macOS Why do I need the "*" ?

[CmdrLaForge]

Sometimes in programming the little things are the ones that throw me off. The big concepts I always get as they are usually explained very well. I have the following code and I just don't understand why I need the *colorname before the function name. Does it have to do with the function call within NSLog?

This is not the whole code, just an excerpts:

typedef enum {
	kRedColor,
	kGreenColor,
	kBlueColor
} ShapeColor;

NSString *colorName (ShapeColor color)
{
	switch (color) {
		case kRedColor:
			return (@"red");
			break;
		case kGreenColor:
			return @"green";
			break;
		case kBlueColor:
			return @"blue";
			break;
	}
	
	return @"no clue";
	
} // colorName

@implementation Rectangle

- (void) setFillColor: (ShapeColor) c
{
	fillColor = c;
} // setFillColor


- (void) setBounds: (ShapeRect) b
{
	bounds = b;
} // setBounds


- (void) draw
{
	NSLog (@"drawing a rectangle at (%d %d %d %d) in %@",
		   bounds.x, bounds.y, 
		   bounds.width, bounds.height,
		   colorName(fillColor));
} // draw

@end // Rectangle

 

[sammich]

oops, too quick, hang on, lemme have another look

EDIT: is it because you return the NSString by reference?

 

[garethlewis2]

All Objective-C classes have to be allocated on the heap, so you effectively have a pointer to your object, that is what the star is, a pointer. Nothing more, nothing less.

In your example function you are returning an NSString object which are all pointers to a char array wrapped up in CFoundation.

Usually all classes with NS are pointers, but not all of them, some are structs and aren't declared with a *.

 

[CmdrLaForge]

All Objective-C classes have to be allocated on the heap, so you effectively have a pointer to your object, that is what the star is, a pointer. Nothing more, nothing less.

In your example function you are returning an NSString object which are all pointers to a char array wrapped up in CFoundation.

Usually all classes with NS are pointers, but not all of them, some are structs and aren't declared with a *.

Hi

thanks a lot. If the return type would be a int the function wouldn't be declared as a pointer?

 

[GorillaPaws]

This tutorial should help explain pointers better. Having a solid understanding of pointers is critical to C programming (objective-c too of course).

 

[garethlewis2]

To answer your question, yes if a function only returned an int,

it could have the function prototype

int funcA(int, int);

or it could have the function prototype

void funcA(int, int, int *);

Where the third parameter gets the result.

If the function is called

int* funcA(int, int);

All it means is you are returning a pointer to an int, you have no idea how many ints, you may have one, or you may have millions.

Once you understand pointers, C is one of the easiest languages in the world to program in.

 

[notjustjay]

I don't know about easiest -- Java and Python are both conceptually a little easier than C, though of course every language has its flaws.

But definitely take the time to learn about pointers and how they work. They can be a real headache, but they are very powerful.

 

[ranguvar]

Sometimes in programming the little things are the ones that throw me off. The big concepts I always get as they are usually explained very well.

Way to go, grasshopper. Pointers aren't exactly a "little thing".

 

[whooleytoo]

Hi

thanks a lot. If the return type would be a int the function wouldn't be declared as a pointer?

Precisely, if the return type was "int", the function header would be:

int colorName (ShapeColor color)

Although this is just a 'cosmetic' point, I'd always write the original function this way; to me it makes more sense:

NSString* colorName (ShapeColor color)

That to me reads more clearly. "This is a function called colorName which returns a pointer to an NSString".

I think it's more consistent too with how you'd write the equivalent Objective-C method something like this :

- (NSString*) colorName: (ShapeColor) color