Why is `std::this_thread::yield()` 10x slower than `std::this_thread::sleep_for(0s)`?

Question

Just testing the two small programs,

#include <thread>

int main()
{
    for (int i = 0; i < 10000000; i++)
    {
        std::this_thread::yield();
    }

    return 0;
}

and:

#include <thread>
#include <chrono>

int main()
{
    using namespace std::literals;

    for (int i = 0; i < 10000000; i++)
    {
        std::this_thread::sleep_for(0s);
    }

    return 0;
}

I get the respective timings on my system (Ubuntu 22.04 LTS, kernel version 5.19.0-43-generic),

./a.out  0,33s user 1,36s system 99% cpu 1,687 total

and:

./a.out  0,14s user 0,00s system 99% cpu 0,148 total

Why is std::this_thread::yield() 10x slower than std::this_thread::sleep_for(0s) ?

N.B. Timing is similar between g++ and clang++.

edit: As pointed out in the answer this is an optimization of the STL implementation, calling sleep(0) is in fact 300x slower (50us vs 150ns).

Answer 1

Taking a quick look at the source for this_thread::sleep_for

template<typename _Rep, typename _Period>
inline void
sleep_for(const chrono::duration<_Rep, _Period>& __rtime)
{
    if (__rtime <= __rtime.zero())
      return;
    ...

So sleep_for(0s) does nothing, in fact your test program uses 0.0s of system time, basically an empty loop that runs entirely in user space (in fact I suspect that if you compile with optimizations, it will be completely removed)

On the other hand, yield calls^* sched_yield which in turns will call schedule() in kernel space, thus at least executing some logic to check if there is another thread to schedule.

I believe that your 0.33s of user space time is basically syscall overhead.

^{* _{Actually jumps to __libcpp_thread_yield which then calls sched_yield, at least on linux}}