31.03.2014 17:32, Martin Frb пишет:
Show quoted text 5 lines
Just asking this again, as I did not get any answer yet?
Is that the indented behaviour, and if so what is the reasoning for it?
This behavior is hardly intended.
To get clues about what's going on, you can compile the compiler with OPT=-dEXTDEBUG, then use it with -vd to compile your tests.
So, in case of subrange of integer/byte, the overloaded procedures are matched as follows:
Bar(TFoo);
invalid
Bar(LongInt);
ex: 0 eq: 1 l1: 0 l2: 0 l3: 0 l4: 0 l5: 0 l6: 0 oper: 0 ord: 4.29496729300000047684E+0009
- LongInt : equal
Bar(Variant);
ex: 0 eq: 0 l1: 0 l2: 0 l3: 0 l4: 0 l5: 0 l6: 0 oper: 1 ord: 0.00000000000000000000E+0000
- Variant : convert_operator
And in case of subrange of enum, it becomes as follows:
Bar(TFoo);
ex: 0 eq: 0 l1: 1 l2: 0 l3: 0 l4: 0 l5: 0 l6: 0 oper: 0 ord: 0.00000000000000000000E+0000
- TFoo : convert_l1
Bar(LongInt);
invalid
Bar(Variant);
ex: 0 eq: 0 l1: 1 l2: 0 l3: 0 l4: 0 l5: 0 l6: 0 oper: 0 ord: 0.00000000000000000000E+0000
- Variant : convert_l1
Here Variant gets the same degree of compatibility with TFoo1 as TFoo, that's probably a bug in compiler (Variant should probably have same compatibility as in first case, i.e. convert_operator, if not incompatible at all).
Also, the subrange may receive the same treatment as in the first case, i.e. considered equal to its base type but with some "ordinal distance".
Regards,
Sergei
Show quoted text 65 lines
-------- Original Message --------
Subject: overload question (variant vs enum subrange)
Date: Sun, 26 Jan 2014 19:48:42 +0000
From: Martin Frb
To: FPC-Pascal users discussions
TFoo1 is a sub range of FFoo
And it seems to match both TFoo and variant.
> project1.lpr(24,3) Error: Can't determine which overloaded function to
> call
> project1.lpr(15,11) Hint: Found declaration: Bar(TFoo);
> project1.lpr(11,11) Hint: Found declaration: Bar(Variant);
No other type seems to be bothered by variant.
Also variant is the only type, that I found that conflicts with TFoo1
Why?
I know I can declare
procedure Bar(a: TFoo1); overload;
and it will solve it (even if I leave all the others, because it is an
exact match).
But why does a subrange of integer/byte work? Subrange of integer does
not give the error, even so it could match both.
program Project1;
{$mode objfpc}
{// $mode delpti}
type
TFoo = (a1,a2,a3,a4,a5);
TFoo1 = a2..a4;
TFoo2 = 1..3;
TFoo3 = byte(1)..(3);
procedure Bar(a: Variant); overload;
begin end;
procedure Bar(a: Integer); overload;
begin end;
procedure Bar(a: TFoo); overload;
begin end;
var
f1: TFoo;
f2: TFoo1;
f3: TFoo2;
f4: TFoo3;
begin
Bar(f2);
end.
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