[Bug c++/43932] conditional operator can't convert 0 to pointer

[schwab at linux-m68k dot org]

--- Comment #1 from schwab at linux-m68k dot org  2010-04-29 12:09 ---
A throw-expression cannot be part of a integral constant expression, thus it
cannot be a null pointer constant.


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schwab at linux-m68k dot org changed:

   What|Removed |Added

 Status|UNCONFIRMED |RESOLVED
 Resolution||INVALID


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=43932

[Bug c++/43932] conditional operator can't convert 0 to pointer

[joerg dot richter at pdv-fs dot de]

--- Comment #2 from joerg dot richter at pdv-fs dot de  2010-04-29 12:54 
---
But throw can be part of ?: operator:

$ cat t.cc
int func( bool b )
{
  return b ? 0 : throw 5;
}

$ g++ -c t.cc

works


-- 

joerg dot richter at pdv-fs dot de changed:

   What|Removed |Added

 Status|RESOLVED|UNCONFIRMED
 Resolution|INVALID |


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=43932

[Bug c++/43932] conditional operator can't convert 0 to pointer

[redi at gcc dot gnu dot org]

--- Comment #3 from redi at gcc dot gnu dot org  2010-04-29 13:31 ---
See [expr.cond]p2
if one operand is a throw-expression the result of the conditional-expression
is an rvalue of the type of the other operand (i.e. int)

I think it's actually a bug that this compiles:
int* f(bool b) { return b ? 0 : 0; }

Again, the result of the conditional expression is an rvalue of type int, it is
not a null pointer constant


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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=43932

[Bug c++/43932] conditional operator can't convert 0 to pointer

[redi at gcc dot gnu dot org]

--- Comment #4 from redi at gcc dot gnu dot org  2010-04-29 13:34 ---
That example 'f' is equivalent to this, which also compiles:

int* g(bool b) { return int(b ? 0 : 0); }

that's /definitely/ wrong


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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=43932

[Bug c++/43932] conditional operator can't convert 0 to pointer

[redi at gcc dot gnu dot org]

--- Comment #5 from redi at gcc dot gnu dot org  2010-04-29 14:11 ---
the definition of integral constant expression has changed between C++03 and
C++0x, so that int() is a valid integral constant expression, as is int(b?0:0)

So that makes my 'f' and 'g' examples valid.

However, by [expr.const] a conditional-expression is not a constant expression
if it contains a throw-expression, so Andreas was 100% correct in comment 2 and
the original report is invalid. It should be

   return b ? (void*)0 : throw 5


-- 

redi at gcc dot gnu dot org changed:

   What|Removed |Added

 Status|UNCONFIRMED |RESOLVED
 Resolution||INVALID


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=43932

[Bug c++/43932] conditional operator can't convert 0 to pointer

[redi at gcc dot gnu dot org]

--- Comment #6 from redi at gcc dot gnu dot org  2010-04-29 14:17 ---
for further confirmation, the original case is covered by:
http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_defects.html#367

and my cases are covered by:
http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_active.html#236


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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=43932