#Conversion between double and int

11 messages · Page 1 of 1 (latest)

lone rampart
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I was trying to solve a simple question for my homework as below.

// variable dm represents deciemeter and has been initialized earlier in //the code, e.g as
//int dm = 1;

// Correct the formula down below so it will pass all test cases
double meter =  dm/10;
System.out.print( meter );

At first, I thought nothing was wrong and hit "check answer", which returned value 0, the expected value was 0.1. After some testing I concluded that the correct way was to declare meter as java double meter = (double) dm/10
But why is this the correct way? Is it because dm was declared as an int, so 0.1/10 is calculated by the computer, assuming that the answer should be an integer? If this is the case, why doesn't the computer already figure it out when I declared meter as a double?

stone lavaBOT
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plain tide
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The incorrect example was doing 1/10, which is strictly integer division, and any decimal places are just truncated.

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Java (and most languages) allow you to assign a double variable to an integer expression, and it'll implicitly convert it. That's why the compiler doesn't automatically change the expression to a double calculation

lone rampart
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So if I understand you correctly, the compiler evaluates each term and deduces the type, by comparing the involved types e.g dm/10 should have an integer answer since dm and 10 are integers, but double r = 1.0 and r/10 should have a double answer since r is a double and 10 is an integer

plain tide
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Yes

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Essentially, the resultant type of a whole mathematical expression is just a result of the types that make it up

lone rampart
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Ahh, now I understand!

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Thank you so much Andrew for taking your time and explaining everything in depth : )

stone lavaBOT