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johnacosta
275 Points%s
i was told to replace my name with the string formatter. I am assuming that it is %s. but I keep getting an error. Please help.
// I have setup a java.io.Console object for you named console
String firstName = "John";
console.printf("%s");
console.printf("%s can code in Java");
2 Answers
Steven Parker
229,744 PointsFor every format specifier (like "%s") in the format string, there should be an additional argument that supplies the value that will be rendered in that place.
So in this case, the specifier represents where your name will go, so you must supply the variable that contains your name as a second argument:
console.printf("%s can code in Java", firstName);
johnacosta
275 PointsGotcha! Thanks! Its hard being new lol
Steven Parker
229,744 PointsKeep plugging, you'll get there. Meanwhile, you know where to get to help now.
Happy coding!
James Stride
Courses Plus Student 621 PointsJames Stride
Courses Plus Student 621 PointsWhen you use the string formatter. You have to put the string name after the end quote and a coma. So the compiler knows which variable you're talking about. Otherwise it comes up with nothing and therefore it doesnt know what you mean by %s. Hope this helps. Let me know. Have fun coding bro! For ex.
String firstName = "John"; console.printf(" %s can code in Java", firstName);