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Function passed as template argument

Web Technologies Web Development 2 years ago

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_x000D_ _x000D_ I'm looking for the rules involving passing C++ templates functions as arguments. This is supported by C++ as shown by an example here: #include void add1(int &v) { v+=1; } void add2(int &v) { v+=2; } template void doOperation() { int temp=0; T(temp); std::cout << "Result is " << temp << std::endl; } int main() { doOperation(); doOperation(); } Learning about this technique is difficult, however. Googling for "function as a template argument" doesn't lead to much. And the classic C++ Templates The Complete Guide surprisingly also doesn't discuss it (at least not from my search). The questions I have are whether this is valid C++ (or just some widely supported extension). Also, is there a way to allow a functor with the same signature to be used interchangeably with explicit functions during this kind of template invocation? The following does not work in the above program, at least in Visual C++, because the syntax is obviously wrong. It'd be nice to be able to switch out a function for a functor and vice versa, similar to the way you can pass a function pointer or functor to the std::sort algorithm if you want to define a custom comparison operation. struct add3 { void operator() (int &v) {v+=3;} }; ... doOperation(); Pointers to a web link or two, or a page in the C++ Templates book would be appreciated!
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Posted on 16 Aug 2022, this text provides information on Web Development related to Web Technologies. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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manpreet Tuteehub forum best answer Best Answer 2 years ago
_x000D_ Yes, it is valid. As for making it work with functors as well, the usual solution is something like this instead: template void doOperation(F f) { int temp=0; f(temp); std::cout << "Result is " << temp << std::endl; } which can now be called as either: doOperation(add2); doOperation(add3()); The problem with this is that if it makes it tricky for the compiler to inline the call to add2, since all the compiler knows is that a function pointer type void (*)(int &) is being passed to doOperation. (But add3, being a functor, can be inlined easily. Here, the compiler knows that an object of type add3 is passed to the function, which means that the function to call is add3::operator(), and not just some unknown function pointer.)
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