《汇编语言的艺术》一书中说：
fprem1 指令计算部分余数。 fprem1 计算 st0/st1 的部分余数。如果 ST0 和 ST1 的指数之差小于 64，fprem1 可以计算出精确的 一次操作中余数。
ST0和ST1是FPU数据寄存器。所以我想测试一下它的性能。 我的代码是这样的：
program exercise;
#include( "stdlib.hhf" );
static
x: real80:=156.5e18;//first number
y: real80:=14.36; //second number
z:real80; // partial remainder
begin exercise;
fld(y); //y goes into st0
fld(x); //now x goes into st0 but previous content of st0
//(that is st1) will move into st1, so x is into st0 and y is into st1
fprem1(); //calculates the remainder of st0/st1 (here x/y )but
// for getting the remainder, the quotient will round to the nearest value and then
// remainder = st0 - rounded quotient*st1==>>that is
// equivalent by remainder= x-rounded quotient*y
// ===>> then the remainder can be positive or negative.
fstp(z); //this will store the partial remainder in the z real80 variable.
stdout.put("z: ",z);
end exercise;
但是当我运行这个程序时，意识到对于较小的功率差异，结果是可以预期的，但是对于较大的差异，结果远远达不到预期。
I have summarized my result below:
dividing expected remainder program remainder
*****************************************************************
1) 156.5/14.36 -1.46 -1.4599..9e000
2) 156.5e1/14.36 -.24 -2.39e-1
3) 156.5e2/14.36 -2.4 -2.39
4) 156.5e3/14.36 4.72 4.7200...619
5) 156.5e4/14.36 4.12 4.1200...6195
6) 156.5e5/14.36 -1.88 -1.8799...805
7) 156.5e6/14.36 -4.44 -4.4399..3805
8) 156.5e7/14.36 -1.32 -1.3199...5017
9) 156.5e8/14.36 1.16 1.16006..826
10) 156.5e9/14.36 -2.76 -2.7599..737
11) 156.5e10/14.36 1.12 1.1200619..627
12) 156.5e11/14.36 -3.16 -3.159380..739
13) 156.5e12/14.36 -2.88 -2.873805..346
14) 156.5e13/14.36 -0.08 -7.99..e-2 ~= -0.079
15) 156.5e14/14.36 -0.8 -7.99..e-1 ~= -0.79
16) 156.5e15/14.36 6.36 6.36340...924
17) 156.5e16/14.36 6.16 6.1940300..120
18) 156.5e17/14.36 4.16 4.500300559
/***** unacceptable results (from program) at bellow *****/
19) 156.5e18/14.36 (expected) 1.48 5.13274..e10
20) 156.5e19/14.36 (expected) 0.44 5.13274..e10
****************************************************************
The Results 19) and 20) is unacceptable, because the remainder can not
be greater than divisor, which is incorrect.
>>Am I doing wrong? Or is there a problem with this instruction(fprem1)?
Can someone explain my problem? What can I do? Thanks.
FPREM 不准确：
#include <iostream>
#include <random>
#include <cstdint>
#include <cmath>
#include <bit>
extern double x87Fprem( double x, double y );
using namespace std;
int main()
{
mt19937_64 mt;
uniform_int_distribution<uint64_t> uid( 0, -1 );
constexpr size_t N = 10'000'000;
double accuracySum = 0;
for( size_t r = N; r; )
{
double
a = bit_cast<double>( uid( mt ) ),
b = bit_cast<double>( uid( mt ) );
if( isnan( a ) || isinf( a ) || isnan( b ) )
continue;
double
std = fmod( a, b ),
x87 = x87Fprem( a, b );
if( signbit( std ) == signbit( x87 ) || !x87 && !std )
std = abs( std ),
x87 = abs( x87 ),
accuracySum += x87 != std ? log2( x87 > std ? x87 : std ) - log2( abs( x87 - std ) ) : 53;
--r;
}
cout << accuracySum / 1.0e7 << endl;
return 0;
}
MASM：
PUBLIC ?x87Fprem@@YANNN@Z
_TEXT SEGMENT
?x87Fprem@@YANNN@Z PROC
movsd qword ptr [rsp - 16], xmm0
movsd qword ptr [rsp - 8], xmm1
fld qword ptr [rsp - 8]
fld qword ptr [rsp - 16]
fprem
fstp qword ptr [rsp -8]
fstp st
movsd xmm0, qword ptr [rsp - 8]
ret
?x87Fprem@@YANNN@Z ENDP
_TEXT ENDS
END
上面的代码在我的 Zen4-CPU 的结果中打印 28 个共享位。