee=f*(2-f) 
MACRO e=SQR(ee)  
MACRO eea = ee/(1.0-ee)  

FUNCTION PBMAIN
  MSGBOX STR$(e)+STR$(eea)
END FUNCTION

%name is OK for integer constants. For floating point you need to use a macro. But why not simply declare a variable whose value you set only once at the start of the code? Is there a reason it has to be a constant?

[Cliff Nichols]

See "Floating Point Data Types" in the PB Help File for floating point

(Unless of course if someone tells me I can not use the following in PB to replace C)

Dim ee AS DOUBLE
DIM f AS DOUBLE
DIM e AS DOUBLE
DIM eeaa AS DOUBLE
ee=f*(2-f)  
e=sqrt(ee)  
eea = ee/(1.0-ee)

OR

#ee=#f*(2-#f)  
#e=sqrt(#ee)
#eea = #ee/(1.0-#ee)

sort of thing.


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[Bob Zale]

Julian --

There are a variety of ways to deal with this type of issue, but the simplest is likely to be use of a MACRO as described before. As complexity of expressions grow, you just need to be aware that precedence of operators could come into play. Therefore, if you simply enclose the terms in parentheses, it should force evaluation in the sequence originally expected.

MACRO ee = (f*(2-f))
MACRO e = (sqrt(ee))
MACRO eea = (ee/(1.0-ee))

Best regards,

Bob Zale
PowerBASIC Inc.


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