问题
A couple of developers and I were wondering why:
std::cout<<std::time<<std::endl;
prints out a value of 1. What does the value represents, and why the value is of 1.
Answer to what happened: How to print function pointers with cout?
The C++ Standard specifies:
4.12 Boolean conversions
1 An rvalue of arithmetic, enumeration, pointer, or pointer to member type can be converted to an rvalue of type bool.
Quote from anon:
This is the only conversion specified for function pointers.
Edit: The answer below nicely presents the solution as to why 1 was printed rather than just any bool and explained when 1 would not occur.
回答1:
The cppreference says that:
There are no overload for pointers to non-static member, pointers to volatile, or function pointers (other than the ones with signatures accepted by the (10-12) overloads). Attempting to output such objects invokes implicit conversion to bool, and, for any non-null pointer value, the value 1 is printed (unless boolalpha was set, in which case true is printed).
So you get function pointer std::time converted to bool and it is always true that is without boolalpha set output as 1.
来源：https://stackoverflow.com/questions/52478974/printing-time-function-to-console-produces-1