WDI Fundamentals Unit 7
In this section we'll be talking about expressions. What are expressions? Watch this video to find out.
Chances are, you've probably played around with a scientific calculator at some point in the past. You punch in a big number (for example, 9876435),
x, and then another big number (say, 373848221), and hit the
= button. Then,
the calculator spits back a result (in this case, 3692287654572135).
That information we type into the calculator is called an expression: a
collection of values (12345) and operators (like
The process of reducing this expression down to a single value is called evaluation.
How do we combine numbers and operators to come up with more complex expressions in JS?
It's simple — we use arithmetic operators.
All of the standard arithmetic operators learned in grade school (addition, subtraction, division, and multiplication) are supported in JS. These should look familiar.
But if you don't have a background in programming, that last operator — the modulus operator — might be new.
The modulus operator shows the remainder of a division problem.
For example, 9 divided by 4 equals 2 with a remainder of 1. The modulus operator takes two numbers as inputs and returns what's leftover from the division.
The modulus operator
% is particularly useful in programming if we want to find out if a number is even or odd.
If we divide by 2 and have a remainder of 1, we know the number is odd. If we have a remainder of 0, then we know that the number is even.
Let's look at some examples:
5 % 2; => 1 7 % 2; => 1
4 % 2; => 0 2 % 2; => 0
This may seem tedious now, but it'll come in handy later on.
Look at the following five problems. Type each line of code into the JS Bin Console and see what is returned.
45 % 6;
10 - 20;
7 / 2;
3 * 2;
10 % 4;
Now, let's see how you can use string values (textual information) in JS.
When given string values, the
+ operator actually behaves differently — it concatenates, or combines, two strings together to make one big string.
Take a look in this brief video.
As you can see, putting single or double quotation marks around a value turns it into a string.
So, even though both "6" and "8" look like numbers to us humans, JS sees that they're in quotation marks and therefore treats them as strings.
var number1 = "6"; var number2 = "8"; number1 + number2; // => "68"
+ operator to put the two strings together literally puts them next to each other, instead of evaluating their total.
This is called concatenation (when strings are glued together).
Here's another example of concatenation.
JS glued the two strings together, but do you notice anything wrong?
var firstName = "Han"; var lastName = "Solo"; firstName + lastName; // => "HanSolo"
There is no space between the two words!
This is because we didn't add the spaces in ourselves. It's just one of many reasons why we have to carefully watch our spacing and grammar.
To fix this, we'll have to add in the space ourselves.
var firstName = "Han"; var lastName = "Solo"; firstName + " " + lastName; // => "Han Solo"
Now, let's get back to some math and look at assignment operators.
You're already familiar with the
= assignment operator, but there are also ones we can use to add or subtract value from a variable. Take a look:
+= operator adds a value to an existing variable.
-= operator subtracts a value from an existing variable.
Note: Keep in mind that we'll always need an
=somewhere in the line of code when we want to either assign or update the value of a variable, as in the above chart.
Type each of the following lines of code in the JS Bin Console and hit return to run each line of code.
var myNumber = 8;
myNumber += 3;
myNumber -= 5;
What is the final value of
myNumber; into the console and hit return to check!
Answer: The final value of
While we've covered what seems like a lot of math in this section, don't worry — you're not going to be doing calculus in this course. It's important that we review these concepts, because there will be many times when you'll solve a problem by using one of their basic principles. When it comes down to it, computers operate with a simple, straightforward logic.
Sometimes, we find variables on both sides of the
=. Suppose we have two variables,
y, like the example below:
var x = 5; var y = 10; x = y + 10;
What happens in that third line? For starters, everything to the right of the
= must be evaluated before any kind of assignment can happen. This is why we like to use the phrase "assignment always happens right to left!"
y + 10; evaluates to 20, so what we're left with is the expression
x = 20;. This assigns the value 20 to
x, and the entire expression evaluates to 20.
Let's look at one more example using the same two variables,
var x = 1; var y = 10; x = y * 2; y = x + 1; x = y + 1; y = 2 * x;
Feeling dizzy? Don't worry, we'll step through this one together.
Line 1: We declare a new variable
x and assign it the value
Line 2: We declare another new variable
y and assign it the value
Line 3: As of this point in the code,
y has a value of
10. We multiply that by 2, resulting in
20. We assign that resulting value to
x now has a value of
y then gets assigned a new value of
20 + 1).
y was just changed to 21, so
x becomes 22 (
21 + 1).
x is now 22, so
2 * 22, or 44.
One important thing to mention here is that at no point is any lasting relationship established between x and y (unlike math equations). We are simply evaluating the expression on the right and assigning the result to the variable on the left.
Give these a try — see if you can predict the final values of
Check your answers in JS Bin by copying the entire chunk of code into the editor window, running it, and then checking
z in the JS Bin console by typing out each variable name and hitting the return key.
If you don't see the JS Bin, please refresh the page.
var x = 1; var y = 2; var z = 3; x = y; y = z; z = x;
var x = 1; var y = 0; var z = -1; x = y + z; y = z * x; z = x - y; x = y * y; y = z * z; z = z - 1;
Whoa! That last one's pretty weird — how can
z be on both sides of the
=? What do you think happens there?
The key is remembering how the
= operator works. Before it assigns anything to the variable on the left, it first evaluates the expression on the right. This means that, if we have any expression like
x = x + 1;, what we are doing is taking the old value of
x, adding 1 to it, and storing this new result back in
x. In short, we are "incrementing"
x: increasing its value by 1, no matter its original value.
Think you've got it? Let's cement what we've learned with an exercise.