void
coroutine_yield(struct schedule * S) {
int id = S->running;
assert(id >= 0);
struct coroutine * C = S->co[id];
assert((char *)&C > S->stack);
_save_stack(C, S->stack + STACK_SIZE); // 用于保存堆栈的信息
C->status = COROUTINE_SUSPEND; // 状态变成了挂起
S->running = -1;
swapcontext(&C->ctx, &S->main);
}
static void
_run(struct coroutine *co) {
_set_running_coroutine(co);
_resume(co);
_set_running_coroutine(NULL);
if (co->status == STATUS_EXITING) {
coroutine_delete(co);
return;
}
_save_stack(co);
}
void
coroutine_yield(struct schedule * sched) {
int id = sched->running;
assert(id >= 0);
struct coroutine * C = sched->co[id];
assert((char *)&C > sched->stack);
_save_stack(C,sched->stack + STACK_SIZE);
C->status = COROUTINE_SUSPEND;
sched->running = -1;
swapcontext(&C->ctx , &sched->main);
// 这里resume/delete的时候一定要把原来的函数context执行完，不然函数里声明类的析构函数不会被执行！
}
void
coroutine_yield(struct schedule * S) {
printf("%s:%s\n", __FILE__, __FUNCTION__);
int id = S->running;
//这个assert(id>=0) 即暗示了此函数只会在coroutine中调用。
assert(id >= 0);
//取出正在运行的croutine
struct coroutine * C = S->co[id];
//这个assert()是何意？ 这里的S->stack为何是栈上的地址？不是从堆上分配的吗
//上面的疑问解答：因为swapcontext()时，指定了ss_sp这S->stack,这样当context执行时，它的栈是放在S->stack上的。
//所以C, S->stack 它们的地址都是在相近地
printf("&C:%p S->stack:%p\n", &C, S->stack);
assert((char *)&C > S->stack);
//保存这个coroutine的stack
_save_stack(C,S->stack + STACK_SIZE);
//使进入SUSPEND状态
C->status = COROUTINE_SUSPEND;
S->running = -1;
//切换回S->main, 就是当初从coroutine_resume()中过来的那里
swapcontext(&C->ctx , &S->main);
}
