Problems about function pointers in C/C++ [duplicate]

Question

I am reading Linux kernel recently. I find that in many cases they use the struct "typedef xxx f(xxx)", but I cannot understand how it works. (something like function pointer?)

Here is my test code.

#include<stdio.h>
typedef int Myfunc(int);
typedef int (*point_to_myfunc)(int);
static Myfunc example;
static int example(int a){
    printf("example a=%d\n", a);
    return 1;
}
static void example2(Myfunc* f){
    printf("example2\n");
    f(2);
}
static void example3(int (*)(int));
static void example3(int (*point_to_Myfunc)(int)){
    printf("example3\n");
    point_to_Myfunc(3);
}
int main(){
    point_to_myfunc f=&example;
    example2(f);
    example3(f);
    return 0;
}

Can anyone provide a brief explanation for me? Thx~

@Jueecy Why wouldn't it be? It has a very obvious meaning, so why would the Standard not allow it? if you want to know whether a typedef is legal, just remove "typedef" and see if it's a legal declaration. int MyFunc(int); ... yup. — Jim Balter
– Jim Balter, Commented Mar 22, 2013 at 5:56

Jonathan Leffler · Accepted Answer · 2013-03-22 06:16:12Z

#include <stdio.h>
typedef int Myfunc(int);

Myfunc is the name of a type; it is a function taking an int argument and returning an int.

typedef int (*point_to_myfunc)(int);

point_to_myfunc is a pointer to a function taking an int argument and returning an int. You could also have: typedef Myfunc *ptr_to_myfunc; if you wished (another name for the same type).

static Myfunc example;

This says 'there exists a function called example of type Myfunc'.

static int example(int a)
{
    printf("example a=%d\n", a);
    return 1;
}

This is a possible implementation of example. You can't use a typedef name to like Myfunc in the definition of a function of that type.

static void example2(Myfunc *f)
{
    printf("example2\n");
    f(2);
}

This is a function that takes a pointer to a Myfunc. The line f(2); invokes the function pointed at with the argument 2 and ignores the returned value.

static void example3(int (*)(int));

This declares example3 as a function taking a pointer to a function that takes an int argument and returns an int result. It could have been written as static void example3(point_to_myfunc); or static void example3(ptr_to_myfunc); or static void example3(Myfunc *);.

static void example3(int (*point_to_Myfunc)(int))
{
    printf("example3\n");
    point_to_Myfunc(3);
}

This is an implementation of example3.

int main(void)
{
    point_to_myfunc f = &example;
    example2(f);
    example3(f);
    return 0;
}

This program has a variable f that's a pointer to a function. Interestingly, you could have:

    point_to_myfunc f2 = example;
    point_to_myfunc f3 = *example;

Etc. And they all mean the same thing.

You could also invoke them using:

    (*f2)(101);
    (**f3)(103);

The standard notation for the initialization would use neither the & nor the *. If you're an old school C programmer, you may well invoke the function pointer using the (*f2)(101) notation; before the C89 standard, that was the only way to invoke function pointers. Modern style tends to be f2(101); instead.

Thanks a lot for your clear explanation. This usage is really powerful.

TerryZJ · Accepted Answer · 2013-03-22 06:01:54Z

1

Vaughn Cato is correct, In addition,

typedef int (*point_to_myfunc)(int);

defines a function pointer, it means point_to_myfunc is a type,we can use it like this:

point_to_myfunc f=&example;

now f is just like example(), we could f() to call method example

answered Mar 22, 2013 at 6:01

TerryZJ

235 bronze badges

Comments

Jim Balter · Accepted Answer · 2013-03-22 06:59:53Z

1

typedef int Myfunc(int);

This means that Myfunc is the type of a function which takes an int parameter and returns an int.

This line:

static Myfunc example;

is the same as saying

static int example(int);

which forward-declares the example function.

One use for this would be to make it clearer that a particular set of functions are used for a particular purpose.

typedef char CharacterConverter(char);

extern CharacterConverter make_upper_case;
extern CharacterConverter make_lower_case;

extern void process_string(char *s,CharacterConverter *f);
    // easier to see that make_upper_case and make_lower_case are valid arguments.

edited Mar 22, 2013 at 6:59

Jim Balter

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answered Mar 22, 2013 at 5:49

Vaughn Cato

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3 Comments

Koushik Shetty Over a year ago

what would be the practical use of such typedef?

Vaughn Cato Over a year ago

@Koushik: I've added one possible use.

Koushik Shetty Over a year ago

+1. ah nice. there is a typo. its characterConverte(r) not (d).

Bing · Accepted Answer · 2013-03-22 06:19:43Z

0

typedef is useful when define a type.

For example: char *a, b; defined a pointer "a", and a char b. char *a, *b defined two char pointers. If use typedef, it will be clear:

typedef char* PCHAR;
PCHAR a,b;

Now, both a and b is a char pointer.

typedef int Myfunc(int);
typedef int (*point_to_myfunc)(int);

the two lines defined a pair, a function format and a type of pointer which can point to the function, so it will be clear and more obvious when using them.

edited Mar 22, 2013 at 6:19

answered Mar 22, 2013 at 5:53

Bing

964 bronze badges

2 Comments

Jim Balter Over a year ago

The OP didn't ask what typedef is or for an example of what it's good for.

Bing Over a year ago

You're right, My mistake.

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