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Java += operator


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) ?



I've tried googling for it but couldn't find anything relevant.


Source: Tips4allCCNA FINAL EXAM

Comments

  1. 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.


    And an example:


    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);



    In other words, your assumption is correct.

    ReplyDelete
  2. 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'

    ReplyDelete
  3. 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);

    ReplyDelete
  4. 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

    ReplyDelete
  5. 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.

    ReplyDelete
  6. 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)

    ReplyDelete

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