Question

I read this Wikipedia text slice:

Because a cooperatively multitasked system relies on each process regularly giving up time to other processes on the system, one poorly designed program can consume all of the CPU time for itself or cause the whole system to hang.

Out of curiosity, how does one give up that time? Is this some sort of OS call? Let’s think about non-preemptive cases like fibers or evented IO that do cooperative multitasking. How do they give up that time?

Take this NodeJS example:

var fs = require('fs');
fs.readFile('/path/to/file', function(err, data) {});

It is obvious to me that the process does nothing while it’s waiting for the data, but how does V8 in this case give up time for other processes?

Let’s assume Linux/Windows as our OS.

Edit: I found out how Google is doing this with their V8.

On Windows they basically sleep zero time:

void Thread::YieldCPU() {
  Sleep(0);
}

And on Linux they make an OS call:

void Thread::YieldCPU() {
  sched_yield();
}

of sched.h.

Answer 1

Yes, every program participates in the scheduling decisions of the OS, so you have to call a particular syscall that tells the kernel to take back over. Often this was called yield(). If you imagine how difficult it is to guarantee that a paticular line of code is called at regular, short intervals, or even at all, you get an idea of why cooperative multitasking is a suboptimal solution.

In your example, it is the javascript engine itself is interrupted by the OS scheduler, if it’s a preemptive OS. If it’s a cooperative one, then no, the engine gets no work done, and neither does any other process. As a result, such systems are usually not suitable for real-time (or even serious) workloads.