Understanding function pointers

Question

int f1(){}
int* f2(){}

int main()
{
    int *a;//1
    int b;
    int *c;
    a=c; 
    a=&b; 

    int (*p)(); //2
    p=f2; //showing error
    p=&f1;
}

I expected that in my program '2' must behave similar to '1'. Why function pointer are behaving differently. Or am I missing something?

`f2` is a function returning a pointer to `int` (and would be converted into an `int * (*)()` in the line `p = f2;` if it compiled). `p` is a pointer to a function returning `int`. Incompatible types. — Daniel Fischer, Jul 01 '13 at 12:44
Here's a tip to help declare function pointer types easily: `typedef int type_of_f1();`. Now just add a star and you have `type_of_f1* p;` to declare a function pointer suitable for f1. — R. Martinho Fernandes, Jul 01 '13 at 12:54
possible duplicate of [Why do all these crazy function pointer definitions all work? What is really going on?](http://stackoverflow.com/questions/6893285/why-do-all-these-crazy-function-pointer-definitions-all-work-what-is-really-goi) — David Rodríguez - dribeas, Jul 01 '13 at 13:23

score 3 · Answer 1 · answered Jul 01 '13 at 12:44

int* f2(){}

A function that accepts nothing, and returns a pointer to an int.

int (*p)();

A pointer to: A function that accepts nothing and returns an int

You have a type mismatch. p is not a pointer to the type of f2

If you have trouble understanding such definitions, like all mortal do, use the spiral rule of thumb

score 3 · Accepted Answer · edited May 23 '17 at 10:26

3

p=f2; error because incompatible types. f2 is a function that can return a int* whereas p is pointer to function that can point to a function returns int e.g. f1()

for int* f2(), you can defined a pointer to function as below:

 int* (*p2)();   // pointers to function f2
 p2 = f2;

Additionally, you don't need to use & before function name just function name is enough. Here is a good link to read: Why do all these crazy function pointer definitions all work? What is really going on?

Edit:
Some time &functionname and functionname are not same e.g. sizeof(functionname) is not valid whereas sizeof(&functionname) is perfectly valid.

edited May 23 '17 at 10:26

Community

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answered Jul 01 '13 at 12:44

Grijesh Chauhan

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Thanks. But why p2=f2;? why not p2=&f2; – Alex Jul 01 '13 at 12:50
1

@Alex, those two forms are interchangeable when it comes to functions. – StoryTeller - Unslander Monica Jul 01 '13 at 12:50
1

@Alex You can try `p=*******************************************f2;` – johnchen902 Jul 01 '13 at 12:51
@Alex you don't need `&` for function address, just function name is enough (but you can & it not wrong), there is good link wait – Grijesh Chauhan Jul 01 '13 at 12:54
@Alex I'm not sure, but I heard that `*f2` is equals to `f2`. ( Only if `f2` is a pointer function ) – johnchen902 Jul 01 '13 at 12:56
@Alex and `&f2` is equal to `f2` :) in function, read: [Function pointers and address of a function](http://stackoverflow.com/questions/9552663/function-pointers-and-address-of-a-function) – Grijesh Chauhan Jul 01 '13 at 12:57
@G But you can't do `p=&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&f2` nor `p=& & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & &f2` – johnchen902 Jul 01 '13 at 12:59
@Alex ok read this: [Why do all these crazy function pointer definitions all work? What is really going on?](http://stackoverflow.com/questions/6893285/why-do-all-these-crazy-function-pointer-definitions-all-work-what-is-really-goi?lq=1) – Grijesh Chauhan Jul 01 '13 at 13:00
@Alex [this link](http://yosefk.com/c++fqa/function.html) may also be helpful some time because you also added c++ tag. – Grijesh Chauhan Jul 01 '13 at 13:12

score 2 · Answer 3 · answered Jul 01 '13 at 12:44

2

This is a function that returns int*

int* f2(){}

So you need:

(int*) (*q)();
q = f2;

answered Jul 01 '13 at 12:44

johnchen902

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score 1 · Answer 4 · answered Jul 01 '13 at 12:45

p is pointer to function taking void argument and returning integer.

f1 is a function taking void argument and returning integer.

f2 is a function taking void argument and returning pointer to integer.

Now as per definitions you can see f1 can be assigned to p but f2 can't.To assign f2 in p, declaration of p should be int *(*p)();

score 1 · Answer 5 · answered Jul 01 '13 at 12:45

1

You have made wrong assignments i.e incompatible types

For the functions

int f1(){}
int* f2(){}

The correct assignments would be

int (*p)(); 
p = f1;

int* (*p)();
p = f2;

answered Jul 01 '13 at 12:45

GeekFactory

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Understanding function pointers

5 Answers5