Code is not initializing after

[cloud heron]

so our code is not working after being asked if a name is correct in our code , getting this error message
java.lang.NullPointerException
at java.util.regex.Matcher.getTextLength(Matcher.java:1283)
at java.util.regex.Matcher.reset(Matcher.java:309)
at java.util.regex.Matcher.<init>(Matcher.java:229)
at java.util.regex.Pattern.matcher(Pattern.java:1093)
at java.util.Formatter.parse(Formatter.java:2547)
at java.util.Formatter.format(Formatter.java:2501)
at java.io.PrintStream.format(PrintStream.java:970)
at java.io.PrintStream.printf(PrintStream.java:871)
at HernandezRileyTerrazas001PA2.setLaptopChoice(HernandezRileyTerrazas001PA2.java:154)
at HernandezRileyTerrazas001PA2.processLaptopOrder(HernandezRileyTerrazas001PA2.java:110)
at HernandezRileyTerrazas001PA2.main(HernandezRileyTerrazas001PA2.java:51)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:498)
at edu.rice.cs.drjava.model.compiler.JavacCompiler.runCommand(JavacCompiler.java:259)
could really use some help if possible

[solid flicker]

<@&987246399047479336> please have a look, thanks.

[minor pumice]

Which line is 154 of HernandezRileyTerrazas001PA2?

[cloud heron]

it would be int choice = 0;

[minor pumice]

It wasn't when the exception occurred. That line would have a printf on it.

[cloud heron]

In line 158 we had our printf that was still in the do while loop

[minor pumice]

If the file has changed... maybe the error has too.

[cloud heron]

gotcha gotcha

right now this is where my group is getting stuck at

[minor pumice]

So what is on line 161?

[cloud heron]

while(!input.hasNextInt()) //As long as the input is "NOT" an integer

[solid flicker]

Mixing any nextXXX method with nextLine from the Scanner class for user input, will not ask you for input again but instead result in an empty line read by nextLine.

To prevent this, when reading user input, always only use nextLine. If you need an int, do

int value = Integer.parseInt(scanner.nextLine());

instead of using nextInt.

Assume the following:

Scanner scanner = new Scanner(System.in);

System.out.println("Enter your age:");
int age = scanner.nextInt();
System.out.println("Enter your name:");
String name = scanner.nextLine();

System.out.println("Hello " + name + ", you are " + age + " years old");

When executing this code, you will be asked to enter an age, suppose you enter 20.
However, the code will not ask you to actually input a name and the output will be:

Hello , you are 20 years old.

The reason why is that when you hit the enter button, your actual input is

20\n

and not just 20. A call to nextInt will now consume the 20 and leave the newline symbol \n in the internal input buffer of System.in. The call to nextLine will now not lead to a new input, since there is still unread input left in System.in. So it will read the \n, leading to an empty input.

So every user input is not only a number, but a full line. As such, it makes much more sense to also use nextLine(), even if reading just an age. The corrected code which works as intended is:

Scanner scanner = new Scanner(System.in);

System.out.println("Enter your age:");
// Now nextLine, not nextInt anymore
int age = Integer.parseInt(scanner.nextLine());
System.out.println("Enter your name:");
String name = scanner.nextLine();

System.out.println("Hello " + name + ", you are " + age + " years old");

The nextXXX methods, such as nextInt can be useful when reading multi-input from a single line. For example when you enter 20 John in a single line.

simon_verhoeven • used /tag id: scanner · Mar 25, 2024, 4:03 AM