Loongnix操作系统

三个c语言测试性能的程序发现龙芯性能极差

第一个求圆周率的程序，目前发现，龙芯到目前为止无法算完999999位数的圆周率，因为我们测试的龙芯3a3000搞坏了，目前只测了99999圆周率，源码如下：
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main(int argc, char * argv[])
{
clock_t start, finish;
double duration;
long * pi, * t, m, n, r, s;
int t0[][3] = {48, 32, 20, 24, 8, 4}, k0[][3] = {1, 1, 0, 1, 1, 1};
int n0[][3] = {18, 57, 239, 8, 57, 239}, d, i, j, k, p, q;

d = (argc > 1) ? (((i = atoi(argv[1])) < 0) ? 0 : i) : 1000;
q = (argc > 2) ? 1 : 0;
printf("%s\n\n", "Nature (R) Pi value compute Program (C) Tue 1999.11.30");
printf("pi= %s%d * arctg(1/%d) %s %d * arctg(1/%d) %s %d * arctg(1/%d) [%s]\n",
k0[q][0] ? "" : "-", t0[q][0], n0[q][0], k0[q][1] ? "+" : "-", t0[q][1],
n0[q][1], k0[q][2] ? "+" : "-", t0[q][2], n0[q][2], q ? "Stomer" : "Gauss");
if ((t = (long *)calloc((d += 5) + 1, sizeof(long))) == NULL) return 1;
if ((pi = (long *)calloc(d + 1, sizeof(long))) == NULL) return 2;

start = clock();

for (i = d; i >= 0; i--) pi[i] = 0;
for (p = 0; p < 3; p++) {
for (k=k0[q][p], n=n0[q][p], t[i=j=d]=t0[q][p], i--; i >= 0; i--) t[i] = 0;
for (r = 0, i = j; i >= 0; i--) {
r = (m = 10 * r + t[i]) % n;
t[i] = m / n;
k ? (pi[i] += t[i]) : (pi[i] -= t[i]);
}
while (j > 0 && t[j] == 0) j--;
for (k = !k, s = 3, n *= n; j > 0; k = !k, s += 2) {
for (r = 0, i = j; i >= 0; i--) {
r = (m = 10 * r + t[i]) % n;
t[i] = m / n;
}
while (j > 0 && t[j] == 0) j--;
for (r = 0, i = j; i >= 0; i--) {
r = (m = 10 * r + t[i]) % s;
m /= s;
k ? (pi[i] += m) : (pi[i] -= m);
}
}
}
for (n = i = 0; i <= d; pi[i++] = r) {
n = (m = pi[i] + n) / 10;
if ((r = m % 10) < 0) r += 10, n--;
}
finish = clock();
printf("pi= %ld.", pi[d]);
for (i = d - 1; i >= 5; i--)
printf("%ld%s", pi[i], ((m = d - i + 5) % 65) ? ((m % 5) ? "" : " ") : "\n");
printf("%sDIGITS: %d\n", (m % 65) ? "\n" : "", d - 5);
duration = (double)(finish - start) / CLOCKS_PER_SEC;
printf("time %f seconds \n", duration);
return 0;
}

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8 个回复

jiangtao9999

赞同来自: zzz19760225 、神龙覆云

这种大量循环的算法……
其实本身算法内对于不同的架构就有不同的影响。

admin

赞同来自: zzz19760225 、神龙覆云

如果确实测出性能差，那么给出调优方案，才是真正的给力，一味地指责带不来进步:)

天高地厚

赞同来自:

第二段是一个求积分的过程，在测试过程中，如果不对库功能单独代替，龙芯的情能是极差极差的：
本代码中前部注释的地方，就是对库功能代替实现部分：
默认用gcc编释后，龙芯实在太垃圾了：
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
// Function to be integrated
// Define and prototype it here
// | sin(x) |
#define INTEG_FUNC(x) fabs(sin(x))
// Prototype timing function
double dclock(void);
int main(void)
{
// Loop counters and number of interior points
unsigned int i, j, N;
// Stepsize, independent variable x, and accumulated sum
double step, x_i, sum;
// Timing variables for evaluation
clock_t start, finish;
// Start integral from
double interval_begin = 0.0;
// Complete integral at
double interval_end = 2.0 * 3.141592653589793238;
// Start timing for the entire application
printf(" \n");
printf(" Number of | Computed Integral | \n");
printf(" Interior Points | | \n");
start = clock();
for (j=2;j<27;j++)
{
printf("------------------------------------- \n");
// Compute the number of (internal rectangles + 1)
N = 1 << j;
// Compute stepsize for N-1 internal rectangles
step = (interval_end - interval_begin) / N;
// Approx. 1/2 area in first rectangle: f(x0) * [step/2]
sum = INTEG_FUNC(interval_begin) * step / 2.0;
// Apply midpoint rule:
// Given length = f(x), compute the area of the
// rectangle of width step
// Sum areas of internal rectangle: f(xi + step) * step
for (i=1;i<N;i++)
{
x_i = i * step;
sum += INTEG_FUNC(x_i) * step;
}
// Approx. 1/2 area in last rectangle: f(xN) * [step/2]
sum += INTEG_FUNC(interval_end) * step / 2.0;
printf(" d | e | \n", N, sum);
}
finish = clock();
printf(" \n");
printf(" Application Clocks = %d \n", (finish - start)/CLOCKS_PER_SEC);
printf(" \n");
return 0;
}

天高地厚

赞同来自:

第三段是龙芯资深爱好者写的混合代码，这个程序我们打算在龙芯上跑时，龙芯让我们弄坏了，结果没跑过，只在x86上跑过，
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>

int main(){

clock_t start, finish;
double duration;
double testa;
double testb;
int sum=0;
int num=1;
int sum2=0;
int num2=2;
double sqrt(double x);
start = clock();
while(num<=90000000){
sum=sum+num;
num=num+2;
testa=sqrt(sum);
testa=testa*num;
}
printf("奇数和为:%d\n",sum);
printf("testa最终值 %.52lf\n", testa);
while(num2<=90000000){
sum2=sum2+num2;
num2=num2+2;
testb=sqrt(sum2);
testb=testb*num2;
}
printf("偶数和为：%d\n",sum2);
printf("testb最终值 %.52lf\n", testb);
finish = clock();
duration = (double)(finish - start) / CLOCKS_PER_SEC;
printf("time %f seconds \n", duration);

return 0;
}