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Yes, just look at it as:<br />
<br />
x = a ++ + y ++ ;<br />
<br />
becomes<br />
<br />
x = ( a ++ ) + ( y ++ ) ;<br />
<br />
So a ++ and y ++ are evaluated first (as the current values of a and y respectively) and then added to be assigned to x. The values of a and y are then incremented after the statement.
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Yes, just look at it as:

x = a ++ + y ++ ;

becomes

x = ( a ++ ) + ( y ++ ) ;

So a ++ and y ++ are evaluated first (as the current values of a and y respectively) and then added to be assigned to x. The values of a and y are then incremented after the statement.

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mayankeagle:<br />
<br />
it seems yu added the comment during I was closing it, so I missed yours, I am sorry. how to add points to you now?
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mayankeagle:

it seems yu added the comment during I was closing it, so I missed yours, I am sorry. how to add points to you now?

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<div class="content wysiwyg-content">
int x, a=6, y=7;<br />
x = a++ + y++;<br />
<br />
x will be 13, <br />
but I think ++ has high precendence than + ,<br />
so a++, y++ will be exercuted before +, <br />
so I think x should be 15.<br />
<br />
please let me know what is wrong in my think.<br />
<br />

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int x, a=6, y=7;
x = a++ + y++;

x will be 13, 
but I think ++ has high precendence than + ,
so a++, y++ will be exercuted before +, 
so I think x should be 15.

please let me know what is wrong in my think.



x = a++ + y++; what is x?

Java is a platform-independent, object-oriented programming language and run-time environment, designed to have as few implementation dependencies as possible such that developers can write one set of code across all platforms using libraries. Most devices will not run Java natively, and require a run-time component to be installed in order to execute a Java program.