std::vector emplace_back 和 push_back 是线程安全的吗

Question

If I have a function that only do emplace_back or push_back on a global vector with unique_ptr as a values, will it be thread-safe or I have to use mutexes? Is mutex the only way to make it thread safe?

void (T param)
{
globalVector.emplace_back(std::make_unique<T>(param));
//or globalVector.push_back (std::make_unique<T>(param));
}

And if it only be vector of T?

void (T param)
{
globalVector.emplace_back(param);
//or globalVector.push_back (param);
}

Answer 1

will it be thread-safe

No

or I have to use mutexes?

Yes

See cppreference for thread-safety of standard containers.

Answer 2

Consider this toy example that illustrates the critical part that eg a vector<int>::push_back might do:

struct broken_toy_vector {
    size_t size;
    size_t capacity;
    int* data;
    void push_back(int x){
        if (size+1 > capacity) {
            int* temp = new int[size+1];
            // copy from data to temp
            temp[size] = x;
            size += 1;
            delete data;
            data = temp;
        } else {
            throw "not implemented";
        }
    }
};
             

A std::vector is surely not a broken_toy_vector , but when the new size exceeds current capacity then a vector needs to reallocate. There is no good reason for std::vector to make push_back thread-safe, because that would incur overhead in any single-threaded usage. Further, also with more than one thread, often you don't want synchronisation on the lowest level. Consider:

 for (int i=0; i<1000; ++i) {
      vect.push_back(i);
 }

Locking a mutex in every iteration would be extremely wasteful.

TL;DR

No. It is not safe to call std::vector::push_back concurrently from different threads. A container that protects any acces to its data, would be of limited use, because the user has no control over granularity of the locking.