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Joseph Davis
Courses Plus Student 2,022 PointsWhy do you need to convert a String into an int if you enter an int? int answer=Integer.parsel(answerToLife);
Why do you need to convert a String into an int if you enter an int? int answer=Integer.parsel(answerToLife);
Also, does java recogonise "answer" as separate as long as its lower case and connects them?
String answerToLife = "42";
int answer= Integer.parseInt(answerToLife);
3 Answers
Alexis Beliveau
8,029 Points"42" is of type String. Though it has a number in it, it is still a String data type. In order for the "answer" variable to hold the number 42, you must use Integer.parseInt(answerToLife); This will make "answer" equal to the int parsed from "answerToLife".
Joseph Davis
Courses Plus Student 2,022 Pointsbecause if its an integer why does it need to be a String? I'm sorry, I'm just lost.
Alexis Beliveau
8,029 PointsSo a string is a sequence of characters, the compilers doesn't necessarily care if it has numbers or letters in it. "H3llo" would still be a string. Think of it like characters being strung across the screen. So the answerToLife object is 42, but that's a string. If you were to try to combine answerToLife twice, it would not equal 84, because they are a sequence of characters, and not ints.
I would read the documentation a little
https://docs.oracle.com/javase/7/docs/api/java/lang/String.html
And if you don't want to read that here is my very loosely based example!
``` int answerToLife = 42; answerToLife + answerToLife = 84
String answerToLife = 42; answerToLife + answerToLife = "42 42" ```
Joseph Davis
Courses Plus Student 2,022 Pointsint is always mathematical. String is said and literal? Thank you very much. I was so pissed.
Joseph Davis
Courses Plus Student 2,022 PointsJoseph Davis
Courses Plus Student 2,022 PointsSo 42 is an integer but because its being typed in its a String?