 Milo is a high school senior who is very involved politically and is currently volunteering for Elizabeth Warren's campaign. He is an active leader in a variety of extracurricular activities both at his school and in his community. Milo enjoys swimming, performing community service, learning about computer science, studying history, and reading news articles.

Once you recognize the frequent symbols used in Java, you’ll be a pro at understanding what’s happening in the code!

## Java Symbols to Know: 🏆

• ==  compares the value on the left and the value on the right
• % gives you the remainder when the value on the left is divided by the right
• ! means “not”, so != means “not equal to”
• i++ is the same as i += 1 or i = i+1
• i– is the same as i -=1 or i = i-1
• i*= 2 is the same as i = i*2

## Java Examples

### Example #1 🤔

```for (int i=0; i<5; i++)
{
System.out.print(i);
}```

What is printed as a result of executing this code?

• The value of i starts off at 0 when the for loop begins, so when i is printed, the computer prints “0”
• Since the loop increments by i++, the value of is now 1, and the computer prints “1”
• increases in value until i equals 4 and the computer prints “4”
• After that, i increases in value to 5, but 5 is not less than 5, so the computer exits the for loop and does not print anything else
• Since the code uses “print”, all of the numbers are printed on the same line

01234

### Example #2 😃

```for (int i=0; i<5; i+=1)
{
if (i%2 == 0)
{
System.out.print(i);
}
​
else if (i!=3)
{
System.out.println(i*2);
}

else
{
System.out.print("Three");
}
}```

What is printed as a result of executing this code?

• The value of i starts off at 0 when the for loop begins
• Since 0%2 equals 0 (since the remainder when 0 is divided by 2 is 0), the first if statement is true and the code inside executes, printing “0”
• The value of i then increases by 1 to become 1
• Since 1%2 is not equal to 0, the computer checks the else-if statement
• 1 is not equal to 3, so the computer prints 1*2, or “2”
• BE CAREFUL!!! Since the code says “println”, the computer prints the “2” next to the “0”, then starts a new line!
• If you continue tracing the code, when becomes 2, the computer prints “2”
• When i equals 3, it fails the if statement and the else-if statement, which means the code in the else statement runs, printing “Three”
• Continue tracing the code for equals 4 and the computer prints “4”

02
2Three4

### Example #3 😎

```int count = 1;
while (count<5)
{
count*=2;
}
System.out.println(count);```

What is printed as a result of executing this code?

• When the code begins, count equals 1
• Since 1 is less than 5, the computer enters the while loop and count is doubled, making count equal to 2
• Since 2 is still less than 5, the computer re-enters the while loop and count is doubled again, making count equal to 4
• Since 4 is still less than 5, the computer re-enters the while loop and count is doubled and now equals 8
• Now that 8 is not less than 5, the computer moves on and prints count, or “8”

8

### Example #4: 👍

```public int addNums (int x, int y)
{
return x+y;
}```

What does the method return with after the method call addNums (5,10)?

The method returns the integer value 15. (Since 5+10 = 15.)

### Example #5 🤓

```for (int y=5; y>0; y--)
{
for (int x=0; x<5; x++)
{
System.out.print(y);
}
System.out.println();
}```

What is the output when you execute this code?

First of all, don’t panic when you see this snippet, it’s easy to solve if you follow it step-by-step and write every step down. 🙌

Start when y equals 5:

• Now that you’re on the inside of the double for loop, just focus on the inner for loop
• The inner for loop prints y, which is currently 5, five times
• Now the the inner for loop is finished, the computer exits that loop and moves on to “System.out.println()”, which causes the computer to skip a line
• Since there’s nothing inside the parentheses, nothing is printed

The value of y then decreases by 1, so y now equals 4. Repeat the same process as above for y equals 4, then 3, then 2, then 1. The code stops running after that because y ends up being 0 and does not meet the condition that y is greater than 0, causing the computer to exit all of the loops.