We all have used syntax’s like
i += j and
i = i + j thousands of times in our day to day programming. In first sight, they both look similar. In fact, they will result in same output in almost all of the cases in practical cases. But, to surprise you they are not similar. In run-time, they are treated differently when i and j are of different types. Let’s look at the example below:
int i = 5; double d1 = (double)i + 4.5; //necessary to satisfy compiler i += 4.5; System.out.println(i); System.out.println(d1); Output: 9 9.5
Weird. Isn’t it. Both are expected to be same as operation is same. Why they have different values? Let’s find out.
Java language specification says following:
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.
So effectively our original example code can be re-written as below:
int i = 5; double d1 = (double) i + 4.5; i = (int)(i + 4.5); //Result converted to int System.out.println(i); System.out.println(d1); Output: 9 9.5
So the value 9 is nothing but a result of loss of precision while converting from double to int.
Always use the compound assignment operator [i += j] very carefully. You should use only when you are dealing with similar data types. In different data types, result can be incorrect.
Happy Learning !!
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