Checking compiler mode
lazarus / 62-General

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MarkMLl · 2021-07-31 10:26

Is there a compile-time check which will show whether the current mode
is one in which the mod operator will work in accordance with the ISO
Pascal definition?

I'm aware of -Miso and {$modeswitch ISOMOD } , but at a particular
place in the code I want to know whether one of these is active and if
not use a custom operator:

Code: Pascal  [Select][+][-]window.onload = function(){var x1 =
document.getElementById("main_content_section"); if (x1) { var x =
document.getElementsByClassName("geshi");for (var i = 0; i (* "Number
theory compatible" implementation of the modulo operator, hopefully 
to match the Python operator's characterisitics.*)operator >< (a, b:
int64): int64; (* See https://forum.lazarus.freepascal.org/index.php/t
opic,48549.msg349691.html#msg349691 *)(* In principle $modeswitch
ISOMOD should fix the problem, but I don't know     *)(* what version
of the compiler introduced it (I suspect later than 3.0.4).    
*) begin  Result := abs(a) mod abs(b);  if (a < 0) and (Result<>0)
then    Result := abs(b) - Resultend { >< } ; 
MarkMLl

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Kays · 2021-07-31 10:44

Quote from: MarkMLl on July 31, 2021, 12:26:43 pmIs there a compile-
time check which will show whether the current mode is one in which
the mod operator will work in accordance with the ISO Pascal
definition?No. As you probably already know, the {$ifOpt} directive
only works on settings a short/single-letter alternative exists for.

Quote from: MarkMLl on July 31, 2021, 12:26:43 pmI'm aware of -Miso
and {$modeswitch ISOMOD } , but at a particular place in the code I
want to know whether one of these is active and if not use a custom
operator: […]Simply always use the “custom” operator. Write a separate
unit in which you “overload”, for instance, the >< operator in any
case, yet this unit has {$modeSwitch ISOMod+} at the top.Code: Pascal
 [Select][+][-]window.onload = function(){var x1 =
document.getElementById("main_content_section"); if (x1) { var x =
document.getElementsByClassName("geshi");for (var i = 0; i unit
foo;{$modeSwitch ISOMod+}{$modeSwitch result+}interface  operator ><
(const x: ALUSint; const y: ALUUint): ALUUInt;implementation  operator
>< (const x: ALUSint; const y: ALUUint): ALUUInt;    begin      result
:= x mod y;    end;end.The unit can be compiled in any compiler
compatibility mode (that supports units and operator overloading), but
will always use an ISO-compliant definition of mod.

The ISO-compliant definition of mod exists for almost ten years now,
basically since {$mode ISO} was supported in 2.6.0. The {$modeSwitch
ISOMod+}, however, exists for over 5 years now. It was not available
in 3.0.0. The most recent git test conversion stops tagging at
3.0.0RC1, so I can’t make any definite statement, whether it was
available in 3.0.4.

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PascalDragon · 2021-08-01 11:05

Quote from: MarkMLl on July 31, 2021, 12:26:43 pmIs there a compile-
time check which will show whether the current mode is one in which
the mod operator will work in accordance with the ISO Pascal
definition?
The current mode of the unit is available as FPC_<mode> define.
Modeswitches themselves have no defines however (so you would not be
able to detect {$MODE ISO}{$MODESWITCH ISOMOD-}.

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MarkMLl · 2021-08-01 12:56

Quote from: PascalDragon on August 01, 2021, 01:05:18 pmQuote from:
MarkMLl on July 31, 2021, 12:26:43 pmIs there a compile-time check
which will show whether the current mode is one in which the mod
operator will work in accordance with the ISO Pascal definition?
The current mode of the unit is available as FPC_<mode> define.
Modeswitches themselves have no defines however (so you would not be
able to detect {$MODE ISO}{$MODESWITCH ISOMOD-}.

Thanks for that, still useful.

MarkMLl