typedef function pointers and extern keyword

Question

I'm having problems understanding the syntax of a pointer to a function using typedef. I've read a lot of answers but still couldn't understand something. I'll try to explain how I see things so you could understand my thinking.

So we use typedef to give aliases to existing types for example :

typedef int number;

Will make it so we could use a number same as an integer (similar to preprocessor directives - I know there're some differences like when making a typedef of a pointer). Another example :

typedef struct
{
    int num;
} MyStruct;

Will give the unnamed structure an alias named MyStruct.

So Here's the syntax of a pointer to function typedef:

typedef int (*pFunc)(int, int);

Maybe I'm having hard time to understand this since typedef is like it's name giving aliases to TYPES and a function is not exactly type but anyway, from my understanding this is more of a pointer to some sort of a function signature, so the first int is the returned type, the second parenthesis are to indicate what types are the arguments being passed to the function. Now what I don't quite understand is this part :

(*pFunc)

• What I think it is, is that we create a new type (using typedef) named pFunc that is a pointer and that's the role of the *. Now we could create variables of this type that will point to ANY function with the signature we described. Am I correct ?

Ok, say I'm correct, usually pointers to some memory are declared as follow :

int *p;
double *p;
.
.
.

So wouldn't it make more sense to do it as follow :

(pFunc*)

Because to me it looks like if the asterisk is before the name it looks like pFunc is a variable name of type pointer of some type and not an actual type pointer.

• Can we do that ^ ? If so, is it commonly used to put the asterisk after and not before ? If it's more common to put it before then why is that ? Because like I said when we define a pointer type we always put the asterisk after the name itself like the examples above so why is that exactly ?
• Another question regarding to this, I don't quite understand what are the job of the parenthesis around *pFunc. I think that they're used to indicate that pFunc type pointer of something and if we wouldn't do put parenthesis then the return type of the signature will be of type int* instead of just int, am I correct here ?

Ok, another thing that bothers me about it as the order of the syntax. So far, in all typedef definitions we had the type on the left and the alias(es) on the right.

typedef int number;

typedef struct
{
    int num;
} MyStruct;

We see that the int and the struct are the types which are being on the left and the alias we gave them are on the right.

Now, in pointers to function typedef it doesn't follow this convention. We have the type returned of the function on the right then the typename in parenthesis then the type of the arguments in parenthesis, this order makes me confused after looking on how the other typedef are working on the same order.

• Wouldn't it make more sens to do something like this ? :

  typedef int (int,int) Func; So we have a typedef first, the type we want to give alias(es) to, which in this case is a function signature which takes 2 ints and return an int, and then on the right we have the alias name. Won't it make more sense ? this follows the other typedef order, I just don't get the function pointer order that much ..

• Another question of mine : When we make a pointer to a function, what does it actually mean ? I understand we can call this function using our alias but pointers like variables are stored in a memory address ? What is there to store for a function ?
• Lastly, I've read that the keyword extern has something to do with pointers to function but couldn't understand what this keyword does, could someone explain to me what it does ?

Answer 1

The typedef uses the same syntax for declaring types as would normally be used for declaring values.

For instance, if we declare an int called myInt, we do:

int myInt;

If we want to declare a type called myIntType to be an int, we simply add typedef:

typedef int myIntType;

We can declare a function myFunc, as follows:

int myFunc(int a, int b);

Which tells the compiler that there is an actual function with that name and signature that we can call.

We can also declare a function type myFuncType by doing:

typedef int myFuncType(int a, int b);

And we could do:

myFuncType myFunc;

Which is equivalent to the previous declaration of myFunc (although this form would rarely be used).

