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Start your free trialHassan Osman
628 Pointsstring firstname="hassan";console.printf("my name is hassan",firstname);
whats wrong
// I have setup a java.io.Console object for you named console
string firstname= "HAssan";console.printf("my name is hassan");
3 Answers
Ryan Ruscett
23,309 PointsString first name = "hassan"; < --- That's not right. This says create a String variable named first with .... ?
String firstName = "hassan"; <--- Says create a String variable with firtName as it's reference equal to Hassan
Perhaps you should watch the video again.
Ishay Frenkel
1,620 PointsYour code should look like this:
String firstName = "Hassan";
console.printf("My name is %s", firstName);
Okay, now the explanation:
First of all, make sure you type String
with capital S
, this is important.
variables should not contain spaces first name
is wrong. Also best practice is to write with camel cases (thisIsAnExampleForTypingInCamelCase), FYI, you can't name variables with numbers in the beginning (2ndName is bad, but name2 is okay)
After a semicolon, move down a line, that's also a best practice and makes your code look better and more understandable.
if you type "my name is Hassan" you are not using the variable so use the variable, not the plain text of your name.
Ryan Ruscett
23,309 PointsThe first thing I see is that you didn't capitalize the s in String. You must capitalize String for reasons to complicated to explain here.
Then your printf is a formatting print. So you need TWO arguments. So you need a string to print and a variable in which to use inside the string.
%s is a place holder within a string for some variable. For example
String variable = "cool";
console.printf("Something is %s", variable);
This returns
Something is cool.
You can put them on the same line but I wouldn't do that at this point.
Let me know if that clears it up for you!
Hassan Osman
628 Pointshere is my code not working String first name = "hassan"; console.printf("my name is %s",firstname);