Adding int to short [duplicate]

Question

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?

For interest try this char ch = '0'; ch *= 1.2; now ch is '8' ;) — Peter Lawrey
– Peter Lawrey, Commented Sep 22, 2012 at 6:38

Community · Accepted Answer · 2020-06-20 09:12:55Z

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}

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Sep 21, 2012 at 20:27

arshajii

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Comments

Gilberto Torrezan · Accepted Answer · 2012-09-21 20:45:30Z

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.

edited Sep 21, 2012 at 20:45

answered Sep 21, 2012 at 20:29

Gilberto Torrezan

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Comments

jefflunt · Accepted Answer · 2012-09-21 20:26:51Z

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.

edited Sep 21, 2012 at 20:26

jefflunt

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answered Sep 21, 2012 at 20:25

asteri

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2 Comments

Em Ae Over a year ago

I wrote wrong equation. I updated my question. See its x = x + i.

lleite Over a year ago

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.

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