The Solutions:
Solution 1: Implied Cast in Compound Assignment Operators
The Java Language Specification (JLS) defines the behavior of compound assignment operators such as *=
, +=
, and -=
. According to the JLS, when using these operators, an implicit cast is performed to convert the right-hand operand to the type of the left-hand operand.
This means that in the expression x *= 0.90;
, the double
value 0.90
is implicitly cast to int
before being multiplied by x
. This results in the correct result, as x
is an int
variable and can only hold integer values.
On the other hand, in the expression x = x * 0.90;
, the double
value 0.90
is not implicitly cast to int
, and the result of the multiplication is a double
. Since x
is an int
variable, assigning a double
to it results in a lossy conversion error, as the fractional part of the double
will be lost.
To fix this error, you can explicitly cast the double
value 0.90
to int
using the expression (int)0.90
before performing the multiplication. This will ensure that the result of the multiplication is an int
and can be assigned to x
without a lossy conversion error.
Solution 2: Java’s implicit conversion
Java’s implicit conversion rules sometimes lead to unexpected results. This is especially true when working with different data types, such as int
and double
.
In the first example, the expression x *= 0.90;
is valid because the result of the multiplication is a double
, which is then implicitly converted to an int
before being assigned to x
. This is allowed because Java automatically widens primitive data types to accommodate larger values.
In the second example, the expression x = x * 0.90;
is invalid because the result of the multiplication is a double
, and there is no implicit conversion from double
to int
. To fix this error, you would need to explicitly cast the result of the multiplication to an int
using a cast operator, such as (int)
.
Here is a modified version of your code that uses an explicit cast:
int x = 18;
x = (int) (x * 0.90);
System.out.println(x);
This code will print 16
, just like the first example.
The lesson here is to be aware of Java’s implicit conversion rules and to use explicit casts when necessary to avoid errors.
Solution 3: Concise Assignment without Conversion Errors
The crux of the issue is in the difference between the explicit assignment operator =
and the compound assignment operator *=
. The explicit assignment operator, x = x * 0.90;
, leads to a lossy conversion error because it requires an explicit conversion from double
(the result of x * 0.90
) to int
(the type of x
). This conversion is risky as it may result in information loss when the fractional part of the double
value is truncated.
In contrast, the compound assignment operator, x *= 0.90;
, elegantly handles this situation. It first implicitly converts the constant 0.90
from double
to int
. Since 0.90
is an exact integer representation, this conversion is safe and lossless. The product of x
and 0.90
is then computed using integer arithmetic, eliminating any lossy conversion.
To highlight this, consider the following examples:
- Lossy Conversion Error with Explicit Assignment:
int x = 5;
x = x * 0.90; // Error: lossy conversion from double to int
Here, the attempt to assign the result of x * 0.90
directly to x
triggers a lossy conversion error because x * 0.90
is a double
and x
is an int
.
- No Error with Compound Assignment:
int x = 5;
x *= 0.90; // No error: implicit conversion handled by compound assignment
With the compound assignment *=
, the conversion from double
to int
happens implicitly, and it’s guaranteed to be lossless because the constant 0.90
can be precisely represented as an integer.
Q&A
Why using compound assignment x *= 0.90;
does not give a lossy conversion
error but x = x * 0.90;
does?
The Java Language Specification implicitly casts the result of a compound assignment operation.
What is the reason behind this design decision in Java?
To maintain consistency in casting rules across different types and operators.
Can you provide an example to illustrate the difference?
With x *= 0.90;
, the result is implicitly converted to an int
, while with x = x * 0.90;
, the explicit assignment requires a cast to avoid a lossy conversion
error.