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Start your free trialGustavo Borges
309 PointsI cant pass this exercise!
In my perspective, I'm not doing anything wrong.
But this message keeps coming, every time I ask to check my work:
"Did you forget to pass the lastName
parameter to the printf function?"
I think I a;ready passed the lastName
parameter to the printf function, right?
Could someone help me?!
// I have imported java.io.Console for you. It is a variable called console.
String firstName = console.readLine("What is your first name? ");
String lastName = console.readLine("What is your last name? ");
console.printf("First name", firstName);
console.printf("%s", firstName);
console.printf("Last name: ", lastName);
console.printf("%s", lastName);
3 Answers
Gustavo Borges
309 PointsThanks a lot!
That's weird... because before the lastName step, I used two lines (as you could in my code) and it worked.
Once I tried to do the same thing in the last step, the system didn't allow me to finish it.
But I understand your point!
Again, thanks a lot!!!
Grigorij Schleifer
10,365 PointsHey,
glad I could help :)
Grigorij Schleifer
10,365 PointsHey Gustavo,
look at my code suggestion. I added just a few modifications.
String firstName = console.readLine("What is your first name? ");
String lastName = console.readLine("What is your last name? ");
console.printf("First name: %s", firstName);
console.printf("Last name: %s", lastName);
Grigorij
David Lacedonia
13,627 PointsThis works:
console.printf("First name: %s", firstName);
console.printf("Last name: %s", lastName);
Don't pass any parameter when you dont have %s.
If you want the printf methods in diferent lines, also you can do...
console.printf("First name: ");
console.printf("%s",firstName);
console.printf("Last name: ");
console.printf("%s",lastName);
Grigorij Schleifer
10,365 PointsGrigorij Schleifer
10,365 PointsYour code is almost fine, but the environment is complaining because two lines are missing the string formatters.