Bruce Cran wrote:
cpghost wrote:
There's a mismatch here: scanf("%d", ...) expects a pointer to int,
while &nnote is a pointer to a short. Normally, an int occupies more
bytes in memory than a short (typically sizeof(int) == 4 on 32bit
platforms, and sizeof(int) == 8 on 64bit platforms; while typically
sizeof(short) == 2).
I think short and int stay the same on both 32 and 64 bit platforms,
while it's only long that gets bumped to 8 bytes. At least that seems
to be what happens on FreeBSD amd64.
--
Bruce
No... you're only safe using int32, int64, etc. Just for grins try
compiling a program like this:
#include <stdio.h>
int main() {
printf("%d\n", sizeof(int));
return 0;
}
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