Creating an instance of shared_ptr<std::thread> with make_shared<std::thread>

Question

Consider the following code:

class A
{
    ....
    shared_ptr<std::thread> mThread;
    void Step();
    void LaunchTrhead();
}

void A::LaunchThread()
{
    ...
    mThread=make_shared<std::thread>(Step); // This line gives an error
    ...
}

void A::Step()
{
    ...
}

I'm trying to initialise the shared pointer mThread so that it calls the function Step. However, the compiler gives me the error "invalid initialization of reference of type ... from expression of type 'unresolved overloaded function type'". Obviously I'm doing something stupid, but I can't put my finger on it. Can anybody help? Thanks in advance!

Praetorian · Accepted Answer · 2014-02-14 16:25:34Z

8

Step() is a non-static member function, so it has an implicit first parameter of type A*. You need to bind the current instance of A when invoking it.

mThread = std::make_shared<std::thread>(std::bind(&A::Step, this));

You can also use a lambda instead of bind

mThread = std::make_shared<std::thread>([this]{ Step(); });

As @Casey points out in the comments, std::thread's constructor has special treatment for pointer to member functions, and will assume the first following argument is a pointer or reference to the instance on which to call the member function. This means you can avoid bind and directly pass this as the second argument.

mThread = std::make_shared<std::thread>(&A::Step, this);

edited Feb 14, 2014 at 16:25

answered Feb 14, 2014 at 14:05

Praetorian

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5 Comments

DrD Over a year ago

On the spot! The compiler was telling me ("unresolved function"), but I just couldn't see it, I was completely blinded (I should go home...). Also, in line with what you pointed out, I could declare Test static. But your solution is more practical in my case, because it avoids having to pass a pointer to the owner object. Thank you!

Casey Over a year ago

You can pass a pointer-to-member and pointer-to-object directly to the std::thread constructor (and std::bind, and std::async): mThread = std::make_shared<std::thread>(&A::Step, this);

DrD Over a year ago

Thanks a lot Casey, that's an even more elegant solution :)

Vatsal Aggarwal Over a year ago

@Casey How do we pass functional with 1 or more arguments? Suppose the function step had a "function pointer" as an argument then what would be the syntax?

Praetorian Over a year ago

@VatsalAggarwal Pass the additional arguments to thread's constructor (or make_shared in the example above)

Mike Seymour · Accepted Answer · 2014-02-14 14:24:19Z

1

Try (use labda instead of the free function):

mThread=make_shared<std::thread>([this](){ Step(); });

The way it is, you're not passing a reference to this to the constructor inspite of it being a member function.

This solution uses a lambda to create a function-object that takes no parameters but has a reference to this.

If you want to use the global function do this instead, and move void Step() to before it's usage:

mThread=make_shared<std::thread>(::Step());

the :: removes the ambiguity over the scope of the function.

edited Feb 14, 2014 at 14:24

Mike Seymour

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answered Feb 14, 2014 at 13:59

nishantjr

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2 Comments

DrD Over a year ago

::Step() doesn;t work, sorry. The lambda solution does. Thanks!

DrD Over a year ago

Step() is a member function, only that I forgot to include A:: in its declaration. Now it's edited. My apologies.

Rory McCrossan · Accepted Answer · 2015-05-27 08:46:34Z

0

You should use shared_from_this() replace this

edited May 27, 2015 at 8:46

Rory McCrossan

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answered May 27, 2015 at 8:25

Smith-WangBing

1

1 Comment

rsilva4 Over a year ago

Could you provide a more complete example of your suggestion?

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