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Postings from September 2012
On Sun, Sep 02, 2012 at 08:50:33AM +1000, Tony Cook wrote:
> The code is built with usemorebits, hence uselongdouble, so the NV
> element would force 16-byte alignment.
>
> So the compiler is correct to assume 16-byte alignment (assuming I
> remember the x64 ABI correctly).
# based on Configure
bash-3.2$ cc -oalignbytes alignbytes.c
bash-3.2$ ./alignbytes
16
bash-3.2$ cat alignbytes.c
#include <stdio.h>
typedef long double NV;
struct foobar {
char foo;
NV bar;
} try_algn;
int main()
{
printf("%d\n", (int)((char *)&try_algn.bar - (char *)&try_algn.foo));
return(0);
}
But:
neptune:perl tony$ grep alignbytes config.sh
alignbytes='8'
Urk:
wctomb() found.
writev() found.
You seem to be either cross-compiling or doing a multiarchitecture build,
skipping the memory alignment check.
Checking how long a character is (in bits)...
What is the length of a character (in bits)? [8]
Presumably because of:
: Check if is a multiplatform env
case "$osname" in
next|rhapsody|darwin) multiarch="$define" ;;
esac
case "$multiarch" in
''|[nN]*) multiarch="$undef" ;;
esac
Tony
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