15

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?

Raedwald
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Em Ae
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3 Answers3

20
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

Community
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arshajii
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4

The + operator of integral types (int, short, char and byte) always returns an int as result.

You can see that with this code:

//char x = 0;
//short x = 0;
//byte x = 0;
int x = 0;
x = x + x;

It won't compile unless x is an int.

Gilberto Torrezan
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2

Numbers are treated as int unless you specifically cast them otherwise. So in the second statement when you use a literal number instead of a variable, it doesn't automatically cast it to the appropriate type.

x = x + (short)1;

...should work.

jefflunt
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asteri
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  • I wrote wrong equation. I updated my question. See its `x = x + i`. – Em Ae Sep 21 '12 at 20:32
  • Ok, then why doesn't x = x + x; work? The problem isn`t on the number 1 but on the + operator that returns an int. – lleite Sep 21 '12 at 20:34