Home C++ Thread taking reference argument failed compile

# C++ Thread taking reference argument failed compile

Rich
1#
Rich Published in 2016-03-31 19:10:11Z
 #include #include using namespace std; void f1(double& ret) { ret=5.; } int main() { double ret=0.; thread t1(f1, ret); t1.join(); cout << "ret=" << ret << endl; }  The above code fails compilation with the following error message: g++ -std=c++14 -O2 -Wall -pedantic -pthread main.cpp && ./a.out In file included from /usr/local/include/c++/5.3.0/thread:39:0, from main.cpp:2: /usr/local/include/c++/5.3.0/functional: In instantiation of 'struct std::_Bind_simple': /usr/local/include/c++/5.3.0/thread:137:59: required from 'std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (&)(double&); _Args = {double&}]' main.cpp:11:21: required from here /usr/local/include/c++/5.3.0/functional:1505:61: error: no type named 'type' in 'class std::result_of' typedef typename result_of<_Callable(_Args...)>::type result_type; ^ /usr/local/include/c++/5.3.0/functional:1526:9: error: no type named 'type' in 'class std::result_of' _M_invoke(_Index_tuple<_Indices...>) ^  I understand that I can use std::ref() to pass the argument. But if I pass by value, why is it an error since thread should just copy the argument by value and pass some object stored inside thread to bind with the reference argument of function f1. I feel that if I can understand what this result_of is doing and why it is giving error, I can better understand the reason. So could anyone walk me through the error msg? Especially the meanings of std::_Bind_simple and std::result_of. EDIT: I know if I pass a value, the thread will only work on the copy and has no effect after the thread returns. That is not my concern. I want to know why it is giving error now, but it was not giving error to other posts on SO like the following:Difference between pointer and reference as thread parameter
Jonathan Wakely
2#
Jonathan Wakely Reply to 2016-03-31 19:49:19Z
 I know if I pass a value, the thread will only work on the copy and has no effect after the thread returns. No, that's not correct. The code should not silently make a copy and work on the copy, the standard says it must not even compile. The standard requires that the arguments to the called function are copied (into storage managed by the C++ runtime) and then the copies are forwarded as rvalues. So in your example f1 gets passed an rvalue of type double and the parameter of type double& cannot bind to that rvalue. The reason the standard requires that is so there is no silent copying and loss of data: if the function requires a modifiable reference then it won't compile unless you pass a reference explicitly using a reference_wrapper. The compiler error you get involves result_of because that's how I made GCC's std::thread check if the function can be called with the supplied arguments. I use result_of to check if the function pointer &f1 (which is of type void(*)(double&)) can be called with an rvalue of type double. It can't be called with an argument of that type, so the nested type result_of::type is not defined, so the compiler says: error: no type named 'type' in 'class std::result_of'  The error is a bit confusing because the C++ declarator rules mean that decltype(&f1)(double) gets displayed as void(*(double))(double&). That is not my concern. I want to know why it is giving error now, but it was not giving error to other posts on SO Those posts were using an old pre-C++11 or non-conforming compiler which didn't meet the requirements of the C++11 standard and incorrectly compiled the code.
Richard Hodges
3#
Richard Hodges Reply to 2016-03-31 20:19:49Z
 Jonathan's answer is definitive. Time spent studying it would be time well spent. In the meantime, modifying the code thus will do what you want - namely send a reference into the thread function: thread t1(f1, std::ref(ret)); 
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