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


Answers

A good example of this casting is using *= or /=

byte b = 10;
b *= 5.7;
System.out.println(b); // prints 57

or

byte b = 100;
b /= 2.5;
System.out.println(b); // prints 40

or

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'
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)?




Subtle point here...

There is an implicit typecast for i+j when j is a double and i is an int. Java ALWAYS converts an integer into a double when there is an operation between them.

To clarify i+=j where i is an integer and j is a double can be described as

i = <int>(<double>i + j)

See: this description of implicit casting

You might want to typecast j to (int) in this case for clarity.




Sometimes, such a question can be asked at an interview.

For example, when you write:

int a = 2;
long b = 3;
a = a + b;

there is no automatic typecasting. In C++ there will not be any error compiling the above code, but in Java you will get something like Incompatible type exception.

So to avoid it, you must write your code like this:

int a = 2;
long b = 3;
a += b;// No compilation error or any exception due to the auto typecasting



Yes,

basically when we write

i += l; 

the compiler converts this to

i = (int)(i + l);

I just checked the .class file code.

Really a good thing to know




The problem here involves type casting.

When you add int and long,

  1. The int object is casted to long & both are added and you get long object.
  2. but long object cannot be implicitly casted to int. So, you have to do that explicitly.

But += is coded in such a way that it does type casting. i=(int)(i+m)




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