Primitives float and double: why does f+=d not result in Type Mismatch Error?

Question

Possible Duplicate:
Java += operator

Code example:

    double d = 1;
    float f = 2;
    f += d;  // no error?
    f = f+d; // type mismatch error, should be f = (float) (f+d);

So why does f+=d not produce an error (not even at runtime), although this would decrease the accuracy of d?

The linked dupe is for `int` and `long`, but it's the same situation here with `float` and `double`. — Mysticial, Dec 11 '12 at 14:51
Thank you for pointing out the duplicate. I did a lot of searching, but "+=" is a bad term to search for. My apologies. — sulai, Dec 11 '12 at 15:04

kosa · Answer 1 · 2012-12-11T15:01:30.577

5

As per JLS 15.26.2

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.

That means:

f += d;

would become as

f = (float) (f+d);

edited Dec 11 '12 at 15:01

answered Dec 11 '12 at 14:51

kosa

65,990
13
130
167

I think you mean `f = (float)(f + d);` ;) – Peter Lawrey Dec 11 '12 at 14:53
I hope you don't mind if I merge your excellent answer into Peter Lawrey's answer? – sulai Dec 11 '12 at 15:18
@sulai: Sure whatever the way you wish. – kosa Dec 11 '12 at 15:19

Peter Lawrey · Accepted Answer · 2012-12-11T15:27:27.117

4

The compount assignment does an implicit cast.

a #= b;

is equivalent to

a = (cast to type of a) (a # b);

Another example

char ch = '0';
ch *= 1.1; // same as ch = (char)(ch * 1.1);
// ch is now '4'

edited Dec 11 '12 at 15:27

answered Dec 11 '12 at 14:52

Peter Lawrey

525,659
79
751
1,130

1

Merged in [Nambari's anwser](http://stackoverflow.com/a/13822265/1208581), both answers excellent. – sulai Dec 11 '12 at 15:26
1

@sulai Improving answers is always welcome. ;) – Peter Lawrey Dec 11 '12 at 15:26

Primitives float and double: why does f+=d not result in Type Mismatch Error?

2 Answers2