Lifetime of std::thread arguments

Question

When parameters are passed to std::thread(), does the thread from which the new thread is being spawned wait until all the parameters are completely copied into new thread local storage?

Simple example:

void f()
{  
  int array[10];
  ........ //done something with array  
  std::thread th(someF, array); //assuming that someF accepts int[]  
  th.detach();  
}

Should I automatically assume that all the data is safely copied before f() has ended? One of the scenarios I see, assuming f() doesn't wait, and plows full steam ahead, is that th is attempting to copy array that is being in destroyed.

Answer 1

Yes. If it fails to copy it will throw in the constructing thread.

§30.3.1.2 thread constructors

template explicit thread(F&& f, Args&&... args);

Requires: F and each Ti in Args shall satisfy the MoveConstructible requirements. INVOKE (DECAY_- COPY ( std::forward(f)), DECAY_COPY (std::forward(args))...) (20.8.2) shall be a valid expression. 4

Effects: Constructs an object of type thread. The new thread of execution executes INVOKE (DECAY_- COPY ( std::forward(f)), DECAY_COPY (std::forward(args))...) with the calls to DECAY_COPY being evaluated in the constructing thread. Any return value from this invocation is ignored. [ Note: This implies that any exceptions not thrown from the invocation of the copy of f will be thrown in the constructing thread, not the new thread. —end note ] If the invocation of INVOKE (DECAY_COPY ( std::forward(f)), DECAY_COPY (std::forward(args))...) terminates with an uncaught exception, std::terminate shall be called.

Synchronization: The completion of the invocation of the constructor synchronizes with the beginning of the invocation of the copy of f.

Answer 2

Yes, if the arguments cannot be copied, to storage available to the new thread, it will throw an exception.

From http://en.cppreference.com/w/cpp/thread/thread/thread

Any exceptions thrown during evaluation and copying/moving of the arguments are thrown in the current thread, not the new thread.