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public class q2 { public double d; public q2(double d) { this.d = d; } public double calculate(int i) { if (i < 0) { double mThe following classes named q2 and test are given. What is the value of the variable resA after successfully running the prog

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Answer #1

When an object of q2 is created , the value stored in it's d = 36.0

In the next statement , the scope of variable d declared is in only in the main .

When calculate(16) is called then the value 16 is passed is (i) in the function named calculate.

Since this value of i which is 16, is greater than 0 , therefore else statement will be executed resulting in d = i * d = 16*36 = 576.0 and the next statement will print d(calculate) == 576.0

Thus calculate(16) returns 576.0 and there is typecasting of int therefore resA variable will store 576

Hope this solves your doubt , please give an upvote if it does .

THANK YOU.

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