为什么UIButton不需要alloc和init？

Question

Why UIButton does not require to alloc and init call instead we use only UIButton *myBtn = [UIButton buttonWithType:UIButtonTypeCustom]; .

1. Does the above line automatically alloc s and init the uibutton??
2. Is it necessary to release myBtn?,since i'm not using alloc and init explicitly.

This may be a simple question but i don't know the correct answer for this, does anybody can help? Any help is appreciated,thanks in advance.

Answer 1

Well the buttonWithType will return a type of UIButton which is alloc and init for you. It is also autoreleased.

You can als alloc and init a UIButton your self, this will give you an UIButton of type UIButtonTypeCustom .

Answer 2

buttonWithType returns an auto-released object, which you don't have to release. As you did not alloc , no need to release .

Answer 3

There's a set of rules that the method names follow - read this link ;)

Basically, names beginning with alloc , new , copy or mutableCopy require you to call release (or autorelease ).

Any other methods that return an object will return autoreleased objects.

For example :

// You need to release these
NSString *myString = [[NSString alloc] init];
NSString *nextString = [myString copy];
UIbutton *button = [[UIButton alloc] initWithType:UIButtonTypeCustom];

// You don't need to release these
NSString *thirdString = [NSString stringWithString:@"Hello"];
UIButton *button = [UIButton buttonWithType:UIButtonTypeCustom];

Hope that helps!

Answer 4

The implementation is Hidden for every UIClass objects the class level alloc method comes from NSObject in case of UIButton also we get the alloc method if we write But apple Guy has given one more method called buttonwithtype class level method where he does something like this

(id)buttonWithType:(UIButtonType)buttonType
{

UIButton=[NSObject alloc];

 code for button type specification

 return id;

}

thats why we need not do alloc again