Re^2: address of function

[Maarten Litmaath]

karl at haddock.ima.isc.com (Karl Heuer) writes:
\...
\	int  this[100], (*that)[100] = &this;
\which is analogous to
\	int  foo(void), (*bar)(void) = &foo;
\
\This particular symmetry holds only if the explicit "&" is used.  Writing
\"this" without the ampersand yields the same as "&this", but writing "foo"
\without the ampersand yields "&foo[0]", which is not the same as "&foo".
\(Though they will compare equal if brought to a common type.)

Ahum Karl, typo: this <-> foo.
-- 
"I HATE arbitrary limits, especially when |Maarten Litmaath @ VU Amsterdam:
   they're small."  (Stephen Savitzky)    |maart at cs.vu.nl, mcvax!botter!maart