co_yield 实现的一点小问题

[liuly0322]

NJUOSLab-M2-libco/libco/co.c

Lines 137 to 145 in cb3c3d4

	if (co_next->status == CO_NEW)
	{
	assert(co_next->status == CO_NEW);
	((struct co volatile *)current)->status = CO_RUNNING; // fogot!!!
	stack_switch_call(&current->stack[STACK_SIZE], current->func, (uintptr_t)current->arg);
	((struct co volatile *)current)->status = CO_DEAD;
	if (current->waiter)
	current = current->waiter;
	}

这里在 stack_switch_call 返回后应该重新调度协程，而不是简单的返回到调用 yield 处。

对于现在的实现，若 test2 改为:

struct co *thd3 = co_start("consumer-1", consumer, queue);
struct co *thd4 = co_start("consumer-2", consumer, queue);
struct co *thd1 = co_start("producer-1", producer, queue);
struct co *thd2 = co_start("producer-2", producer, queue);

就会导致 main 函数的执行环境，即 test2() 函数调用无法 co_wait(thd1);，测试会陷入死循环。

一个（可能）可行的修改：

void co_schedule() {
    // switch to co_next
    struct co *co_next = current;
    do {
        co_next = co_next->next;
    } while (co_next->status == CO_DEAD || co_next->status == CO_WAITING);
    current = co_next;

    if (current->status == CO_NEW) {
        ((struct co volatile *)current)->status = CO_RUNNING;
        stack_switch_call(&current->stack[STACK_SIZE], current->func,
                          (uintptr_t)current->arg);
        ((struct co volatile *)current)->status = CO_DEAD;
        if (current->waiter) {
            current = current->waiter;
            longjmp(current->context, 1);
        } else {
            co_schedule();
        }
    } else if (current->status == CO_RUNNING) {
        longjmp(current->context, 1);
    }
}

void co_yield () {
    if (current == NULL)
        co_current_init();

    if (setjmp(current->context) == 0)
        co_schedule();
    else
        return;
}