Why does Java compiler prints a value, while a C++ compiler outputs another in the next expression? | Sololearn: Learn to code for FREE!
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Why does Java compiler prints a value, while a C++ compiler outputs another in the next expression?

If we have: int x ; x = x + (x = 3); I know that what a C++ compiler would do first is to evaluate the expression from right to left, so the expression x = x + (x = 3) becomes x = 3 + 3. But, what about Java? I read that Java compiler uses right to left associaticity as well, but Java will print 4 in that expression. I'd like to know why this happens.

29th May 2020, 9:08 AM
Sergio Enrique ☕
Sergio Enrique ☕ - avatar
1 Answer
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Bro, you have got the wrong information!! Java uses left to right order of evaluation. So , firstly variable x needs to be initialized else it will throw an error message. Therefore int x=0; x=x+(x=3); x=0+3 x=3 Hence, it will output 3. Hope u understood !! 😃
1st Jun 2020, 7:22 AM
Reca Resu
Reca Resu - avatar