A function is not a conventional value; it represents a block of code with an entry point address. Function declarations like those above are implicitly extern; they tell the compiler that the named thing exists somewhere else. However, you can take the address of a function, which is called a function pointer. A function pointer can point to any function with the correct signature. A pointer is declared by prefixing the name of the type/value with a *, so, we might try:

int *myFuncPtr(int a, int b);

But this would be incorrect because the * binds more tightly with the int, so we have declared that myFuncPtr is a function that returns a pointer to an int. We must put parens around the pointer and name to change the binding order:

int (*myFuncPtr)(int a, int b);

And to declare a type, we simply add typedef to the front:

typedef int (*myFuncPtrType)(int a, int b);

In the declaration of myInt above, the compiler allocated some memory for the variable. However, if we were writing some code in a different compilation unit, and wanted to reference myInt, we would need to declare it as extern (in the referencing compilation unit) so that we reference the same memory. Without the extern, the compiler would allocate a second myInt, which would result in a linker error (actually that's not quite true because C allows tentative definitions, which you shouldn't use).

As noted above, functions are not normal values, and are always implicitly extern. However, function pointers are normal values, and need the extern if you are trying to reference a global function pointer from a separate compilation unit.

Normally, you would put externs for your global variables (and functions) into a header file. You would then include the header into the compilation units that contain the definitions of those variables and functions so that the compiler can make sure the types match.

Answer 2

The syntax for a variable definition or declaration is the type followed by one or more variables possibly with modifiers. So some examples would be:

 int  a, b;    // two int variables a and b
 int  *a, b;   // pointer to an int variable a and an int variable b
 int  a, *b, **c;   // int variable a, pointer to an int variable b, and pointer to a pointer to an int variable c

Notice that in all of these the asterisk modifies the variable to the right of the asterisk changing it from an int into a pointer to an int or a pointer to a pointer to an int. So these might be used like:

int a, *b, **c, d;

a = 5;     // set a == 5
b = &a;    // set b == address of a
c = &b;    // set c == address of b which in this case has the address of int variable a

d = **c;   // put value of a into d using pointer to point to an int variable a
d = *b;    // put value of a into d using pointer to an int variable a
d = a;     // put value of a into d using the variable a directly

The extern statement is used to indicate that the definition of a variable is located in some other file and that the variable has global visibility. So you can declare a variable using the extern keyword to be explicit about a variable so that the C compiler will have the information it needs to do a good level of checking when compiling. The extern indicates that the variable is actually defined with its memory allocation somewhere other than the file where the source using the variable is located.

Since a function pointer is a variable then if you have a function pointer that needs global visibility when you declare the variable for files other than the source file where it is actually defined and its memory allocated you would use the extern keyword as part of the function pointer variable declaration. However if it is a variable that is allocated on the stack within a function or if it is used within a struct to create a member of the struct then you would not use the extern modifier on the variable.

The typedef is a very nice feature of modern C because it allows you to create an alias that amounts to a kind of halfway new type. To have the full capability of creating a new type really requires the class type features of C++ which allows the definition of operators for the new type as well. However typedef does provide a good way of allowing a programmer to create an alias for a type.

Years ago in the older C compiler days before typedef was added to the C standard, people would often use the C Preprocessor to define macros to create an alias for a complex type. typedef is a much cleaner way of doing it!

Most uses of typedef are to provide a way to make it shorter and cleaner to write a variable definition. It is used a lot with struct definitions for that reason. So you might have a struct definition like the following:

typdef struct {
    int iA;
    int iB;
} MyStruct, *PMyStruct;

This will create two new aliases for the struct, one for the struct itself and one for a pointer to the struct and these might be used like:

MyStruct  exampleStruct;
PMyStruct pExampleStrut;

pExampleStruct = &exampleStruct;

This example has the basic structure of typedef keyword, definition of the new type in terms of existing types, and name of the new type.

Before typedef was added to the C standard, you would specify a tag for the struct and the result would be code that looked like this:

struct myTagStruct {     // create a struct declaration with the tag of myTagStruct
    int a;
    int b;
};

struct myTagStruct myStruct;      // create a variable myStruct of the struct

At which pointer people would usually add a C Preprocessor define to make it easier to write as in:

#define MYTAGSTRUCT  struct myTagStruct

and then use it something like:

MYTAGSTRUCT myStruct;

However the syntax of typedef for function pointers is a bit different. It is more along the lines of a function declaration where the name of the function is the actual name of the typedef or the alias that is actually used. The basic structure of a typedef for a function pointer is typedef keyword followed by the same syntax as a function prototype where the name of the function is used as the typedef name.

