一个51单片机的性能测试源码

ID:384624 · 发表于 2018-8-9 15:18

目前我无论如何改代码都无法在keil中编释通过，特来请高人指点，
源码下载网址：https://gitee.com/test386/test
以下是源码：

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
#define INTEG_FUNC(x) fabs(sin(x))
double dclock(void);
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) : 9999;
q = (argc > 2) ? 1 : 0;
printf("%s\n\n", "pi calc..................................");
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--;
}
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);
//_______________________________________
// Loop counters and number of interior points
unsigned int ia, ja, Na;
// Stepsize, independent variable x, and accumulated sum
double duration2;
double stepa, x_i, suma;
// Timing variables for evaluation
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");
for (ja=2;ja<27;ja++)
{
printf("------------------------------------- \n");
// Compute the number of (internal rectangles + 1)
Na = 1 << ja;
// Compute stepsize for Na-1 internal rectangles
stepa = (interval_end - interval_begin) / Na;
// Approx. 1/2 area in first rectangle: f(x0) * [stepa/2]
suma= INTEG_FUNC(interval_begin) * stepa / 2.0;
// Apply midpoint rule:
// Given length = f(x), compute the area of the
// rectangle of width stepa
// Suma areas of internal rectangle: f(xi + stepa) * stepa
for (ia=1;ia<Na;ia++)
{
x_i = ia * stepa;
suma += INTEG_FUNC(x_i) * stepa;
}
// Approx. 1/2 area in last rectangle: f(xN) * [stepa/2]
suma += INTEG_FUNC(interval_end) * stepa / 2.0;
printf(" %10d | %14e | \n", Na, suma);
}
printf(" \n");
//_______________________________________
double duration1;
double testa;
double testb;
int sum=0;
int num=1;
int sum2=0;
int num2=2;
double sqrt(double x);
while(num<=90000000){
sum=sum+num;
num=num+2;
testa=sqrt(sum);
testa=testa*num;
}
printf("Anti cheating verification code is=:%d\n",sum);
printf("Floating-point precision %.38lf\n", testa);
while(num2<=90000000){
sum2=sum2+num2;
num2=num2+2;
testb=sqrt(sum2);
testb=testb*num2;
}
printf("Anti cheating verification code is=：%d\n",sum2);
printf("Floating-point precision %.38lf\n", testb);
finish = clock();
duration = (double)(finish - start) / CLOCKS_PER_SEC;
printf("Total time %f seconds \n", duration);
printf("The less the time it is used, the more the performance becomes strongert!\n");
printf("Press Enter to exit\n");
sum = getchar( );
return 0;
}

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ID:384624 · 发表于 2018-8-9 15:57

希望管理员帮我看看，这是我第一个开源项目，能帮我实现小目标吗。

ID:384624 · 发表于 2018-8-9 15:59

能不能帮我看看，这是我第一个开源项目，如果无法在51上跑起来，龙芯群里的人会笑话我，我以前牛吹大了，说这程序能在51上跑，结果不行。

ID:380391 · 发表于 2018-8-10 08:01

51上面time.h这些c标准库里的东西能随便用吗？

ID:155507 · 发表于 2018-8-10 08:43

c51里面好像没有time.h，这个应该是标准c的函数。

没听说过自带的库里有这个啊，一般都是自己写的，毕竟大家用的CPU，晶振，精度要求都不一样，你要求KEIL写个通用的函数，这也太难为人家了。

我感觉至少clock_t clock(void)函数是需要你自己实现的。

ID:384624 · 发表于 2018-8-11 06:08

angmall 发表于 2018-8-10 08:43
c51里面好像没有time.h，这个应该是标准c的函数。

没听说过自带的库里有这个啊，一般都是自己写的，毕竟 ...

打算用定时器计时，现在遇到变量非法问题

ID:155507 · 发表于 2018-8-11 09:13

Time.h

// Time.h for C51
// Prototypes for time clock functions.
#ifndef _Time_h
#define _Time_h
#ifndef _CLOCK_T_DEFINED
#define _CLOCK_T_DEFINED
typedef unsigned long clock_t;
#endif
#define CLOCKS_PER_SEC 1000
#include<reg52.h> //包含头文件，一般情况不需要改动，头文件包含特殊功能寄存器的定义
static clock_t Clock_Tick=0;
clock_t clock( void );
/*------------------------------------------------
定时器初始化子程序
------------------------------------------------*/
void Init_Timer0(void)
{
TMOD &= 0xF0; //设置定时器模式
TMOD |= 0x01; //使用模式1，16位定时器，使用"|"符号可以在使用多个定时器时不受影响
TH0=0xFC; //给定初值，这里使用定时器0 1毫秒@12.000MHz
TL0=0x18; //设置定时初值
EA=1; //总中断打开
ET0=1; //定时器中断打开
TF0 = 0; //清除TF0标志
TR0=1; //定时器开关打开定时器0开始计时
}
/*------------------------------------------------
定时器中断子程序
------------------------------------------------*/
void Timer0_isr(void) interrupt 1 using 1
{
TH0=0xFC; //重新赋值 1毫秒
TL0=0x18;
Clock_Tick++;
}
clock_t clock( void )
{
return Clock_Tick;
}
#endif /* _Time_h */

