How to Point to a Function in C++

Question

I want to know, how to create a pointer that points to the address of a function.

Supose that we have the following function:

int doublex(int a)
{
    return a*2;
}

I already know that & is used to get the address. How could I point to this function?

I don't understand the question. Are you asking how to call a function via a pointer? — HolyBlackCat, Dec 15 '18 at 16:59
@Some programmer dude Yes, but I do not know how to assign this address to the pointer. I just have do "pointer = & function_name"? — Grandtour, Dec 15 '18 at 17:01
Also note that you don't even need the `&`. Example: `int foo(int x) {return x*2;} int main(int (*ptr)(int); ptr = foo; std::cout << ptr(1);)`. — HolyBlackCat, Dec 15 '18 at 17:04
stackoverflow.com is not a substitute for a C++ book. Someone who wants to learn how to use function pointers will find far more information in a good C++ book, then from a brief comment on stackoverflow.com — Sam Varshavchik, Dec 15 '18 at 17:04
@HolyBlackCat I didn't knew that. I thought that "&" was the only way to get the address. — Grandtour, Dec 15 '18 at 17:06
@HolyBlackCat The `&` operator can only be omitted for non-member functions. — Some programmer dude, Dec 15 '18 at 17:07
Specifically for functions it's optional. (For non-member functions, as Some programmer dude mentions.) — HolyBlackCat, Dec 15 '18 at 17:07
Actually one should always believe that "&" is the only way to get the address; it's something automagically help you turn the function name into address, just like array name. Happy programming. — Kindred, Dec 15 '18 at 17:08
Why don't you guys put the answer? Is there anything in this question that doesn't make it clear? — Grandtour, Dec 15 '18 at 17:10
Cause @Someprogrammerdude thought that the solution was `auto pointer_function = &double;` — Grandtour, Dec 15 '18 at 17:11
For "Why don't you guys put the answer?", hmm, [it's a bug I think](https://meta.stackexchange.com/q/98950/421325). (the first comment) — Kindred, Dec 15 '18 at 17:13
Handy reading: [Is the type of “pointer-to-member-function” different from “pointer-to-function”?](https://isocpp.org/wiki/faq/pointers-to-members#fnptr-vs-memfnptr-types) — user4581301, Dec 15 '18 at 17:21

Moia · Accepted Answer · 2018-12-15T17:32:13.963

3

As said, you can just take the address with the & operator. Then the easiest way is to assign it with a auto variable to store it, now you can use your variable like a function itself.

int doubleNumber(int x)
{
    return x*2;
}

int main()
{
  auto func = &doubleNumber;
  std::cout << func(3);
}

See a live example here

edited Dec 15 '18 at 17:32

answered Dec 15 '18 at 17:15

Moia

2,216
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1

This is right, but in my case, the pointer was declared using "malloc" – Grandtour Dec 15 '18 at 17:21
and what's your point? You should give a more precise context of what you're asking – Moia Dec 15 '18 at 17:24
That's because I'm always confused in how to use malloc, so I didn't put the declaration of the pointer in my answer. – Grandtour Dec 15 '18 at 17:25
1

@Grandtour begin with not using malloc if you're writing C++. – Moia Dec 15 '18 at 17:27
Oh, yes, I tought that I was still programming C, thanks for reminding me. – Grandtour Dec 15 '18 at 17:28
Can you explain what is the "auto" type in your answer? This is only the remaining thing to I accept your answer. – Grandtour Dec 15 '18 at 17:32
It's quite OT, but anyway. **auto specifier**: _For variables, specifies that the type of the variable that is being declared will be automatically deduced from its initializer._ https://en.cppreference.com/w/cpp/language/auto – Moia Dec 15 '18 at 17:36
To do this in C, is it the same way? – Grandtour Dec 15 '18 at 17:44
@Grandtour auto here is deduced as (an unconfortable) `int (*)(int) func;` so the same as a C function pointer. see here https://cppinsights.io/ – Moia Sep 11 '19 at 13:00

IncredibleCoding · Answer 2 · 2018-12-15T17:37:12.560

2

Just do:

auto function_pointer = &doublex;

And what is the auto type?

The auto keyword specifies that the type of the variable that is being declared will be automatically deducted from its initializer. In case of functions, if their return type is auto then that will be evaluated by return type expression at runtime. Source here

This will help you: C++ auto keyword. Why is it magic?

edited Dec 15 '18 at 17:37

answered Dec 15 '18 at 17:05

IncredibleCoding

191
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Judging by the comments, OP wants to know how to call a function pointer. – HolyBlackCat Dec 15 '18 at 17:06
@ptr_user7813604 so how would I get the address? – Grandtour Dec 15 '18 at 17:08
@Grandtour The point is that `double` is not a valid function name. Neither are `int`, `return`, etc. – HolyBlackCat Dec 15 '18 at 17:10
1

For modern C++, I believe we should always use what @Some programmer dude said. But you may also try `int (*ptr)(int) = funct_name_and_Im_not_keyword;` (or adding a `&` before it) to be a cool guy. – Kindred Dec 15 '18 at 17:10
1

So the program will return compilation error, right? – Grandtour Dec 15 '18 at 17:11
@Grandtour Why won't you try it? It will. – HolyBlackCat Dec 15 '18 at 17:11
So, I think I will change the function_name to be more right – Grandtour Dec 15 '18 at 17:12

score 0 · Answer 3 · answered Dec 15 '18 at 17:11

You can do something like this and store references to compatible functions or pass your function as parameter in another function.

typedef int (*my_function_type)(int a);
int double_number(int a)
{
    return a*2;
}
my_function_type my_function_pointer = double_number;
int transform_number(int target, my_function_type transform_function) {
    return transform_function(target);
}

I hope it helps you

How to Point to a Function in C++

3 Answers3