Single Precision和Double Precision IEEE 754基2浮点值可以表示整数范围,而不会丢失。
给定乘积A = BC
,其中B
和C
是表示为无损的浮点值的整数,如果乘积A
在数学上落在浮点类型的无损范围内,它是否总是无损?
更具体地说,我们是否知道普通的现代处理器是否可以确保对乘积进行计算,以使整数乘积的行为如上所述?
编辑:要澄清每个链接上方可以无损表示的整数范围,双精度为+ -253,单精度为+ -16777216。
编辑:IEEE-754要求将运算四舍五入到最接近的可表示精度,但是我特别想了解现代处理器的行为
最佳答案
对于任何基本运算,IEEE-754要求,如果数学结果是可表示的,那么它就是结果。
这个问题没有用IEEE-754标记,因此通常只询问浮点数。当精确的结果可表示时,没有一个明智的系统会给出不准确的结果,但是仍然有可能创建一个结果。
补充
这是一个测试float
案例的程序。
#include <math.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
static void Test(float x, float y, float z)
{
float o = x*y;
if (o == z) return;
printf("Error, %.99g * %.99g != %.99g.\n", x, y, z);
exit(EXIT_FAILURE);
}
static void TestSigns(float x, float y, float z)
{
Test(-x, -y, +z);
Test(-x, +y, -z);
Test(+x, -y, -z);
Test(+x, +y, +z);
}
int main(void)
{
static const int32_t SignificandBits = 24;
static const int32_t Bound = 1 << SignificandBits;
// Test all x * y where x or y is zero.
TestSigns(0, 0, 0);
for (int32_t y = 1; y <= Bound; ++y)
{
TestSigns(0, y, 0);
TestSigns(y, 0, 0);
}
/* Iterate x through all non-zero significands but right-adjusted instead
of left-adjusted (hence making the low bit set, so the odd numbers).
*/
for (int32_t x = 1; x <= Bound; x += 2)
{
/* Iterate y through all non-zero significands such that x * y is
representable. Observe that since x and y each have their low bits
set, x * y has its low bit set. Then, if Bound <= x * y, there is
a also bit set outside the representable significand, so the
product is not representable.
*/
for (int32_t y = 1; (int64_t) x * y < Bound; y += 2)
{
/* Test all pairs of numbers with these significands, but varying
exponents, as long as they are in bounds.
*/
for (int xs = x; xs <= Bound; xs *= 2)
for (int ys = y; ys <= Bound; ys *= 2)
TestSigns(xs, ys, (int64_t) xs * ys);
}
}
}