Assigning function pointer to function pointer

Question

I've got a problem when trying to assign a function pointer to a function pointer in a struct.

I have a struct command, which holds an invocation string value, a message to acknowledge its activation, and a function to be called upon activation.

However, I am having trouble assigning the function pointer in the struct's constructor (might make the struct into a class later, not sure) below.

struct Command
{
    Command(string _code, string _message, void *_func(void))
        : code(_code), message(_message) { /* ERROR: */ func = _func; }

    string code;        // The string that invokes a console response
    string message;     // The response that is printed to acknowledge its activation
    void *func(void);   // The function that is run when the string is called
};

In the above code, marked by /* ERROR: */ I get the error, "expression must be a modifiable value". How can I fix this? I just want to pass a reference to a function to the struct.

void *func(void) is not a function pointer, it's a declaration of a function named func which takes no arguments and return a void*. void (*func)(void) is a variable named func which is a pointer to a function taking no arguments and returning nothing. — Some programmer dude
– Some programmer dude, Commented Oct 15, 2014 at 13:06
Also, in C++ you should be careful with pointers to functions, as those can only be used for non-member functions, static member function or one type of lambda expression. If you want a more generic function-type, read about std::function. — Some programmer dude
– Some programmer dude, Commented Oct 15, 2014 at 13:09
@Joachim Pileborg Ugh, you're right. Cool man, thanks for the answer! Works now :) — JordanBell
– JordanBell, Commented Oct 15, 2014 at 13:09

Community · Accepted Answer · 2017-05-23 10:25:58Z

5

As mentioned by @Joachim Pileborg you don't declare a pointer to a function.

To declare a function pointer you need to add parentheses around the asterisk and identifier part, e.g.

// 'func' is a pointer to function taking parameter void and returning void.
void (*func)(void);

Since C++11 you can also declare a function pointer like this which is less terse:

std::add_pointer_t<void()> func;

std::add_pointer_t<void(int, int)> func; // Pointer to func taking 2 ints.

edited May 23, 2017 at 10:25

CommunityBot

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answered Oct 15, 2014 at 13:14

Felix Glas

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2 Comments

JordanBell Over a year ago

After a quick messaround I don't fully understand add_pointer nor how to use std::function for my purposes. I'll have to look into that further. Until then, thanks for the quick fix :D

Felix Glas Over a year ago

You can simply replace all occurences of void (*func)(void) with std::add_pointer_t<void()> func with no subsequent change in behaviour. Also remember that in C++ you don't have to declare a parameter list that takes nothing with (void), instead you can simply do ().

Adrian May · Accepted Answer · 2014-10-15 13:09:21Z

2

You need brackets around *func and *_func

answered Oct 15, 2014 at 13:09

Adrian May

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