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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