On Wednesday, 1 February 2017 at 10:05:49 UTC, Richard Delorme
wrote:
On Tuesday, 31 January 2017 at 23:30:04 UTC, Walter Bright
wrote:
On 1/31/2017 3:00 PM, Richard Delorme wrote:
The thing about memcpy is compilers build in a LOT of
information about it that simply is not there in the
declaration. I suggest retrying your example for gcc/clang,
but use your own memcpy, i.e.:
void* mymemcpy(void * restrict s1, const void * restrict
s2, size_t n);
Let us know what the results are!
//-----8<-------------------------------------------------------
#include <string.h>
#include <stdio.h>
void* mymemcpy(void* restrict dest, const void* restrict src,
size_t n) {
const char *s = src;
char *d = dest;
for (size_t i = 0; i < n; ++i) d[i] = s[i];
return d;
}
void *copy(const void *c, size_t n) {
char d[16];
return mymemcpy(d, c, n);
}
int main(void) {
char a[16] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14, 15};
char *b = copy(a, 8);
for (int i = 0; i < 16; ++i) printf("%d ", b[i]);
putchar('\n');
}
//-----8<-------------------------------------------------------
$ gcc mymemcpy.c -O2 -W
mymemcpy.c: In function 'copy':
mymemcpy.c:13:9: warning: function returns address of local
variable [-Wreturn-local-addr]
return mymemcpy(d, c, n);
^~~~~~~~~~~~~~~~~
memcpy4.c:12:7: note: declared here
char d[16];
clang (version 3.8.1) failed to find error in this code.
You have to define the mymemcpy() in another source file and only
put the prototype in this module. If the compiler sees the code
it can do the complete data flow analyses. With only the
declaration it can't and that is Walter's point. The annotations
allow to give to the declaration the information the compiler can
not deduce itself from the code, because the code is in another
module (object file, library).
