Android Javascript Confusion

In programming language initialisation literally means giving a value while declaration of a type.

To understand this lets pay attention to this line

TextView label = (TextView) findViewById(R.id.label1);

There are two parts here

On the left had side (TextView label) is the declaration. A general format of the declaration is data_type variable_name.

Data_type (which represents what kind of data) can be of any data type for e.g. int, float, etc. and in our case TextView and variable_name could be any of your choice.

Now when we declare a variable, we are simply saying that label is of a type TextView. That's all. label as of now has no value stored in it.

(A lot many time when you are programming you get a compile time error that some_variable was not initialised. Which actually means that you declared the variable but never assigned any value to it)

[Excuse me for the build up but it was required for better understanding]

Now coming to the right hand side (findViewById) of the statement above. This statement puts a reference to the id value of the TextView label1 into the variable label. So after the statement executes, label will hold a reference value to the label1 TexView. You can read about differences between primitive types and reference types in Java for more clarity .

So in short "Initialize the getAnswerButton variable using the findViewById() method" ** means assigning a value (in this case a reference to id value of the TextView) through findViewById method and store it in the getAnswerLabel variable which can then be used elsewhere in the program with ease **

Hope this helps. Ask again if you have doubts.