pc上用C语言模拟51多任务的案例程序

#include <stdio.h>

#include <stdlib.h>

//任务槽个数.在本例中并未考虑任务换入换出,所以实际运行的任务有多少个,

//就定义多少个任务槽,不可多定义或少定义

#define MAX_TASKS 5

//任务的栈指针

unsigned char  *task_sp[MAX_TASKS];

//最大栈深.最低不得少于2个,保守值为12.

//预估方法:以2为基数,每增加一层函数调用,加2字节.

//如果其间可能发生中断,则还要再加上中断需要的栈深.

//减小栈深的方法:1.尽量少嵌套子程序 2.调子程序前关中断.

#define MAX_TASK_DEP 12

unsigned char  task_stack[MAX_TASKS][MAX_TASK_DEP] =

{

    0,1,2,3,4,5,6,7,8,9,'a',9,8,7,6,5,4,3,2,1,0,11,12,13,113,31,4

};//任务堆栈.

//任务装入函数.将指定的函数(参数1)装入指定(参数2)的任务槽中.

//如果该槽中原来就有任务,则原任务丢失,但系统本身不会发生错误.

//将各任务的函数地址的低字节和高字节分别入在

//task_stack[任务号][0]和task_stack[任务号][1]中

void task_load(unsigned int fn, unsigned char tid)

{

	task_sp[tid] = task_stack[tid] + 1;

	task_stack[tid][0] = (unsigned int)fn & 0xff;

	task_stack[tid][1] = (unsigned int)fn >> 8;

}

void func1()

{

	static unsigned char i;

	i = 0;

	while(1)

	{

		if(i<250)

		{

			i++;

		}

		if(i>=250)

		{

			printf("task1\n");

			i = 0;

		}

		//task_switch();

	}

}

void func2()

{

	static unsigned int j;

	j = 0;

	while(1)

	{

		if(j<654)

		{

			j++;

		}

		if(j>=654)

		{

			printf("task2\n");

			j = 0;

		}

		//task_switch();

	}

}

int main()

{

    printf("Hello world!\n");

    printf("task_stack[0] = %p\n",task_stack[0]);

    printf("task_stack+0 = %p\n",task_stack+0);

    printf("task_stack+1 = %p\n",task_stack+1);

    printf("task_stack+2 = %p\n",task_stack+2);

    printf("task_stack+3 = %p\n",task_stack+3);

    printf("task_stack+4 = %p\n",task_stack+4);

    printf("task_stack[0] = %p\n",task_stack[0]);

    printf("*(task_stack+0) = %p\n",*(task_stack+0));

    printf("*(task_stack+1) = %p\n",*(task_stack+1));

    printf("*(task_stack+2) = %p\n",*(task_stack+2));

    printf("*(task_stack+3) = %p\n",*(task_stack+3));

    printf("*(task_stack+4) = %p\n",*(task_stack+4));

    printf("task_stack[0] = %p\n",task_stack[0]);

    printf("*(task_stack+0)+0 = %p\n",*(task_stack+0)+0);

    printf("*(task_stack+1)+1 = %p\n",*(task_stack+1)+1);

    printf("*(task_stack+2)+2 = %p\n",*(task_stack+2)+2);

    printf("*(task_stack+3)+3 = %p\n",*(task_stack+3)+3);

    printf("*(task_stack+4)+4 = %p\n",*(task_stack+4)+4);

    task_sp[0] = (task_stack[0]+1);

    task_sp[1] = (task_stack[0]+2);

    task_sp[2] = (task_stack[0]+3);

    task_sp[3] = (task_stack[0]+4);

    task_sp[4] = (task_stack[0]+5);

    //task_sp[0] = *(task_stack+0)+9;

    printf("task_stack = %p\n",task_stack);

    printf("task_stack[0] + 5 = %p\n",(task_stack[0] + 5));

    printf("task_sp = %p\n",task_sp);

    printf("\ntask_sp[0] = %d\n",task_sp[0]);

    printf("task_sp[1] = %d\n",task_sp[1]);

    printf("task_sp[2] = %d\n",task_sp[2]);

    printf("task_sp[3] = %d\n",task_sp[3]);

    printf("task_sp[4] = %d\n",task_sp[4]);

    //task_load(func1, 0);//将func1函数装入0号槽

	//task_load(func2, 1);//将func2函数装入1号槽

    return 0;

}

Hello world!

task_stack[0] = 00403000

task_stack+0 = 00403000

task_stack+1 = 0040300C

task_stack+2 = 00403018

task_stack+3 = 00403024

task_stack+4 = 00403030

task_stack[0] = 00403000

*(task_stack+0) = 00403000

*(task_stack+1) = 0040300C

*(task_stack+2) = 00403018

*(task_stack+3) = 00403024

*(task_stack+4) = 00403030

task_stack[0] = 00403000

*(task_stack+0)+0 = 00403000

*(task_stack+1)+1 = 0040300D

*(task_stack+2)+2 = 0040301A

*(task_stack+3)+3 = 00403027

*(task_stack+4)+4 = 00403034

task_stack = 00403000

task_stack[0] + 5 = 00403005

task_sp = 004050B0

task_sp[0] = 4206593

task_sp[1] = 4206594

task_sp[2] = 4206595

task_sp[3] = 4206596

task_sp[4] = 4206597

Terminated with return code 0

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巴特西

pc上用C语言模拟51多任务的案例程序

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