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Java Java Basics Getting Started with Java Strings and Variables

Cesar Daniel Gonzalez Estrada
Cesar Daniel Gonzalez Estrada
2,546 Points
Posted by Cesar Daniel Gonzalez Estrada
Cesar Daniel Gonzalez Estrada
2,546 Points

Step 3 console.printf ("s% can code in Java!/n", firstName); What's wrong with my code?

String firstName = "Cesar";

console.printf ("s% can code in Java!/n", firstName);

What's wrong with my code?

3 Answers

bzfchen
bzfchen
6,484 Points
bzfchen
bzfchen
6,484 Points

It should be %s instead of s%

Also, \n instead of /n

Jakob Wozniak
Jakob Wozniak
17,896 Points
Jakob Wozniak
Jakob Wozniak
17,896 Points

You need %s for the string and %n for the new line. Good luck!

abmsc86
abmsc86
4,949 Points
abmsc86
abmsc86
4,949 Points

\n not %n for new line. :)

Jakob Wozniak
Jakob Wozniak
17,896 Points
Jakob Wozniak
Jakob Wozniak
17,896 Points

Actually, %n is a command for System.getProperty("line.seperator"), and is more consistent than \n. The \n works, but it gives you a Unix style line ending, just like \r\n gives you a Windows line ending.

%n will always find the system's new line function, so it's a best practice for this.