指向函数的C ++指针导致seg错误

Question

I'm having a bit of trouble. I can't seem to figure out why my main function can't call the function pointed to by intFunction without a seg fault.

Also, this is code that I'm using for testing purposes. I'm still fairly new to C++.

Thanks for your help.

#include <iostream>

int tester(int* input){
    std::cout << "\n\n" << *input << "\n\n";
}

int (*intFunction)(int*);

template<typename FT>
int passFunction(int type, FT function){
    if(type == 1){
        function = tester;
        //Direct call...
        tester(&type);
        int type2 = 3;
        //Works from here...
        function(&type2);
    }
    return 0;
}

int main(int argc, char* argv[]){
    passFunction(1,intFunction);
    int alert = 3;
    //But not from here...
    intFunction(&alert);
    return 0;
}

Answer 1

When passing function pointers as parameters, they are not any different than other variables in that you are passing a copy of the value (ie whichever function address it has at the time).

If you are wanting to assign a variable in another function, you have to either pass it by reference or as a pointer to the original variable.

By reference:

int passFunction(int type, FT& function)

Or as a pointer

int passFunction(int type, FT* ppfunction)
{
    if(type == 1)
    {
        *ppfunction = tester;
        //Direct call...
        tester(&type);
        int type2 = 3;
        //Works from here...
        (*ppfunction)(&type2);
    }
    return 0;
}

// which then requires you pass the address of your variable when
// calling `passFunction`

passFunction(1, &intFunction);