www.digitalmars.com         C & C++   DMDScript  

digitalmars.D.bugs - Weird overload resolution (or rather, lack thereof) for literals

reply Stewart Gordon <smjg_1998 yahoo.com> writes:
Using DMD 0.131 (has DMD itself actually changed since then?), Windows 98SE.

----------
enum Qwert { yuiop }

void asdfg(Qwert hjkl) { printf("Called hjkl(Qwert)\n"); }
void asdfg(uint zxcvb) { printf("Called hjkl(uint)\n"); }

void main() {
     int nm = 2;

     asdfg(nm);
     asdfg(cast(int) nm);
     asdfg(3);
     asdfg(cast(int) 3);
}
----------
D:\My Documents\Programming\D\Tests\implicit_enum.d(11): function 
implicit_enum.asdfg called with argument types:
     (int)
matches both:
     implicit_enum.asdfg(Qwert )
and:
     implicit_enum.asdfg(uint)
D:\My Documents\Programming\D\Tests\implicit_enum.d(12): function 
implicit_enum.asdfg called with argument types:
     (int)
matches both:
     implicit_enum.asdfg(Qwert )
and:
     implicit_enum.asdfg(uint)

Tool completed with exit code 1
----------

For the integer variable, it manages to resolve the overload.  Indeed, 
it calls the uint version.

However, with the literal, it doesn't, even if it has been cast.  This 
is absurd.  Just as integer variables don't implicitly convert to enums, 
neither should integer literals.  And that line 10 passes but line 12 
doesn't is even more absurd, as both have the same type set in the same 
amount of stone.

Stewart.

-- 
-----BEGIN GEEK CODE BLOCK-----
Version: 3.1
GCS/M d- s:- C++  a->--- UB  P+ L E  W++  N+++ o K-  w++  O? M V? PS- 
PE- Y? PGP- t- 5? X? R b DI? D G e++>++++ h-- r-- !y
------END GEEK CODE BLOCK------

My e-mail is valid but not my primary mailbox.  Please keep replies on 
the 'group where everyone may benefit.
Oct 13 2005
parent Thomas Kuehne <thomas-dloop kuehne.cn> writes:
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1

Stewart Gordon schrieb am 2005-10-13:
 Using DMD 0.131 (has DMD itself actually changed since then?), Windows 98SE.

 ----------
 enum Qwert { yuiop }

 void asdfg(Qwert hjkl) { printf("Called hjkl(Qwert)\n"); }
 void asdfg(uint zxcvb) { printf("Called hjkl(uint)\n"); }

 void main() {
      int nm = 2;

      asdfg(nm);
      asdfg(cast(int) nm);
      asdfg(3);
      asdfg(cast(int) 3);
 }
 ----------
 D:\My Documents\Programming\D\Tests\implicit_enum.d(11): function 
 implicit_enum.asdfg called with argument types:
      (int)
 matches both:
      implicit_enum.asdfg(Qwert )
 and:
      implicit_enum.asdfg(uint)
 D:\My Documents\Programming\D\Tests\implicit_enum.d(12): function 
 implicit_enum.asdfg called with argument types:
      (int)
 matches both:
      implicit_enum.asdfg(Qwert )
 and:
      implicit_enum.asdfg(uint)

 Tool completed with exit code 1
 ----------

 For the integer variable, it manages to resolve the overload.  Indeed, 
 it calls the uint version.

 However, with the literal, it doesn't, even if it has been cast.  This 
 is absurd.  Just as integer variables don't implicitly convert to enums, 
 neither should integer literals.  And that line 10 passes but line 12 
 doesn't is even more absurd, as both have the same type set in the same 
 amount of stone.

Added to DStress as http://dstress.kuehne.cn/run/o/overload_26_A.d http://dstress.kuehne.cn/run/o/overload_26_B.d Thomas -----BEGIN PGP SIGNATURE----- iD8DBQFEcL9q3w+/yD4P9tIRAvehAJ9qK4MMTukennpThWOTEbYz0o/HNwCgyGQI 655YZjqDsA9BA6QYNPZg4I4= =PhVn -----END PGP SIGNATURE-----
May 21 2006