The usage of overloaded operator standard library function
Question: is int(int, int) a common type??
For example, function: int add(int a, int b); For example, lambda: Auto mod = [] (int a, int b) ; For example, function object class: int operator()(int a, int b);The common characteristics of the above three are: int(int, int), but how to make the above three forms become common???
Answer: use the function class.
std::function<int(int, int)> f1 = add; std::function<int(int, int)> f2 = mod; std::function<int(int, int)> f3 = divide(); std::cout << f1(1,2) << std::endl; std::cout << f2(4,3) << std::endl; std::cout << f3(6,2) << std::endl;
Example: suppose that some processing parameters are always 2 ints and the return value is always int. you want to put these processing in a function table, such as square to std::map.
#include <functional> #include <map> #include <iostream> int add(int a, int b){ return a+ b; } auto mod = [](int a, int b); struct divide{ int operator()(int a, int b){ return a / b; } }; int main(){ /* std::map<std::string, int(*)(int, int)> mp; mp.insert({"+", add}); mp.insert({"%", mod}); divide dv; mp.insert({"/", divide()});//bian yi bu guo std::function<int(int, int)> f1 = add; std::function<int(int, int)> f2 = mod; std::function<int(int, int)> f3 = divide(); std::cout << f1(1,2) << std::endl; std::cout << f2(4,3) << std::endl; std::cout << f3(6,2) << std::endl; */ std::map<std::string, std::function<int(int, int)>> mp; mp.insert({"+", add}); mp.insert({"-", std::minus<int>()}); mp.insert({"*", [](int a, int b)}); mp.insert({"%", mod}); mp.insert({"/", divide()}); std::cout << mp["+"](1, 2) << std::endl; std::cout << mp["-"](3, 2) << std::endl; std::cout << mp["*"](2, 2) << std::endl; std::cout << mp["/"](100, 2) << std::endl; std::cout << mp["%"](31, 15) << std::endl; }QQ group of mutual learning in c/c + +: 877684253