typedef int (*pFunc)(int a1, int b1);

The parenthesizes are important because they force the compiler to interpret the source text in a particular way. The C compiler has rules that it uses to parse the source text and you can change the way that the C compiler interprets the source text by using parenthesizes. The rules have to do with the parsing and how the C compiler locates the variable name and then determines the type of the variable by using rules about left and right associativity.

a = 5 * b + 1;     // 5 times b then add 1
a = 5 * (b + 1);   // 5 times the sum of b and 1

int *pFunc(int a1, int b1);      // function prototype for function pFunc which returns a pointer to an int
int **pFunct(int a1, int b1);    // function prototype for function pFunc which returns a pointer to a pointer to an int
int (*pfunc)(int a1, int b1);    // function pointer variable for pointer to a function which returns an int
int *(*pFunc)(int a1, int b1);  // function pointer variable for pointer to a function which returns a pointer to an int

A function prototype is not a function pointer variable. The syntax of a typedef is similar to the syntax for a variable definition that is not using a typedef.

typedef  int * pInt;    // create typedef for pointer to an int
int *a;                 // create a variable that is a pointer to an int
pInt b;                 // create a variable that is a pointer to an int
typedef int (*pIntFunc)(int a1, int b1);    // create typedef for pointer to a function
int (*pFuncA)(int a1, int b1);           // create a variable pFuncA that is a pointer to a function
pIntFunc  pFuncB;                        // create a variable pFuncB that is a pointer to a function

So what does it mean to have a pointer to a function? A function entry point has an address because a function is machine code that is located at a particular memory area. The address of the function is where the execution of the functions machine code is supposed to start.

When the C source code is compiled, a function call is translated into a series of machine instructions which jump to the address of the function. The actual machine instructions are not really a jump but are instead a call instruction which saves the return address before it makes the jump so that when the called function completes it can do a return back to where it was called from.

A function pointer variable is used like a function statement. The difference between the two is similar to the difference between an array variable and a pointer variable. An array variable is treated like a constant pointer to a variable by most C compilers. A function name is treated like a constant pointer to a function by most C compilers.

What a function pointer does give you though is flexibility though it is flexibility that as with any great power can also lead to great ruin.

One use of function pointer variables is to pass a function address as an argument to another function. For instance the C Standard library has a couple of sort functions that require an argument of a collation function for comparing two elements being sorted. Another example would be a threading library that when you create a thread, you specify the address of the function to be executed as a thread.

Another case is to provide some kind of interface that hides implementation details. Let's say that you have a print function that you want to use for several different output sinks or places where the output to go, say a file, a printer, and a terminal window. This is similar in nature to how virtual functions are implemented by C++ compilers or how COM objects are implemented through a COM interface. So you could do something like the following which is a very simple example missing details:

typedef struct {
    int  (*pOpenSink) (void);
    int  (*pPrintLine) (char *aszLine);
    int  (*pCloseSink) (void);
} DeviceOpsStruct;

DeviceOpsStruct DeviceOps [] = {
   {PrinterOpen, PrinterLine, PrinterClose},
   {FileOpen, FileLine, FileClose},
   {TermOpen, TermLine, TermClose}
};

int OpenDevice (int iDev)
{
    return DeviceOps[iDev].pOpenSink();
}

int LineDevice (int iDev, char *aszLine)
{
    return DeviceOps[iDev].pPrintLine (aszLine);
}
int CloseDevice (int iDev)
{
    return DeviceOps[iDev].pCloseSink();
}

Answer 3

Just to make clear the explanation given by others, in C/C++, the parenthesis are right associative, therefore the following declaration:

typedef int *pFunc(int, int);

is equivalent to:

typedef int *(pFunc(int, int));

which would be the declaration prototype of a function returning a pointer to an integer and not the declaration of a pointer to a function returning an integer.

This is why you need to write the parenthesis around (*pFunc) to break the right association and tell the compiler that pFunc is a pointer to a function and not simply a function.