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Anurag Gracian
315 Pointsi do not know why my string variable is not working
i defined the string variable. but the code is not working
// I have setup a java.io.Console object for you named console
String firstName = "Anurag";
console.printf("%s\n, firstName");
}
}
1 Answer
Michael Liebegott
3,960 PointsHey there Anurag!
Your code is almost perfect.
The console.printf() function requires one extra argument for every formatter you pass into it. Take this as an example:
console.printf("%s %s %s %s");
//This will throw an error.
You have 4 string formatters, which means you must have 4 strings (either as literals or variables) passed into the method.
The code, when properly written, would look like this:
String one = "one";
String two = "two";
String three = "three";
String four = "four";
console.printf("%s %s %s %s", one, two, three, four);
//The console will display: one two three four
For your code specifically, you need to remove "firstName" from the string and rewrite it like so:
// I have setup a java.io.Console object for you named console
String firstName = "Anurag";
console.printf("%s\n", firstName);
}
}
This is telling the printf() function that you have a string ("%s\n") with one variable/literal being passed into it (firstName).
Hopefully this helps. :-)