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Is unique_ptr thread safe? Is it impossible for the code below to print same number twice?

#include <memory>
#include <string>
#include <thread>
#include <cstdio>

using namespace std;

int main()
{
    unique_ptr<int> work;

    thread t1([&] {
        while (true) {
            const unique_ptr<int> localWork = move(work);
            if (localWork)
                printf("thread1: %d
", *localWork);
            this_thread::yield();
        }
    });

    thread t2([&] {
        while (true) {
            const unique_ptr<int> localWork = move(work);
            if (localWork)
                printf("thread2: %d
", *localWork);
            this_thread::yield();
        }
    });

    for (int i = 0; ; i++) {
        work.reset(new int(i));

        while (work)
            this_thread::yield();
    }

    return 0;
}
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1 Answer

unique_ptr is thread safe when used correctly. You broke the unwritten rule: Thou shalt never pass unique_ptr between threads by reference.

The philosophy behind unique_ptr is that it has a single (unique) owner at all times. Because of that, you can always pass it safely between threads without synchronization -- but you have to pass it by value, not by reference. Once you create aliases to a unique_ptr, you lose the uniqueness property and all bets are off. Unfortunately C++ can't guarantee uniqueness, so you are left with a convention that you have to follow religiously. Don't create aliases to a unique_ptr!


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
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