Displaying 3 results from an estimated 3 matches for "tu_int".
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to_int
2018 Apr 25
2
compiler-rt incorrect for this udivmodti4 case?
Here is my test case:
#include <stdio.h>
int main(int argc, char **argv) {
tu_int a = (tu_int)0x1ec273014 << 64 | 0xff7377ffffffffffuLL;
tu_int b = (tu_int)0x8ac7230489e80000uLL;
tu_int r;
tu_int q = __udivmodti4(a, b, &r);
utwords qt;
qt.all = q;
utwords rt;
rt.all = r;
fprintf(stderr, "q=0x%.16llX%.16llX\nr=0x%.16llX%.16llX\n&quo...
2018 Apr 25
0
compiler-rt incorrect for this udivmodti4 case?
> On Apr 25, 2018, at 12:33 AM, Andrew Kelley via llvm-dev <llvm-dev at lists.llvm.org> wrote:
>
> Here is my test case:
>
> #include <stdio.h>
>
> int main(int argc, char **argv) {
> tu_int a = (tu_int)0x1ec273014 << 64 | 0xff7377ffffffffffuLL;
> tu_int b = (tu_int)0x8ac7230489e80000uLL;
> tu_int r;
> tu_int q = __udivmodti4(a, b, &r);
>
> utwords qt;
> qt.all = q;
> utwords rt;
> rt.all = r;
> fprintf(stderr, "...
2020 May 21
2
on division of __int128 bit integer
Hi Team,
I observer that division of __int128 bit is very heavy operation.
It internally call a routine '__udivti3', which internally call '
__udivmodti4'.
Due to it the overall performance is much much slower (almost 15 time
slower than if I do it via a combination of 64-bit or microsoft '_udiv128').
Also what to know if I can directly call below routine directly from