C++ Calling pointer of a struct's function [duplicate]

Question

I have a simple code as below:

int global1(int x)
{
 return x * 5;
}

int global2(int x)
{
 return x + 5;
}

struct some_struct {
  int a;

  int foo1(int x)
  {
     return x * a;
  }

  int foo2(int x)
  {
     return x + a;
  }

  int bar1(int x, bool a)
  {
     auto func = a ? global1 : global2; // works
     // auto func = a ? &global1 : &global2; // also works

     return func(x);
  }

  int problem_function(int x, bool a) // It is my question
  {
     auto func = a ? foo1 : foo2; // gives error
     // auto func = a ? &foo1 : &foo2; // also gives error
     // auto func = a ? &this->foo1 : &this->foo2; // also gives error

     return func(x);
  }
};

It is a very simplified form of the real code. I have no way to carry function on the outside like how I did in this with global1() & global2().

I want to call one of the functions in the struct, but using a pointer to a function, but it gives an error.

Note: I can't use if..else because in the real code it doesn't return func(x). In the real code, I use func(x) as a condition for a function in a loop (I am not joking, it's literaly what I said).

A part from the real code:

void* find_first(ExprToken* (*cond)(bool (*)(ExprToken*))) {...} // yeah

I know if I want to call a function of a struct, I have to tell the compiler which variable (struct) I am using, but how?

Does C++ even support a function pointer in a struct?

Sorry for the complexity, sometimes I forget what the real code is actually doing.

Non-member functions and member functions are two very different things. The difference, of course, is that member functions needs an object to be called with, which non-member functions doesn't. There's plenty of information about this all over the Internet, including quite a lot on this site alone. Use your favorite search engine and search for e.g. c++ call pointer to member function. — Some programmer dude
– Some programmer dude, Commented May 4, 2024 at 22:12
And regarding that snippet from the "real" code, you can't use it for member functions, that's just not possible. You need to change the implementation of your design. Either to use std::function together with std::bind or lambdas; Or use templates (again with std::bind or lambdas), which is how the C++ standard library handles all its callable objects., — Some programmer dude
– Some programmer dude, Commented May 4, 2024 at 22:14
"I can't use if..else because in the real code it doesn't return func(x)." -- I don't buy this. Mostly because you could write a function that returns func(x) and use that new function in place of the loop conditional. Break your messy functions into smaller pieces, and perceived problems like this tend to vanish. A few type aliases could also help. — JaMiT
– JaMiT, Commented May 5, 2024 at 1:36

lost_in_nowhere7 · Accepted Answer · 2024-05-04 22:37:21Z

2

you should set the function pointer to the method like that

auto func = a ? &some_struct::foo1 : &some_struct::foo2;

now you refer to the function of the struct and not of the object.

For call it, now you must use this syntax:

(this->*func)(x);

it mean "I call the function stored at adress of func and pass it the parameter x"

NB: In C++ avoid pointer to function like (*ptr)(void), it better to use the functional include which give you the syntax:

std::function<returnType(parameterType1, parameterType2, ..., parameterTypeN>

edited May 4, 2024 at 22:37

answered May 4, 2024 at 22:33

lost_in_nowhere7

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1 Comment

catnip Over a year ago

"Better to use std::function" - not always, it comes at a price (there's a performance overhead)

Egemen Yalın · Accepted Answer · 2024-05-04 22:23:34Z

1

Resource: https://public.websites.umich.edu/~eecs381/handouts/Pointers_to_memberfuncs.pdf

I found the answer on the internet and I wanted to share it under this question.

Answer:

  int problem_function(int x, bool a) // It is my question
  {
     auto func = a ? &some_struct::foo1 : &some_struct::foo2;

     return (*this).*func(x);
  }

answered May 4, 2024 at 22:23

Egemen Yalın

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2 Comments

Peter Over a year ago

You might want to look at this for alternative (syntactic) options. stackoverflow.com/a/63484420/4706785

Remy Lebeau Over a year ago

Use (this->*func)(x) instead of (*this).*func(x)

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