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ID:155507 · 发表于 2018-8-11 09:17

serial.h

#define XON 0x11
#define XOFF 0x13
sfr AUXR=0x8E;
/*------------------------------------------------
函数声明
------------------------------------------------*/
void SendStr(unsigned char *s);
/*------------------------------------------------
串口初始化
------------------------------------------------*/
void InitUART (void)
{
PCON &= 0x7F; //波特率不倍速
AUXR &= 0xBF; //定时器1时钟为Fosc/12,即12T
AUXR &= 0xFE; //串口1选择定时器1为波特率发生器
SCON = 0x50; // SCON: 模式 1, 8-bit UART, 使能接收
TMOD &= 0x0F; //设定定时器1
TMOD |= 0x20; // TMOD: timer 1, mode 2, 8-bit 重装
TH1 = 0xFD; // TH1: 重装值 9600 波特率晶振 12.000MHz
TL1 = 0xFD; //设定定时初值
TR1 = 1; // TR1: timer 1 打开
ET1 = 0; //禁止定时器1中断
EA = 1; //打开总中断
//ES = 1; //打开串口中断
}
/*------------------------------------------------
主函数
------------------------------------------------*/
/*
Example #include <stdio.h>
void tst_printf (void) {
char a = 1;
int b = 12365;
long c = 0x7FFFFFFF;
unsigned char x = 'A';
unsigned int y = 54321;
unsigned long z = 0x4A6F6E00;
float f = 10.0;
float g = 22.95;
char buf [] = "Test String";
char *p = buf;
printf ("hello world");
printf ("char %bd int %d long %ld\n",a,b,c);
printf ("Uchar %bu Uint %u Ulong %lu\n",x,y,z);
printf ("xchar %bx xint %x xlong %lx\n",x,y,z);
printf ("String %s is at address %p\n",buf,p);
printf ("%f != %g\n", f, g);
printf ("%*f != %*g\n", (int)8, f, (int)8, g);
}
*/
/*
上面代码调试运行后，串口不断输出"hello world"，程序陷入死循环。
问题的解决：在程序末尾应加上while(1);
原因：如果不加while(1); 将执行一条RET指令，这条指令会让程序复
位(软件复位，从堆栈弹出0000H到PC寄存器)，复位后程序重新开始
执行。
思考：习惯了window下编程，主程序执行完了就退出到操作系统，单片机就
不一样了，退到哪里呢？以下是单片机和编译器的对话：
单片机：我没事做了，死了算了~~~
编译器：想死？你还是复位吧~~~RET
单片机：o，shit~~~
单片机串口相关问题：标准51只有一个串口，但是现在很多基于51内核的单
片机都会有两个串口或是更多，那么printf函数针对哪个串口呢？
回答：Keil\C51\LIB目录下有个putchar.c文件，printf函数会调用该文件中的
putchar函数，putchar里面的使用的寄存器都是标准51的(如TI，SBUF)，它
们的地址是固定的，由此可知printf函数只针对串口0。如果想让printf函数支
持串口1或串口2。。。那么修改putchar函数的寄存器即可，比如SBUF改为
SBUF1。但是能不能用printf函数对串口0，串口1。。。都适用呢，自己想办
法吧：）
另外：如果只是用printf输出字符串的话还是自己编个函数吧，因为printf要处
理各类数据的输出，占用代码空间很大，我试了一下，程序里就一条printf语
句，code就1K了，而且printf执行起来也慢。
在C51中，使用printf()函数进行格式化输出时，格式控制符与ANSI C有所不同。
在格式控制字符中，b表示byte
以十进制输出uint8_t : %bu
以十进制输出int8_t : %bd
*/
/*
* putchar (full version): expands '\n' into CR LF and handles
* XON/XOFF (Ctrl+S/Ctrl+Q) protocol
*/
char putchar (char c) {
if (c == '\n') {
if (RI) {
if (SBUF == XOFF) {
do {
RI = 0;
while (!RI);
}
while (SBUF != XON);
RI = 0;
}
}
while (!TI);
TI = 0;
SBUF = 0x0d; /* output CR */
}
if (RI) {
if (SBUF == XOFF) {
do {
RI = 0;
while (!RI);
}
while (SBUF != XON);
RI = 0;
}
}
while (!TI);
TI = 0;
return (SBUF = c);
}
#if 0 // comment out versions below
/*
* putchar (basic version): expands '\n' into CR LF
*/
char putchar (char c) {
if (c == '\n') {
while (!TI);
TI = 0;
SBUF = 0x0d; /* output CR */
}
while (!TI);
TI = 0;
return (SBUF = c);
}
/*
* putchar (mini version): outputs charcter only
*/
char putchar (char c) {
while (!TI);
TI = 0;
return (SBUF = c);
}
#endif

