Issue about casting object brackets

Question

I have noticed that there are two ways to cast objects (the difference is the placement of the outer parenthesis): ``` 1. SimpleType simpleType = ((SimpleType) (property.getType())); 2. SimpleType simpleType = ((SimpleType) property).getType(); ``` Are they doing the same thing ?

Accepted Answer

Are they doing the same thing ?

No they are not.

- The first one is casting your value returned from `property.getType()` to `SimpleType`. (Invocation is done before Casting)

- The second one is first casting your `property` to `SimpleType` and then invoking the `getType()` method on it. (Casting is done before Invocation).

You can also understand it from the precedence of parenthesis. Since it has the highest precedence, it will be evaluated first.

First Case: -

So, in `((SimpleType) (property.getType()));`: -

```
(property.getType())
```

is evaluated first, then the casting is performed. In fact you don't really need a parenthesis around that. (`property` binds tighter to the `dot (.)` operator than the `cast` operator). So, invocation will always be done before casting. Unless you force it to reverse as in the below case: -

Second Case : -

In `((SimpleType) property).getType()`: -

```
((SimpleType) property)
```

is evaluated first, then the invocation is done. As, now you have enclosed `property` inside the brackets, due to which it binds tighter to the `cast` operator, due to higher precedence enforced by parenthesis.

Issue about casting object brackets

Solution

Problem