是的,如果您愿意,您当然可以自己计算这些。 Acouple http://en.wikipedia.org/wiki/Machine_epsilon#How_to_determine_machine_epsilon examples http://en.wikipedia.org/wiki/Machine_epsilon#Approximation_using_C维基百科页面上的 C 语言给出了如何计算机器 epsilon 的信息;类似地,您可以通过除/乘以二来找到最小值/最大值,直到低于/溢出。 (然后,您应该在最后一个有效值和下一个二分之一之间进行搜索,以找到“真实”最小/最大值,但这为您提供了一个很好的起点)。
不过,如果您的设备的计算能力为 2.0 或更高,那么数学主要是 IEEE 754,有一些小偏差(例如,并非支持所有舍入模式),但这些偏差不足以影响像这样的基本数值常数;所以你会得到 5.96e-08 的单倍和 1.11e-16 的双倍的标准 emac; FLT_MIN/MAX 为 1.175494351e-38/3.402823466e+38,DBL_MIN/MAX 为 2.2250738585072014e-308/1.7976931348623158e+308。
在计算能力 1.3 的计算机上,单精度不支持非规范化数字,因此您的 FLT_MIN 将明显大于 CPU 上的数字。
在计算能力 2.0 机器上进行快速测试,对最小值/最大值进行快速而肮脏的计算:
#include <stdio.h>
#include <stdlib.h>
#include <getopt.h>
#include <cuda.h>
#include <sys/time.h>
#include <math.h>
#include <assert.h>
#include <float.h>
#define CHK_CUDA(e) {if (e != cudaSuccess) {fprintf(stderr,"Error: %s\n", cudaGetErrorString(e)); exit(-1);}}
/* from wikipedia page, for machine epsilon calculation */
/* assumes mantissa in final bits */
__device__ double machine_eps_dbl() {
typedef union {
long long i64;
double d64;
} dbl_64;
dbl_64 s;
s.d64 = 1.;
s.i64++;
return (s.d64 - 1.);
}
__device__ float machine_eps_flt() {
typedef union {
int i32;
float f32;
} flt_32;
flt_32 s;
s.f32 = 1.;
s.i32++;
return (s.f32 - 1.);
}
#define EPS 0
#define MIN 1
#define MAX 2
__global__ void calc_consts(float *fvals, double *dvals) {
int i = threadIdx.x + blockIdx.x*blockDim.x;
if (i==0) {
fvals[EPS] = machine_eps_flt();
dvals[EPS]= machine_eps_dbl();
float xf, oldxf;
double xd, oldxd;
xf = 2.; oldxf = 1.;
xd = 2.; oldxd = 1.;
/* double until overflow */
/* Note that real fmax is somewhere between xf and oldxf */
while (!isinf(xf)) {
oldxf *= 2.;
xf *= 2.;
}
while (!isinf(xd)) {
oldxd *= 2.;
xd *= 2.;
}
dvals[MAX] = oldxd;
fvals[MAX] = oldxf;
/* half until overflow */
/* Note that real fmin is somewhere between xf and oldxf */
xf = 1.; oldxf = 2.;
xd = 1.; oldxd = 2.;
while (xf != 0.) {
oldxf /= 2.;
xf /= 2.;
}
while (xd != 0.) {
oldxd /= 2.;
xd /= 2.;
}
dvals[MIN] = oldxd;
fvals[MIN] = oldxf;
}
return;
}
int main(int argc, char **argv) {
float fvals[3];
double dvals[3];
float *fvals_d;
double *dvals_d;
CHK_CUDA( cudaMalloc(&fvals_d, 3*sizeof(float)) );
CHK_CUDA( cudaMalloc(&dvals_d, 3*sizeof(double)) );
calc_consts<<<1,32>>>(fvals_d, dvals_d);
CHK_CUDA( cudaMemcpy(fvals, fvals_d, 3*sizeof(float), cudaMemcpyDeviceToHost) );
CHK_CUDA( cudaMemcpy(dvals, dvals_d, 3*sizeof(double), cudaMemcpyDeviceToHost) );
CHK_CUDA( cudaFree(fvals_d) );
CHK_CUDA( cudaFree(dvals_d) );
printf("Single machine epsilon:\n");
printf("CUDA = %g, CPU = %g\n", fvals[EPS], FLT_EPSILON);
printf("Single min value (CUDA - approx):\n");
printf("CUDA = %g, CPU = %g\n", fvals[MIN], FLT_MIN);
printf("Single max value (CUDA - approx):\n");
printf("CUDA = %g, CPU = %g\n", fvals[MAX], FLT_MAX);
printf("\nDouble machine epsilon:\n");
printf("CUDA = %lg, CPU = %lg\n", dvals[EPS], DBL_EPSILON);
printf("Double min value (CUDA - approx):\n");
printf("CUDA = %lg, CPU = %lg\n", dvals[MIN], DBL_MIN);
printf("Double max value (CUDA - approx):\n");
printf("CUDA = %lg, CPU = %lg\n", dvals[MAX], DBL_MAX);
return 0;
}
编译/运行显示答案与 CPU 版本一致(最小值除外;FLT_MIN 是否给出了最小正常值而不是在 CPU 上进行了标准化?)
$ nvcc -o foo foo.cu -arch=sm_20
$ ./foo
Single machine epsilon:
CUDA = 1.19209e-07, CPU = 1.19209e-07
Single min value (CUDA - approx):
CUDA = 1.4013e-45, CPU = 1.17549e-38
Single max value (CUDA - approx):
CUDA = 1.70141e+38, CPU = 3.40282e+38
Double machine epsilon:
CUDA = 2.22045e-16, CPU = 2.22045e-16
Double min value (CUDA - approx):
CUDA = 4.94066e-324, CPU = 2.22507e-308
Double max value (CUDA - approx):
CUDA = 8.98847e+307, CPU = 1.79769e+308