🚀 go-pugleaf

In article <4jr73o$h87@solutions.solon.com>, jcoffin@rmii.com (Jerry Coffin) writes:

|> While `char *' and `void *' are not the same, they are required to have the 
|> same representation.  (The general idea seems to be that if you pass one in 
|> place of the other, it probably shouldn't cause a problem, but they never 
|> quite state that explicitly.  I presume this was an attempt at enforcing 
|> compatibility with older libraries that expected a `char *' where a `void *' 
|> is now indicated.)

My understanding is that a conversion to a (void *) followed by a conversion
to the original pointer type is the identity mapping; in other words, the mapping
a C compilers uses between (void *) and any other pointer type is bijective.
I don't know what you mean with representation; note however that
	T1 *p;
	*(T2 **) &p == (T2 *) p
is not necessarily true, even, I think, if T1 or T2 is void.

This is the reason why casting the comparison function argument
of qsort to a function of two void pointers even if the actual argument
is a function of two pointers to non-void is non-conforming, since the
compiler now has no place to insert the pointer-conversion code.

Konrad Schwarz