i can't understand

Question about the task

Under discussion

TBH i read he lecture 3 times but still didn't understand anyway to solve these programmes. almost every programme i use help. i know its a little wrong but then again if someone takes their time to explain it to me i'll be very thankful.

"Amigo, did you know that lunar gravity is about 17% of gravity on Earth?"
"Nope."
"Neither did I. Now this information will be used a lot. To avoid having to manually calculate it each time, implement a getWeight(int) method that takes a person's body weight on Earth (in newtons), and returns the weight of that person on the moon (in newtons).
The method should return a double."

Consider this example:
The getWeight method is called with the argument 888.

Example output:
150.96

Requirements:

The getWeight(int) method must be public and static.
The getWeight method must return a double.
The getWeight method should not display anything.
The getWeight method should correctly convert the Earth weight in newtons to the lunar weight, and then return this value.

package com.codegym.task.task01.task0136; /* Even to the moon! */ public class Solution { public static void main(String[] args) { System.out.println(getWeight(888)); } public static double getWeight(int earthWeight) { earthWeight(888)*17.0; return earthWeight; } }

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Daniel L. Tovar

Level 9

8 April 2020, 06:42

First of all, you haven't called any variable. The method won't work if you don't declare any variable. Second, remember the conditions. It is asking for double so the type of variable has to be double. Lastly, using the variable that you need to declare and the variable that is already in the method you have to declare a command so the code makes sense.

Gellert Varga

Level 23 , Szekesfehervar, Hungary

7 April 2020, 19:05

useful

A) Line 10: We invoke the getWeight() method. We send to this method the value 888. B) Line 13: the getWeight() method receives the value 888. It puts this value into the variable "earthWeight". C) Line 14: You can work here with variable "earthWeight". This variable contains the value 888 already. Calculate and give a new value to this variable! Line 15: "return" is OK. D) If you want to calculate for example 42 percent of "x", then don't multiply "x" by 42.0!

Jack

Level 4 , Yerevan, Armenia

8 April 2020, 15:34

Gellert Varga thank you for the tip i have 1 question, is ( * ) times and ( / ) dividing, right? if it is then i can't make the programme work let me get this straight... earthWeight is already (888) so if i do (earthWeight) / 17.0 why wont it work? and after that what should i do???

San

Level 20 , Poland

8 April 2020, 17:32

useful

Hi Jack, The variable earthWeight which is in the getWeight method holds the value 888 passed from the main method. The value of earthWeight variable does not change even though you are dividing it by 17.0, To get the result you can try the below: 1. Create a local variable of double data type and assign the value earthWeight / 17.0 and return the local variable. or 2. return earthWeight / 17.0 with no local variables. Hope this helps.

Gellert Varga

Level 23 , Szekesfehervar, Hungary

8 April 2020, 19:52

1) Line10: "earthWeight" is a simple variable. Don't use it this way: "eartWeight(888)". Only without any parentheses! In itself. And without "888". 2) How to assign a new value to a variable that already has an other value? For example, if x=100, and you want the new value of x should be equal to 20% of the old value. You can do it these ways: x = x*0.2; //the new value of x is 20. or x = (x/20.0)*100; //the new value of x is 20. 3) "and after that what should i do???" Nothing else! Because of: the Line 15 is good already.

Jack

Level 4 , Yerevan, Armenia

10 April 2020, 06:31

incompatible types: possible lossy conversion from double to int earthWeight = (earthWeight / 17.0) * 100.0; ^ can i see whats wrong here?

Gellert Varga

Level 23 , Szekesfehervar, Hungary

10 April 2020, 16:48

The earthWeight is a type "int". You want to assign a new value to the earthWeight, but the type of this new value is "double" already. Because if an "int" is divided by a double value (17.0) or multiplied by a double value (100.0), then in this case the result will also be double. And you can't put the "double" value into an "int" variable! Hint: Declare a new variable, with type of "double". Put in it the calculated double value. And "return" this double variable!