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ID:155507 · 发表于 2018-8-11 09:19

// 第三方跨平台cpu性能测试
// https://gitee.com/test386/test#
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
#include <serial.h>
#define INTEG_FUNC(x) fabs(sin(x))
double dclock(void);
//int main(int argc, char * argv[])
void main(void)
{
clock_t start, finish;
double duration;
long pi[400], t[420], 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;
// Loop counters and number of interior points
unsigned int ia, ja, Na;
// Stepsize, independent variable x, and accumulated sum
//double duration2;
double stepa, x_i, suma;
// Timing variables for evaluation
double interval_begin = 0.0;
// Complete integral at
double interval_end = 2.0 * 3.141592653589793238;
//double duration1;
double testa;
double testb;
int sum=0;
int num=1;
int sum2=0;
int num2=2;
double sqrt(double x);
Init_Timer0();
InitUART();
//d = (argc > 1) ? (((i = atoi(argv[1])) < 0) ? 0 : i) : 9999;
d = 99;
//q = (argc > 2) ? 1 : 0;
q = 0;
printf("%s\n\n", "pi calc..................................");
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; // Allocate 40020 bytes
//if ((pi = (long *)calloc(d + 1, sizeof(long))) == NULL) //return 2; // Allocate 40000 byts
// The C library function clock_t clock(void) returns the number of clock ticks elapsed since the program was launched. To get the number of seconds used by the CPU, you will need to divide by CLOCKS_PER_SEC.
// On a 32 bit system where CLOCKS_PER_SEC equals 1000000 this function will return the same value approximately every 72 minutes.
// This function returns the number of clock ticks elapsed since the start of the program. On failure, the function returns a value of -1.
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--;
}
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);
//_______________________________________
// Start timing for the entire application
printf(" \n");
printf(" Number of | Computed Integral | \n");
printf(" Interior Points | | \n");
for (ja=2;ja<27;ja++)
{
printf("------------------------------------- \n");
// Compute the number of (internal rectangles + 1)
Na = 1 << ja;
// Compute stepsize for Na-1 internal rectangles
stepa = (interval_end - interval_begin) / Na;
// Approx. 1/2 area in first rectangle: f(x0) * [stepa/2]
suma= INTEG_FUNC(interval_begin) * stepa / 2.0;
// Apply midpoint rule:
// Given length = f(x), compute the area of the
// rectangle of width stepa
// Suma areas of internal rectangle: f(xi + stepa) * stepa
for (ia=1;ia<Na;ia++)
{
x_i = ia * stepa;
suma += INTEG_FUNC(x_i) * stepa;
}
// Approx. 1/2 area in last rectangle: f(xN) * [stepa/2]
suma += INTEG_FUNC(interval_end) * stepa / 2.0;
printf(" %10d | %14e | \n", Na, suma);
}
printf(" \n");
//_______________________________________
while(num<=90000000){
sum=sum+num;
num=num+2;
testa=sqrt(sum);
testa=testa*num;
}
printf("Anti cheating verification code is=:%d\n",sum);
printf("Floating-point precision %.38lf\n", testa);
while(num2<=90000000){
sum2=sum2+num2;
num2=num2+2;
testb=sqrt(sum2);
testb=testb*num2;
}
finish = clock();
printf("Anti cheating verification code is=%d\n",sum2);
printf("Floating-point precision %.38lf\n", testb);
//On a 32 bit system where CLOCKS_PER_SEC equals 1000000 this function will return the same value approximately every 72 minutes.
duration = (double)(finish - start) / CLOCKS_PER_SEC;
printf("Total time %f seconds \n", duration);
printf("The less the time it is used, the more the performance becomes strongert!\n");
printf("Press Enter to exit\n");
sum = getchar( );
//return 0;
}
/*
Rebuild target 'Target 1'
compiling Test386-51.c...
Test386-51.c(59): error C174: return-expression on void-function
Test386-51.c(159): error C241: 'main': auto segment too large
Target not created
Description
The required space for local objects exceeds the model dependent maximum. The maximum segment sizes are defined as follows: SMALL 128 bytes COMPACT 256 bytes LARGE 65535 bytes
Rebuild target 'Target 1'
compiling Test386-51.c...
linking...
Program Size: data=23.2 xdata=3451 code=8924
"Test386-51" - 0 Error(s), 0 Warning(s).
Rebuild target 'Target 1'
compiling Test386-51.c...
linking...
Program Size: data=23.2 xdata=3451 code=8980
creating hex file from "Test386-51"...
"Test386-51" - 0 Error(s), 0 Warning(s).
*/

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ID:384624 · 发表于 2018-8-13 09:37

感谢高人指点，让我怂回龙芯那群粉。

ID:384624 · 发表于 2018-8-13 10:02

希望高人把程序运行一下，截个图！

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一个51单片机的性能测试源码

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