Difference Between Pointer(*) And Reference(&)

Question

I use Reference(&) here,but the a didn't change.Why? While i use Pointer(*) the value have changed.

int all=0; 
void call_from_thread(int & a)  
{
   a = 5;
}

int main() 
{
    thread s = thread(call_from_thread, all);
    s.join();
    return 0;
}

Another program,in this case, i also use Reference(&),but the value have changed. Why the value a in thread didn't change?

void Func3(int &x)
{
    x = x + 10;
}

int main() {

    int n = 0;
    Func3(n);
    cout << “n = ” << n << endl; // n = 10
}

en.cppreference.com/w/cpp/thread/thread/thread 3) - Yeah, it's annoying. — Karoly Horvath
– Karoly Horvath, Commented Oct 29, 2015 at 13:17

Ben Voigt · Accepted Answer · 2015-10-29 13:21:47Z

7

The std::thread constructor makes copies of its parameters during the thread creation step.... ordinary references don't survive the trip, and the call_from_thread function receives a reference to the copy.

Details can be found in the Standard. The thread::thread constructor behavior is described as

Constructs an object of type thread. The new thread of execution executes INVOKE(DECAY_COPY(std::forward<F>(f)), DECAY_COPY(std::forward<Args>(args))...) with the calls to DECAY_COPY being evaluated in the constructing thread.

The exact definition of DECAY_COPY is quite complex, but as the name suggests, it ignores references and makes a value copy.

A simple workaround is to use std::ref.

thread s = thread(call_from_thread, std::ref(all));

This keeps a pointer through the copies, and only resolves to the actual target when the reference parameter is bound.

edited Oct 29, 2015 at 13:21

answered Oct 29, 2015 at 13:15

Ben Voigt

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