C++ pass function via function and invoke via function address

#include <iostream>
#include <functional>

int callFunction55(int x, int y, function<int(int, int)> func);
void invokeCall56(int x, int y);
int sum57(int x, int y);
int multiply58(int x, int y);

int main(int args, char **argv)
{
    invokeCall56(atoi(argv[1]), atoi(argv[2]));
}

void invokeCall56(int x, int y)
{
    cout << "The sum of x and y is " << callFunction55(x, y, &sum57)<<endl;
    cout << "The prod of x and y is " << callFunction55(x, y, &multiply58)<<endl;
}

int callFunction55(int x, int y, function<int(int, int)> func)
{
    return func(x, y);
}


int multiply58(int x, int y)
{
    return x * y;
}

int sum57(int x, int y)
{
    return x + y;
}

The point located at two aspects, first declare #include <functional>,second declare the function as below via function key word

int callFunc55(int x,int y,function<int(int,int)> func)

Then define it which seems common as general function.

int callFunction55(int x, int y, function<int(int, int)> func)
{
    return func(x, y);
}

Be cautious,it's not pass function address but function keyword with angle bracket.

Compile via g++ and its standard is -std=c++2a, and run as below ./h1 10000 10000

C++ pass function via <functional> function and invoke via function address

原文链接: https://www.cnblogs.com/Fred1987/p/16268808.html

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C++ pass function via <functional> function and invoke via function address

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C++ pass function via function and invoke via function address

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