Pass a protocol as a method argument

Question

First let me explain what I don't mean. I don't want to type an argument to a protocol:

-(void)someMethod:(id<SomeProtocol>)someArgument;

What I do want to is to pass a protocol to a method in the same way I can pass a Class to a method (The following is incorrect, but it hopefully explains what I want to do):

-(void)someMethod:(Protocol)someArgument;

I would then like to be able to use the Protocol to check whether a set of objects implement it.

Solution 1

If you know the name of a protocol at coding-time, use @protocol(SomeProtocol) to get a pointer to that protocol, similar to how you'd use @selector(x).

Beyond that, you just refer to protocols with the class identifier Protocol -- so you're method declaration would look like:

-(void)someMethod:(Protocol*)someArgument

You can see an example in the docs for NSObject conformsToProtocol:

http://developer.apple.com/library/mac/#documentation/Cocoa/Reference/Foundation/Classes/NSObject_Class/Reference/Reference.html#//apple_ref/occ/clm/NSObject/conformsToProtocol:

Solution 2

Protocol is a class, so you just write - (void)someMethod:(Protocol *)someArgument like with any other object type. You can see this in the declaration for conformsToProtocol::

+ (BOOL)conformsToProtocol:(Protocol *)aProtocol

Solution 3

I don't recommend using protocol. It will obscure which interface your code actually relies on. Use id<yourprotocol>*. This is actually how the cocoa frameworks pass protocols. Forgive the use of words if I don't it thinks I'm trying to do HTML.

Solution 4

- (void) executeRequest:(id<Protocol1>)request andCompletion:(id<Protocol2>)response

the only way to pass protocol into an argument

because id<..> means it needs to conform to that protocol before pass throw the argument

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Comments

[Undistraction]

Thanks. So out of interest why does Protocol use a pointer while Class doesn't?

[adpalumbo]

I'm afraid I don't have answer to that, myself. Good question, though.

[jscs]

Protocols and classes are each represented by their own classes. The difference is, the word Class is defined as being a pointer, but Protocol is just the name of the class, like any other (NSString, e.g.); so for a Protocol instance you must have an explicit pointer. @1nd

[Chuck]

@1ndivisible: Protocol is a class, just like NSString or NSArray. Class is not a class, it's a type that signifies a pointer to a class, much the same as id signifies a pointer to objects.

[Solomon]

Also this is a strongly typed language so don't try and make it act loosely typed

[Millie Smith]

Is this down voted because it's not what OP wants? This is what I want... @protocol(MyProtocol)varname didn't even compile for me.

[vishal dharankar]

perfect , this something others not mentioning