What does this mean in C

Question

I know that int *(*func)() means a function pointer which takes no arguments and returns an integer pointer but what does the following syntax means in C:

int *(*func())[]

Any explanations on how do we read such syntax in C would be very helpful.

According to cdecl.org that's not valid syntax. Are you sure that that's what is written? (N.B.: cdecl.org is a site that translates C declarations to English). — AntonH
– AntonH, Commented Mar 23, 2017 at 14:12
Someone correct me, but AFAIK an empty parameter list in C does not mean "takes no arguments", this would be true in C++ not in C. — A.S.H
– A.S.H, Commented Mar 23, 2017 at 14:18
Okay, just realised that func appears to be a reserved keyword on cdecl.org. Anyway, entered int *(*foo())[], and it says declare foo as function returning pointer to array of pointer to int. — AntonH
– AntonH, Commented Mar 23, 2017 at 14:24
@AntonH That would be because that site is pretty bad and fails at numerous cases of valid C code. Instead of linking that bad, irrelevant site, try to compile the code with your standard compliant compiler. Works just fine since it is valid C code. Why 4 people like your comment, I don't know. — Lundin
– Lundin, Commented Mar 23, 2017 at 15:12

msc · Accepted Answer · 2017-03-23 14:33:17Z

3

func is a function and returning a pointer to an array of pointers to int.

reference link : http://gateoverflow.in/35193/regarding-pointers

edited Mar 23, 2017 at 14:33

answered Mar 23, 2017 at 14:13

msc

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3 Comments

dmi Over a year ago

hmm, not really. If no void is explicitely written as the function parameter, this does not mean it takes no arguments in C. Even more, this means the function can have any number of parameters of any type. Here was also a comment about that above from @AntonH.

Lundin Over a year ago

@cmaster No, it is an array pointer. Array pointers don't decay.

cmaster - reinstate monica Over a year ago

@Lundin But this looks dangerously close to the int foo(int bar[]) case, where the declaration is definitely equivalent to int foo(int* bar). Since there is no array variable declared in int* (*func())[], I would guess that should be equivalent to int** (*func()). After all, you can't pass around an array in C, you can only ever pass a pointer to its first element.

Community · Accepted Answer · 2017-05-23 11:46:18Z

It is a function that returns an array pointer. Written more explicit, here is the meaning of the various parts:

item_type (*function_name(parameters))[array bounds];

Where item_type is the type of the array items in the array pointed at.

The array pointer is a pointer to an array of unknown size, which is actually allowed by C, but not overly useful.

Overall, this function doesn't seem very useful and if you should ever need such an odd construct, you shouldn't write it as gibberish like this. Here is the equivalent form with a typedef:

typedef int* array_t[]; array_t* func();

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