what does "+=" mean in Java?

[is55]

what does "+=" mean in Java?

[JosAH]

what does "+=" mean in Java?

It's the 'add assignment' operator; e.g.

x= 20;
x+= 22; // x is 42 now

There's also an operator -=, *=, /= etc. that subtract, multiply, divide etc. a value from a variable.

kind regards,

Jos

[al_Marshy_1981]

e.g x=20;
x+=44;

the above code says x is equal to 20. the next line of code says hey take x current value and add 44 so now x is 64.

+= means take the variable before + current value and add what is on the right of the equals sign to the current value of what is before the + sign.
it essentially could be written as x=x+44; which could be x=20+44;

[Eranga]

In other words both x += 20 and x = x + 20 are exactly the same.

Read the following small article, then you can have a better idea about that.

How to Use Assignment Operators in Java | eHow.com

[JosAH]

In other words both x += 20 and x = x + 20 are exactly the same.

No they are not; compare this:

byte x= 0;
int y= 3;
x+= y;

with this one:

byte x= 0;
int y= 3;
x= x+y;

kind regards,

Jos

[Eranga]

True. But I mentioned it because I don't think that Op really touch on with different data types. If he still confused with compound assignment, I don't want to talk about that.

[al_Marshy_1981]

I dont understand the difference josAh it still seems like 0+3 to me?

[Zack]

byte a = 0;
int b = 3;
b = (int)a + b;
System.out.println(b);

byte x = 0;
int y = 3;
x += y;
System.out.println(x);

That prints 3 and 3. However, if you do not cast a to int (or do a=a+b instead), it will provide an error due to precision. Truly a byte can't have 3 added to it, cause it's... a byte. :P

[al_Marshy_1981]

byte a = 0;
int b = 3;
b = (int)a + b;
System.out.println(b);

byte x = 0;
int y = 3;
x += y;
System.out.println(x);

That prints 3 and 3. However, if you do not cast a to int (or do a=a+b instead), it will provide an error due to precision. Truly a byte can't have 3 added to it, cause it's... a byte. :P

of course yea, thanks got it now

[al_Marshy_1981]

However to be fair I think Eranga was assuming the same data types in her answer to the OP

[Zack]

Yes. Assuming same data types it is always the case.

[Eranga]

I dont understand the difference josAh it still seems like 0+3 to me?

If you've run the code segment at least once, you can see the difference. ;)

[Eranga]

Yes. Assuming same data types it is always the case.

Yep, I assumed that. My point is when answering to a question, only related with the original as much as possible. Because I hope person who post the question will workaround with all the related.

[JosAH]

byte a = 0;
int b = 3;
b = (int)a + b;
System.out.println(b);

byte x = 0;
int y = 3;
x += y;
System.out.println(x);

That prints 3 and 3. However, if you do not cast a to int (or do a=a+b instead), it will provide an error due to precision. Truly a byte can't have 3 added to it, cause it's... a byte. :P

That's a very weak reasoning ;-) In the expression 'l op= r' the righthand operand 'r' is converted to the type of the left hand operand 'l' and the operator 'op' is performed on 'l' and 'r'. The result is stored in 'l'.

In the expression 'l= l op r' both operands are converted to the widest type before the operator 'op' is performed. If the type of the result is wider than the type of 'l' the compiler protests about it because there could be possible loss of precision.

So in my example 'x+= y' the value of 'y' is converted to byte type while in the expression 'x= x+y' both 'x' and 'y' are converted to ints because 'y' is of type int.

kind regards,

Jos