Saturday 8 June 2019

Why don't Java's +=, -=, *=, /= compound assignment operators require casting?




Until today, I thought that for example:



i += j;


Was just a shortcut for:



i = i + j;



But 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)?


Answer



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.




An example cited from §15.26.2





[...] 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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