The equivalent of the offending line would be
foo((double *)(void *) j);
since any access through a void pointer CAN change the value.
'(double *)(void *)' does not keep some sort of remnant of the 'void *'. The
final cast renders the end-result a 'double *' and negates the 'void *'. You
An object may be the type of its last cast -- but it also can't
exactly lose the benefit/cost of being cast to a pointer to an
undefined type. As soon as you undefine the type of a pointer, it
loses the remnant of ever having had the initial type in the first
place.
In KR C, you couldn't cast
I found the rule -- at least for C99. It is ISO 9899:1999 section
6.2.5, rule 26 and footnote 39:
26) A pointer to void shall have the same representation and alignment
requirements as a
pointer to a character type. Similarly, pointers to qualified or unqualified
versions of
An object may be the type of its last cast -- but it also can't
exactly lose the benefit/cost of being cast to a pointer to an
undefined type. As soon as you undefine the type of a pointer, it
loses the remnant of ever having had the initial type in the first
place.
Right, but that doesn't
Good afternoon,
I am installing openssl on a Mac-Intel. I have to introduce the
folllowing patch to have the package compiler with the zlib
configuration option
--- ./work/openssl-0.9.7j/Makefile.org.ori 2006-04-24
15:32:57.0 +0200
+++ ./work/openssl-0.9.7j/Makefile.org
This means you cannot pass an 'X509 **' as a 'char **'.
Unfortunately 'X509 *' and 'char *' are not compatible types because
'X509' and 'char' are not compatible. They are not both unions, they
are not both structures.
But you can pass 'X509 **' as 'void *'. So...
void
But you can pass 'X509 **' as 'void *'. So...
void x509func(void* p) { X509** pp = (X509**)p; ... }
void trampoline(void* p) { x509func(p); }
void caller(void) { X509* p; trampoline(p); }
should (MUST?) work just fine.
/r$
When I change my test programs
David Schwartz wrote:
An object may be the type of its last cast -- but it also can't
exactly lose the benefit/cost of being cast to a pointer to an
undefined type. As soon as you undefine the type of a pointer, it
loses the remnant of ever having had the initial type in the first
place.
Are you sure that problem is in cast ?
$ cat test.c
main() {
int j=2;
double *d=(double*)j;
*d=1.0;
printf(%d %e\n, j, *d);
printf(%d %e\n, j, *d);
}
gcc -O2 test.c ./a.out
2 1.00e+00
0 1.00e+00
Same result in case with line double *d=j; (but expected warning:
