Question

我正在运行启动到终端的linux（没有gui）。我有一个ZyBo电路板，它有一个ArmV7处理器。我写了一个C程序来输出PMOD上的时钟和相应的数据序列。 PMOD的切换速度高达50MHz。但是，我的程序创建的时钟只有115 Hz的最大频率。我需要这个程序尽可能快地输出，因为PMOD I使用的能力是50MHz。我使用以下代码行编译了我的程序： gcc -ofast (c_program)。

#include <stdio.h>
#include <stdlib.h>

#define ARRAYSIZE 511
//________________________________________

//macro for the SIGNAL PMOD
//________________________________________
//DATA 
//ZYBO Use Pin JE1
#define INIT_SIGNAL system("echo 54 > /sys/class/gpio/export"); system("echo out > /sys/class/gpio/gpio54/direction");
#define SIGNAL_ON system("echo 1 > /sys/class/gpio/gpio54/value");
#define SIGNAL_OFF system("echo 0 > /sys/class/gpio/gpio54/value");
//________________________________________

//macro for the "CLOCK" PMOD
//________________________________________
//CLOCK 
//ZYBO Use Pin JE4
#define INIT_MYCLOCK system("echo 57 > /sys/class/gpio/export"); system("echo out > /sys/class/gpio/gpio57/direction");
#define MYCLOCK_ON system("echo 1 > /sys/class/gpio/gpio57/value");
#define MYCLOCK_OFF system("echo 0 > /sys/class/gpio/gpio57/value");

int main(void){

int myarray[ARRAYSIZE] = {//hard coded array for signal data
    1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,0,0,0,0,1,0,0,1,1,1,0,0,1,1,1,0,1,1,1,1,0,0,1,0,0,0,1,0,1,0,0,1,1,1,0,0,1,0,1,0,1,0,0,1,0,1,1,0,1,0,1,1,0,0,1,1,1,1,0,0,1,0,1,0,0,1,1,1,1,1,1,0,0,1,0,0,1,1,0,1,0,0,0,0,1,0,0,0,1,1,0,0,1,0,1,1,1,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,1,0,1,1,1,1,0,1,1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1,0,1,1,0,0,0,0,0,0,0,0,1,0,1,1,0,1,1,1,1,1,0,0,1,1,1,0,0,1,1,0,1,1,0,1,1,1,0,0,1,1,1,1,1,0,0,1,0,1,0,1,0,1,1,0,1,0,0,0,1,1,1,0,1,0,1,0,1,0,1,0,1,0,1,0,0,1,0,0,0,0,1,1,1,0,1,1,1,1,0,1,1,0,1,0,1,0,1,0,0,1,0,1,1,1,0,1,1,1,0,0,1,1,1,0,1,0,0,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,1,0,0,1,0,1,1,0,1,0,1,1,1,0,0,0,0,0,1,0,0,0,1,0,1,1,1,1,1,1,1,0,0,0,0,0,1,1,0,1,1,1,1,1,1,1,1,0,1,1,0,1,0,0,0,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,0,1,1,1,0,1,0,1,1,1,0,0,0,1,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,0,1,0,0,1,0,1,1,1,1,1,1,0,1,1,0,1,0,1,1,1,1,1,1,0,0,1,1,0,1,1,0,0,1,1,0,1,1,0,1,0,1,0,1,0,1,0,0,1,1,1,0,1,1,0,0,0,0,1,1,0,1,1,0,1,1,1,1,1,1,1,0,1,0,1,1,0,0,0,0,0,0,0,0,0,0,0
    };

INIT_SIGNAL
INIT_MYCLOCK;
//infinite loop
int i;
do{
    i = 0;
    do{
        /*
        1020 is chosen because it is twice the size needed allowing for the changes in the clock.
        (511= 0-510, 510*2= 1020 ==> 0-1020 needed, so 1021 it is)
        */
        if((i%2)==0)
        {
            MYCLOCK_ON;
            if(myarray[i/2] == 1){
                SIGNAL_ON;
            }else{
                SIGNAL_OFF;
            }
        }
        else if((i%2)==1)
        {
            MYCLOCK_OFF;
            //dont need to change the signal since it will just stay at whatever it was.
        }
        ++i;
    } while(i < 1021);
} while(1);
return 0;
}

如何使我的可执行程序输出至少达到MegaHertz的大小？

Answer 1

您正试图通过使用system函数（即创建进程）来调用时钟来调用另一个可执行文件（/bin/echo），目的是将值写入{{{ 1}}文件系统。难怪你无法实现超过数百赫兹的速度。

至少应该写入/sys文件系统中的伪文件，直接打开它们并在程序中写入，而不是通过/sys。这可能无法让您进入MHz范围，但它肯定会比您所拥有的速度快数百倍。

Answer 2

使用FILE * fopen ( const char * filename, const char * mode );直接打开文件，然后使用int fprintf ( FILE * stream, const char * format, ... );

打印字符串

http://www.cplusplus.com/reference/cstdio/fopen/

http://www.cplusplus.com/reference/cstdio/fprintf/

如果您需要更高的速度，请查看open（）系统调用和mmap（）

加速可执行程序Linux - arm处理器。比特切换

2 个答案: