How do you declare a pointer to a function that returns a pointer to an array of int values in C / C++?

Question

Is this correct?

int (*(*ptr)())[];

I know this is trivial, but I was looking at an old test about these kind of constructs, and this particular combination wasn't on the test and it's really driving me crazy; I just have to make sure. Is there a clear and solid understandable rule to these kind of declarations? (ie: pointer to... array of.. pointers to... functions that.... etc etc) Thanks!

R

Answer 1

The right-left rule makes it easy.

int (*(*ptr)())[];can be interpreted as

Start from the variable name ------------------------------- ptr

Nothing to right but ) so go left to find * -------------- is a pointer

Jump out of parentheses and encounter () ----------- to a function that takes no arguments(in case of C unspecified number of arguments)

Go left, find * ------------------------------------------------ and returns a pointer

Jump put of parentheses, go right and hit [] ---------- to an array of

Go left again, find int ------------------------------------- ints.

Answer 2

In almost all situations where you want to return a pointer to an array the simplest thing to do is to return a pointer to the first element of the array. This pointer can be used in the same contexts as an array name an provides no more or less indirection than returning a pointer of type "pointer to array", indeed it will hold the same pointer value.

If you follow this you want a pointer to a function returning a pointer to an int. You can build this up (construction of declarations is easier than parsing).

Pointer to int:

int *A;

Function returning pointer to int:

int *fn();

pointer to function returning a pointer to int:

int *(*pfn)();

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If you really want to return a pointer to a function returning a pointer to an array of int you can follow the same process.

Array of int:

int A[];

Pointer to array of int:

int (*p)[];

Function returning pointer ... :

int (*fn())[];

Pointer to fn ... :

int (*(*pfn)())[];

which is what you have.