How can I pass a function using default parameter(s) to std:: thread?

Question

So I have gcc version 4.8.1, g++ version 4.6.4, using flags: -std=c++0x and -pthread. I simplified my problem down to the code shown and still get the original error. What I have below compiles, but when I uncomment the two lines for thread "two", I get the error message shown below the code ``` #include #include using namespace std; void print_int(int x=7); void print_A(){ cout << "A "; } int main(){ thread one (print_int,17); //thread two (print_int); thread three (print_A); one.join(); //two.join(); three.join(); return 0; } void print_int(int x){ cout << x << ' '; } ``` I tried to parse through the error messages but I still have no clue what's going on... ``` In file included from /usr/include/c++/4.6/thread:39:0, from def_params.cpp:2: /usr/include/c++/4.6/functional: In member function ‘void std::_Bind_result<_Result, _Functor(_Bound_args ...)>::__call(std::tuple<_Args ...>&&, std::_Index_tuple<_Indexes ...>, typename std::_Bind_result<_Result, _Functor(_B…

Accepted Answer

```
thread two (print_int);
```

You can't do this, default arguments in C++ are only used at the call site (e.g. when you call `print_int()` yourself) as a syntactic shortcut, but aren't used when e.g. `std::thread` invokes the same function passed to it as an argument. The problem is `std::thread` gets given a pointer to a one-argument function, but is not passed an argument to go with it, so it doesn't know how to call it, and there's no way to look up a function's default argument given a pointer to the function.

The simplest solutin is to use a lambda to call the function, as that call can use the default argument:

```
thread two( []{ print_int(); } );
```

How can I pass a function using default parameter(s) to std:: thread?

Solution

Problem