complicated Typedef in C++

Question

I understand that typedef can be used to define a new custom type, for example:

// simple typedef
typedef unsigned long ulong;

// the following two objects have the same type 
unsigned long l1;
ulong l2;

I recently came across this typedef, and got lost in deciphering what is going on in the declaration:

typedef int16_t CALL_CONVENTION(* product_init_t)(product_descript_t *const description)

Can someone guide me and explain what this is doing?

EDIT: changed NEW_TYPE to CALL_CONVENTION. It's a define. Thanks for spotting that out.

It declares a function pointer type alias. See http://stackoverflow.com/questions/4295432/typedef-function-pointer. — kennytm, May 31 '16 at 20:55
typedefs dont define a new type, they just provide you an alias for an already existing type — 463035818_is_not_an_ai, May 31 '16 at 20:56
Use [cdecl](http://www.cdecl.org/) to decode complex type declarations. — Barmar, May 31 '16 at 20:59
@ForceBru: could you please explain what effect (* stuff)(stuff) has on int16_t. That is, what difference would it make to have only typedef int16_t NEW_TYPE; ? — nikk, May 31 '16 at 21:00
@GregHewgill NEW_TYPE is not a define. It's the 'newly' created alias. — nikk, May 31 '16 at 21:01
Probably because of the `CALL_CONVENTION` macro, since cdecl has no way of knowing that that does. You need to expand the macro first. — Barmar, May 31 '16 at 21:19
@Barmar @tobi303, example `ifdef __amd64__ define CALL_CONVENTION else define CALL_CONVENTION __attribute__((__cdecl__)) endif` — nikk, May 31 '16 at 21:28

AlexD · Accepted Answer · 2016-06-01T19:37:22.083

10

It declares type product_init_t as a pointer to a function which

takes parameter product_descript_t *const description;
returns int16_t;
uses calling convention CALL_CONVENTION (as @M.M suggested even when it was misnamed).

P.S. Since "a complete answer in 2016 should show the modern way to write this type-alias" (@Howard Hinnant), here it is:

using product_init_t = int16_t (CALL_CONVENTION *)(product_descript_t *const description);

edited Jun 01 '16 at 19:37

answered May 31 '16 at 20:55

AlexD

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it's likely that `NEW_TYPE` is a linkage or calling convention specifier; not a part of the return type – M.M May 31 '16 at 21:04
1

@M.M Likely, but then the name is somewhat confusing. Anyway, added to the answer. Thanks! – AlexD May 31 '16 at 21:05
@M.M you are right. I have made the edit to the post. – nikk May 31 '16 at 21:12
Ok thanks. This CLEARLY helps me understand the problem. Accepting as answer. One last thing, I am fairly new to compiler infrastructure, so if you an point me to a web link on dealing with Calling Conventions for these type of statements, I'll truly appreciate it. – nikk May 31 '16 at 21:48
A complete answer in 2016 should show the modern way to write this *type-alias* which imho is much cleaner. – Howard Hinnant Jun 01 '16 at 01:50
@HowardHinnant As you wish :). – AlexD Jun 01 '16 at 19:38

M.M · Answer 2 · 2016-05-31T21:12:59.897

Here is how to make sense of complicated typedefs: First look at the line with typedef taken out:

int16_t NEW_TYPE (* product_init_t)(product_descript_t *const description);

Then work out what this line would declare: it's a variable called product_init_t with a certain type.

Finally, adding typedef means that it declares product_init_t to be an alias for that certain type, instead of a variable of that type.

To work out the above declaration, you can use cdecl , or you can attack it from the outside-in using knowledge of the possible declarators (pointer, array, function).

In this case the outer-most feature is the parameter list at the right, so we suspect this might be a function declarator. Function declarators look like (roughly):

returntype    function_name   (parameters_opt)

although bear in mind that a common compiler extension is to specify additonal properties of the function via __declspec or otherwise, in the same place as the return type; or there might be extern "C" there, and so on.

So far we are at the stage that (*product_init_t) is a function with int16_t NEW_TYPE as return type (and possible declspecs), and parameter list (product_descript_t *const description).

Finally, * product_init_t is just a pointer declarator, so we conclude that product_init_t is a pointer to a function of the above type.

Your comments indicate being unsure about NEW_TYPE. It will be something that is already defined earlier (or a compiler extension keyword); perhaps preprocessing the code with gcc -E would help if your IDE can't find its definition.

The correct line is `typedef int16_t CALL_CONVENTION(* product_init_t)(product_descript_t *const description)` where CALL CONVENTION is defined in `ifdef __amd64__ define CALL_CONVENTION else define CALL_CONVENTION __attribute__((__cdecl__)) endif` how will this now make a difference? — nikk, May 31 '16 at 21:39

score 1 · Answer 3 · edited May 23 '17 at 11:45

It is an interesting example of difference between #define and typedef usage.

#define FX_TYPE void (*)(int)
typedef void (*stdfx)(int);

void fx_typ(stdfx fx); /* ok */
void fx_def(FX_TYPE fx); /* error */

please see here

Therefore in case of following:

typedef int16_t CALL_CONVENTION(* product_init_t)(product_descript_t *const description)

We can use the above typedef as follows:

void fx_typ(product_init_t fx);

fx is a pointer to function which takes product_descript_t *const description as argument and returns int16_t.

The use of CALL_CONVENTION is confusing, however, it can be suppressed by empty macro as follows:

#define  CALL_CONVENTION

Note: the above micro has no body. In my view, CALL_CONVENTION just adds to confusion.

Good point, @vivek. CALL CONVENTION is actually defined as empty (for the most part) in my example. I wonder what is the point. — nikk, Jun 01 '16 at 04:09

complicated Typedef in C++

3 Answers3