[FIXED] In Java, What is the order of precedence for typecasting, ( ) and math operation, like explained in the example below

Issue

int[] A = {1000000007,1000000007,1000000007};

This overflows

System.out.println("Bracket mul: "+(long) (A[0]*A[0]));

Output:

Bracket mul: -371520463

This doesnt - multiplication without brackets

System.out.println("Integer Multiplication: "+ (long) A[0]*A[0]);

Output:

Integer Multiplication: 1000000014000000049

I know how to make use of these now, but want to understand if typecasting is applied to first value and then operation is performed if brackets are not there ?

and if brackets are there, it is treated as integer literal operation, correct me if I am wrong

Solution

Here

(long) (A[0]*A[0])

you're doing a multiplication between integers and then casting the (already overflowed) result to long.

Here

(long) A[0]*A[0]

which is read as

((long)A[0]) * A[0]

you're casting the first operand to long, so the operation will be between longs (because the second operand will be promoted to long as well) and the result will be a long and won't overflow.

Answered By - Federico klez Culloca
Answer Checked By - Candace Johnson (JavaFixing Volunteer)