how to solve this?

Question about the task

Task No. 6 about integer type conversions

Java Syntax, Level 10, Lesson 4

Resolved

my answer is 999.488

Remove unnecessary operators to get the correct answer:
result: 1000.0


double d = (short) 2.50256e2d;
char c = (short) 'd';
short s = (short) 2.22;
int i = (short) 150000;
float f = (short) 0.50f;
double result = f + (i / c) - (d * s) - 500e-3;

Requirements:

The program should display text on the screen.
Don't change the screen output command.
The main() method must contain a char variable c.
The main() method must contain a short variable s.
The main() method must contain an int variable i.
The main() method must contain a float variable f.
The main() method must contain a double variable d.
The main() method must contain a double variable result.
Don't change the initial value of the variables during initialization. You can only change the cast operators.
The program should display "result: 1000.0".

package com.codegym.task.task10.task1006; /* Task No. 6 about integer type conversions */ public class Solution { public static void main(String[] args) { double d = 2.50256e2d; char c = 'd'; short s = (short) 2.22; int i = (int) 150000; float f = 0.50f; double result = f + (i / c) - (d * s) - 500e-3; System.out.println("result: " + result); } }

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Johannes

Level 27 , Centurion, Pretoria, South-Africa

17 March 2020, 14:02

Another of those time-wasters...... Illogical task. I've in 25 years of programming on a financial system NEVER come across manipulating and warping figures to achieve a result. I get that they want to teach casting, but a couple of normal, easy examples should do it, then move on.

+14

polbatona

Level 17 , Omsk, Russian Federation

30 June 2019, 19:35

remove 13 и 14 only.

MecoS Full Stack Developer at Microsoft

17 December 2019, 12:48

Thanks!

Nikesh

Level 15 , Delhi, India

1 June 2019, 06:26

In line 10 do this : double d = (short) 2.50256e2d;

ThomasLC

Level 16 , Singapore, Singapore

25 May 2019, 15:03

solution

A good way to solve these variable conversion questions would be to do a print at every variable to see what exactly are they being turned into: i.e. double d = 2.50256e2d; System.out.println("D: " + d); char c = (short) 'd'; System.out.println("C: " + c); short s = (short) 2.22; It's a great way to learn/understand what's going on Now for your hint: "e" means times 10 to the power of x. so 150e1 = 150 times 10 to the power of 1, or 1,500 330e-4 = 330 times 10 to the power of negative 4, or 0.033 The character "d" is also unicode for 100 in the equation

+12