我有一个简单的模板函数do_something,它返回一个整数：123.
template<typename T>
auto do_something(T input) {
std::this_thread::sleep_for(std::chrono::seconds(1));
return 123;
}
int main(int argc,char *argv[]) {
std::function<int(void)> function = std::bind(do_something<int>,12);
function();
return 0;
}
使用GCC 6.1.1,我收到此错误：
test.cpp: In function ‘int main(int,char**)’:
test.cpp:16:70: error: no matching function for call to ‘bind(<unresolved overloaded function type>,int)’
std::function<int(void)> function = std::bind(do_something<int>,12);
^
In file included from /usr/include/c++/6.1.1/thread:39:0,from test.cpp:4:
/usr/include/c++/6.1.1/functional:1331:5: note: candidate: template<class _Func,class ... _BoundArgs> typename std::_Bind_helper<std::__is_socketlike<_Func>::value,_Func,_BoundArgs ...>::type std::bind(_Func&&,_BoundArgs&& ...)
bind(_Func&& __f,_BoundArgs&&... __args)
^~~~
/usr/include/c++/6.1.1/functional:1331:5: note: template argument deduction/substitution Failed:
test.cpp:16:70: note: couldn't deduce template parameter ‘_Func’
std::function<int(void)> function = std::bind(do_something<int>,from test.cpp:4:
/usr/include/c++/6.1.1/functional:1359:5: note: candidate: template<class _Result,class _Func,class ... _BoundArgs> typename std::_Bindres_helper<_Result,_BoundArgs>::type std::bind(_Func&&,_BoundArgs&&... __args)
^~~~
/usr/include/c++/6.1.1/functional:1359:5: note: template argument deduction/substitution Failed:
test.cpp:16:70: note: couldn't deduce template parameter ‘_Result’
std::function<int(void)> function = std::bind(do_something<int>,12);
如您所见,编译器无法推断出函数的结果类型.
注意：clang 3.8.0可以编译,没有任何错误.
所以我的问题是：有没有办法在这种情况下指定模板函数的预期返回值？
解决方法
看起来编译器不确定do_something< int>的类型. – 我不确定这是编译器问题还是语言问题 – 但你可以强制编译器通过使用do_something< int>来调整其类型.以前相对琐碎的方式.例如,下面用gcc和clang trunk版本编译好(根据godbolt).
#include <functional>
template<typename T>
auto do_something(T input) {
return 123;
}
// Make the compiler workout the type of do_something<int> so we can use it later.
auto f = do_something<int>;
int main(int argc,12);
function();
return 0;
}
