30天自制操作系统之第15天 多任务（1）

farjmp(0, 4 * 8);

这是一个far跳转，先看一下farjmp的实现：

_farjmp:	; void farjmp(int eip, int cs)
		JMP		FAR [ESP+4] ; eip, cs
		RET

set_segmdesc(gdt + 4, 103, (int) &tss_b, AR_TSS32);

void task_b_main(struct SHEET *sht_back)
{
	//for (;;) { io_hlt(); }
	struct FIFO32 fifo;
	struct TIMER *timer_ts, *timer_put;
	int i, fifobuf[128], count = 0;
	char s[11];

	fifo32_init(&fifo, 128, fifobuf);
	timer_ts = timer_alloc();
	timer_init(timer_ts, &fifo, 2);
	timer_settime(timer_ts, 2);
	timer_put = timer_alloc();
	timer_init(timer_put, &fifo, 1);
	timer_settime(timer_put, 1);
	
	for (;;) {
		count++;
		io_cli();
		if (fifo32_status(&fifo) == 0) {
			//io_stihlt();
			io_sti();
		} else {
			i = fifo32_get(&fifo);
			io_sti();
			if (i == 1) { // each 0.01s refreshing the screen, not each count++, enhancing the performance
				sprintf(s, "%11d", count);
				putfonts8_asc_sht(sht_back, 0, 144, COL8_FFFFFF, COL8_0000FF, s, 11);
				timer_settime(timer_put, 1);
			} else if (i == 2) {
				farjmp(0, 3 * 8);
				timer_settime(timer_ts, 2);
			}
		}
	}
}

30天自制操作系统之第15天 多任务（1）,布布扣,bubuko.com

30天自制操作系统之第15天 多任务（1）

原文：http://blog.csdn.net/bluecloudmatrix/article/details/38585053