Java += and implicit casting
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Java += and implicit casting
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This is from two popular answers to the question
Java += operator
Until today I thought that for example:
is just a shortcut for:
But what if we try this:
Then
Does it mean that in fact
I've tried googling for it but couldn't find anything relevant.
A good example of this casting is using *= or /=
or
or
or
Question
Until today I thought that for example:
i += j;is just a shortcut for:
i = i + j;But what if we try this:
int i =5;long j =8;Then
i = i + j; will not compile but
i += j; will compile fine.
Does it mean that in fact
i += j; is a shortcut for something like this
i = (type of i) (i + j)?
I've tried googling for it but couldn't find anything relevant.
Answers
As always with these questions, the JLS holds the answer. In this case §15.26.2 Compound Assignment Operators. An extract:
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.
And an example:
For example, the following code is correct:
short x =3; x +=4.6;and results in x having the value 7 because it is equivalent to:
short x =3; x =(short)(x +4.6);
In other words, your assumption is correct.
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A good example of this casting is using *= or /=
byte b =10;
b *=5.7;System.out.println(b);// prints 57or
byte b =100;
b /=2.5;System.out.println(b);// prints 40or
char ch ='0';
ch *=1.1;System.out.println(ch);// prints '4'or
char ch ='A';
ch *=1.5;System.out.println(ch);// prints 'a'Sensu: workflow automation for monitoring. Learn more—download the whitepaper.
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