I ran into a situation earlier where I tried the following two bits of code:
int score = 100;
score = score * 1.05;
and
int score = 100;
score *= 1.05;
The first one failed (and obviously so, I'm trying to implicitly cast a float to an int). But the second one worked perfectly fine. The compiler didn't complain and I didn't get any runtime errors. Why does the second one work, while the first one doesn't? As far as I'm aware, x *= y
is just a shorthand for x = x * y
.
Compound assignment operators behave a little differently than their "expanded" version. Quoting the JLS, Section 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.
It is implicitly casted back to the type of the variable on the left side, so that there is no error casting a float
to an int
; it's already implicitly casted to an int
.
That does not occur with the =
operator, which is governed by the JLS, Section 5.2, Assignment Conversion:
Assignment contexts allow the use of one of the following:
an identity conversion (§5.1.1)
a widening primitive conversion (§5.1.2)
a widening reference conversion (§5.1.5)
a boxing conversion (§5.1.7) optionally followed by a widening reference conversion
an unboxing conversion (§5.1.8) optionally followed by a widening primitive conversion.
It goes on to talk about a possible narrowing conversion being allowed, but only for constant expressions, and only for the constant expression being a byte
, char
, short
, or int
, neither of which is applicable here.
The first one
int score = 100;
score = score * 1.05;
is basically saying:
int score = 100;
score = (float)(score * 1.05);
This is because if you times a float with an integer then you get a float. This then cannot get assigned to an integer.
However
int score = 100;
score *= 1.05;
Basically means
int score = 100;
score = (int)(score * 1.05);
This is because you are not assigning the float, the calculation is done whilst assigning therefore first converts to an int.
Thats how it makes sense to me. Hope it helps
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