How to ensure wait_for is called before notify_one - c++

Question

Typical usages of condition variables look like this (see code below): http://en.cppreference.com/w/cpp/thread/condition_variable.
However, it seems that the main thread could potentially call notify_one before the worker thread calls wait, which would result in a deadlock. Am I mistaken? If not, what is the usual workaround for this?
#include <iostream>
#include <string>
#include <thread>
#include <mutex>
#include <condition_variable>
std::mutex m;
std::condition_variable cv;
std::string data;
bool ready = false;
bool processed = false;
void worker_thread()
{
// Wait until main() sends data
std::unique_lock<std::mutex> lk(m);
cv.wait(lk, []{return ready;});
// after the wait, we own the lock.
std::cout << "Worker thread is processing data\n";
data += " after processing";
// Send data back to main()
processed = true;
std::cout << "Worker thread signals data processing completed\n";
// Manual unlocking is done before notifying, to avoid waking up
// the waiting thread only to block again (see notify_one for details)
lk.unlock();
cv.notify_one();
}
int main()
{
std::thread worker(worker_thread);
data = "Example data";
// send data to the worker thread
{
std::lock_guard<std::mutex> lk(m);
ready = true;
std::cout << "main() signals data ready for processing\n";
}
cv.notify_one();
// wait for the worker
{
std::unique_lock<std::mutex> lk(m);
cv.wait(lk, []{return processed;});
}
std::cout << "Back in main(), data = " << data << '\n';
worker.join();
}

Answer 1

Note the definition of wait that uses a condition (the only wait you should ever be using):
while (!pred()) {
wait(lock);
}
if notify was already fired it means the condition is already true (that was sequenced before notify_one in the signalling thread). So when the receiver takes the mutex and looks at pred(), it will be true and it will proceed.