Java += operator
76
Until today I thought that for example:

is just a shortcut for:

But what if we try this:

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)?

I've tried googling for it but couldn't find anything relevant.
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Titus Fadel Created at: 2013-11-13 17:07:13 UTC By Titus Fadel
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7 Answers
0
In Java type conversions are performed automatically when the type of the expression on the right hand side of an assignment operation can be safely promoted to the type of the variable on the left hand side of the assignment. Thus we can safely assign:  

 byte -> short -> int ->  float -> long -> double.   

The same will not work the other way round. For example we cannot automatically convert a long to an int because the first requires more storage than the second and consequently information may be lost. To force such a conversion we must carry out an explicit conversion.Type - Conversion
0
Yes,

basically when we are writing 

i+=l; 


compiler is converting this to 

  i = (int)(i + l);


I just checked the .class file code.

really a good thing to know
0
Very good question. The Java Language specification confirms your suggestion.


  For example, 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);


0
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'

0
you need to cast from long to int explicitly in case of i = i + l  then it will compile and give correct output. like 

i = i + (int)l;


or

i = (int)((long)i + l); // this is what happens in case of += , dont need (long) casting since upper casting is done implicitly.


but in case of += it just works fine because the operator implicitly does the type casting from type of right variable to type of left variable so need not cast explicitly.
0
The problem here is of type casting.

When you add int and long, 

The int object is casted to long & both are added and you get long object.
but long object cannot be implicitly casted to int. So, you have to that explicitly.
But += is coded in such a way that it does type casting. i=(int)(i+m)
0
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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