The expression (done | | s.compareTo(t) < 0) is true
MIT 6.00 Intro Computer Science (OCW)
Stacey Warren - Expert brainly.com
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assume that boolean done = false, int x = 10, int y = 11, String s = "Help" and String t = "Goodbye"
This is a Java problem
What does s.compareTo(t) do? I assume it is returning a value less than zero.
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it compares string S with t letter by letter. and returns true if both are equal
Can someone paste the code or problem here please?
This seems like a very strange thing to do and I don't know if there is a question here or what is is but evaluating the first expression since done is false and s is not the same as t we get.
false || false < 0
since || is a logical OR operator we have
false or false < 0
so how does one evaluate false < 0 ? well in some languages, false is just a fancy 0 and true is just a fancy 1 so it could potentially evaluate to false. In some languages it would say boolean start with 'b' and integer starts with 'i' and therefore it could evaluate to true since 'b' is less than 'i' and give us our two options
false or false which is false
false or true which is true
I don't know if this helps or if it answers the question but there it is.