Please explain the following piece of code. I'm a beginner so please help me understand

Question

typedef long (*GuiFunc) (int, int, int, unsigned short*, long, long);

Please help me understand the above line of code

It's a pointer to a funtion taking 6 parameters and returning a long int. — StoryTeller - Unslander Monica, Oct 08 '13 at 14:18
Following up on hackks' question, you seem to be attempting to utilize something which is above your current level of understanding. Consider taking a look at a good C tutorial. — StoryTeller - Unslander Monica, Oct 08 '13 at 14:22
Thanks a lot guys for the prompt input. So I understand that the typedef name is GuiFunc. So whenever I need to define a function pointer, I will simply write: GuiFunc functionPointer = &somefunction; I understand now. Thanks for the help. You guys are great. — user2655793, Oct 08 '13 at 15:04
Why are you calling this a duplicate? The OP has seen that line of code and he does not understand it, hence he cannot know it is about function pointers, don't you think? — Bogdan Alexandru, Aug 28 '14 at 13:26

score 1 · Answer 1 · answered Oct 08 '13 at 14:19

1

you are defining a new type here.

The new type here is a function pointer.

the function pointer has 6 input arguments

answered Oct 08 '13 at 14:19

MOHAMED

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score 1 · Answer 2 · answered Oct 08 '13 at 14:20

1

You define a type GuiFunc which is a pointer (that's that (*GuiFunc) construct) to a function (the stuff in parentheses) which takes 3 ints, a pointer to unsigned short, two longs and returns a long.

answered Oct 08 '13 at 14:20

Alexander L. Belikoff

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I know from text book knowledge that : typedef int dog; dog counter; counter = 5; that's how it goes. Now does the line of code mean that I am using a function pointer in place of long? – user2655793 Oct 08 '13 at 14:27
No, that's a special syntax for typedefs for function pointers. C language is not very regular in its declarations. – Alexander L. Belikoff Oct 08 '13 at 14:51

score 1 · Accepted Answer · answered Oct 08 '13 at 14:27

typedef long (*GuiFunc) (int, int, int, unsigned short*, long, long);

Defines new type GuiFunc.that can declare a function pointer which takes 6 parameters int, int, int, unsigned short*, long, long and returns long.

Assume you have a function like this

long foo(int, int, int, unsigned short*, long, long)
{

}

if you declare

Guifunc callback; //declare a varaible of type Guifunc
callback=foo;

then you can call foo function like this long x=callback(6parameters);

Please explain the following piece of code. I'm a beginner so please help me understand

3 Answers3