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Kenneth Lamere
51 Pointsstring Formatter
Hi. I am having a problrm with: Now replace <YOUR NAME> in the console.printf expression with the firstName variable using the string formatter.
// I have setup a java.io.Console object for you named console
String firstName="ken";
console.printf("ken can code in java!");
console.prinf("my name is %s\n");
2 Answers
Allan Clark
10,810 PointsYou told the formatter where the string variable will go but not what variable to put there. Add the firstName variable to your arguments for printf(). Don't forget the comma! :D
Robert Richey
Courses Plus Student 16,352 PointsHi Kenneth,
It looks like your code was good for the first two tasks. Let's take another look at task 3.
Now replace <YOUR NAME> in the console.printf expression with the firstName variable using the string formatter.
We just need one printf statement for this.
String firstName = "ken";
console.printf("%s can code in Java!", /* what goes here? */);
When passing in a formatted string, printf takes comma separated arguments. The first argument is the string with placeholders, like %s. Then, each additional argument represents the values to insert into the string at those placeholders.
String food = "pizza";
console.printf("I love %s.", food);
// output: I love pizza.
Hope this helps,
Cheers