Pointer to function as parameter, using typedef

Question

I'm trying to create code which uses pointers to functions as parameteres, and I have to use a typedef. I'm not a C pro. It feels like I'm getting there, but I can't seem to find a good explanation of the syntax of pointers to function.

I have a function fillArray:

long fillArray(long *array, int x, int y) {
//
}

then I want to make a typedef of a pointer to this function:

typedef long (*fArray)(long, int, int);

fArray pFill = fillArray;

and I want to give this pFill to a function called doThis():

int doThis (fArray pFill) {
  return 0;
}

and calling it using:

int y = doThis(pFill);

What am I doing wrong?

Oliver Charlesworth · Accepted Answer · 2011-09-17 14:17:55Z

5

Your typedef needs to be:

typedef long (*fArray)(long *, int, int);
                            ^

answered Sep 17, 2011 at 14:17

Oliver Charlesworth

274k34 gold badges591 silver badges687 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

saltandpepper Over a year ago

Great! Kinda lame that I didn't see this myself. Thanks!

Matteo Italia · Accepted Answer · 2011-09-17 14:18:51Z

2

Your fillArray function accepts a long * as first parameter, you forgot a * into the typedef:

typedef long (*fArray)(long *, int, int);

answered Sep 17, 2011 at 14:18

Matteo Italia

128k18 gold badges219 silver badges313 bronze badges

Collectives™ on Stack Overflow

Pointer to function as parameter, using typedef

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related