Using the operator and IDisposable interface

Question

Is it true that the variables declared in the using statement are located together because they are in the scope of the block?

Do I need to do:

using (SomeIdisposableImplementation foo = new SomeIdisposableImplementation()) { using(SomeIdisposableImplementation2 bar = new SomeIdisposableImplementation2()) { } } 

or will that be enough and the "bar" is located along with the "foo"?

 using (SomeIdisposableImplementation foo = new SomeIdisposableImplementation()) { SomeIdisposableImplementation2 bar = new SomeIdisposableImplementation2(); } 

Answer 1

or will that be enough and the "bar" is located along with the "foo"?

No, the panel will not be removed.

Operator

using converted to a try-finally block, so even if an exception occurs, the finally block provides a call to the Dispose method.

After

 using (SomeIdisposableImplementation foo = new SomeIdisposableImplementation()) { SomeIdisposableImplementation2 bar = new SomeIdisposableImplementation2(); } 

Translated to

 { SomeIdisposableImplementation foo; try { foo = new SomeIdisposableImplementation(); SomeIdisposableImplementation2 bar = new SomeIdisposableImplementation2(); } finally { if (foo != null) foo.Dispose(); } } 

Leave bar inactive.

Answer 2

So that both of them are installed using the using statement, you do not need to insert them, but you can write this

 using (SomeIdisposableImplementation foo = new SomeIdisposableImplementation()) { using(SomeIdisposableImplementation2 bar = new SomeIdisposableImplementation2()) { } } 

as

 using (SomeIdisposableImplementation foo = new SomeIdisposableImplementation()) using(SomeIdisposableImplementation2 bar = new SomeIdisposableImplementation2()) { } 

Answer 3

In the second version, the panel will go out of scope, but will not be removed. But you can put both foo and bar in the same command:

 using (SomeIdisposableImplementation foo = new SomeIdisposableImplementation(), SomeIdisposableImplementation2 bar = new SomeIdisposableImplementation2()) { // use foo and bar } 

you can also just put the variables in the use command:

 SomeIdisposableImplementation foo = new SomeIdisposableImplementation(); SomeIdisposableImplementation2 bar = new SomeIdisposableImplementation2(); using (foo, bar) { // use foo and bar }