Friday, 22 July 2016

java - Adding int to short




A colleague of mine asked this question to me and I am kind of confused.



int i = 123456;
short x = 12;


The statement




x += i;


Compiles fine however



x = x + i;


doesn't




What is Java doing here?


Answer



int i = 123456;
short x = 12;
x += i;


is actually



int i = 123456;

short x = 12;
x = (short)(x + i);


Whereas x = x + i is simply x = x + i. It does not automatically cast as a short and hence causes the error (x + i is of type int).







A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.




- JLS §15.26.2



No comments:

Post a Comment

c++ - Does curly brackets matter for empty constructor?

Those brackets declare an empty, inline constructor. In that case, with them, the constructor does exist, it merely does nothing more than t...