Question

Reading in a String with a space character in Java.

Answer 1

The problem is when I use nextLine() to read in strings with spaces, for some reason it skips scanning in gender.

Answer 2

@e.mccormick

Answer 3

Because of:
age = scanner.nextInt();

The changing back and forth between int and line.

Answer 4

http://stackoverflow.com/questions/7056749/scanner-issue-when-using-nextline-after-nextxxx

Answer 5

Ah, thanks, I got it working now.

```
age = 0;

try {
age = Integer.parseInt(scanner.nextLine());
} catch (NumberFormatException e) {
e.printStackTrace();
}

Answer 6

I just switched it to convert the nextLine() to an int. :D

Answer 7

Yah, it has to do with the newline character that gets left in the buffer.

Answer 8

I found out when I ran into the same thing.

Answer 9

Ah, well I just started up a java class this semester, it seems quite similar to C++

Answer 10

Oh yah, but objects are not an option in Java and pass by reference is replaced by use of objects. Otherwise, a lot of the syntax is similar.

Answer 11

so when I do

people[i-1] = new Person(name, age, gender);

I'm not adding an object to an array?

Answer 12

I don't understand what you mean when you say, objects are not an option.

Answer 13

You can do a lot of thigns in C++ without making any objects. Just functions and the main. In Java, the instant you make a Scanner you made an object.

Answer 14

Oh, That makes sense, lol. I was thinking it meant you couldn't use objects in java.

Answer 15

No, no. There is no option to NOT use them. Hehe. They are 100% required. Even the main is treated as an Object. It is one of the fully OOP languages.

Answer 16

Right right, makes sense now, thanks for the help! I'm gonna get back to the programming, I still seem to be getting and java.lang.ArrayIndexOutOfBoundsException: 2

Answer 17

Computer math: things start at 0.

Answer 18

Right, I switched the for loop to start at 1 because I needed my print statement where it prints i to start at 1.

Answer 19

Not that loop. You are adjustng it down at the end. This loop:
for(int i = 0; i <= entries; i++){
That means for 10 items:
0,1,2,3,4,5,6,7,8,9,10

Oops.

Answer 20

With this I still seam to be getting

```
Exception in thread "main" Stats: sdf, 3, sdf
Stats: df, 3, df
java.lang.ArrayIndexOutOfBoundsException: 2
.......................................LabTwoDriver.main(LabTwoDriver.java:48)

Answer 21

seem*

Answer 22

//Write code to print out the array of objects and their attributes
for(int i = 0; i <= entries; i++){

Answer 23

woops, I still left the <= in the print code haha

Answer 24

Boom, working now :D

Answer 25

Yah, the first one was doing entries things. The second loop was entries + 1, so out of range.

Answer 26

Yeps, glad to have that working now, got a lab tomorrow, thanks for all the help!