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2.7 Operators
Many operators are known to us from school. They are addition +, a multiplication *, a subtraction - and so on.
In this chapter we concentrate on aspects that are not covered by school arithmetic.
Before we move on, let’s grasp the common terminology.
-
An operand – is what operators are applied to. For instance in multiplication
5 * 2there are two operands: the left operand is5, and the right operand is2. Sometimes people say “arguments” instead of “operands”. -
An operator is unary if it has a single operand. For example, the unary negation
-reverses the sign of the number:let x = 1; x = -x; alert( x ); // -1, unary negation was applied -
An operator is binary if it has two operands. The same minus exists in the binary form as well:
let x = 1, y = 3; alert( y - x ); // 2, binary minus subtracts values
Formally, we’re talking about two different operators here: the unary negation (single operand, reverses the sign) and the binary subtraction (two operands, subtracts).
Now let’s see special features of JavaScript operators that are beyond school arithmetics.
Usually the plus operator + sums numbers.
But if the binary + is applied to strings, it merges (concatenates) them:
let s = "my" + "string";
alert(s); // mystring
Note that if any of the operands is a string, then the other one is converted to a string too.
For example:
alert( '1' + 2 ); // "12"
alert( 2 + '1' ); // "21"
See, it doesn’t matter whether the first operand is a string or the second one. The rule is simple: if either operand is a string, then convert the other one into a string as well.
However, note that operations run from left to right. If there are two numbers followed by a string, the numbers will be added before being converted to a string:
alert(2 + 2 + '1' ); // "41" and not "221"
String concatenation and conversion is a special feature of the binary plus +. Other arithmetic operators work only with numbers. They always convert their operands to numbers.
For instance, subtraction and division:
alert( 2 - '1' ); // 1
alert( '6' / '2' ); // 3
The plus + exists in two forms. The binary form that we used above and the unary form.
The unary plus or, in other words, the plus operator + applied to a single value, doesn’t do anything with numbers, but if the operand is not a number, then it is converted into it.
For example:
// No effect on numbers
let x = 1;
alert( +x ); // 1
let y = -2;
alert( +y ); // -2
// Converts non-numbers
alert( +true ); // 1
alert( +"" ); // 0
It actually does the same as Number(...), but is shorter.
A need to convert strings to numbers arises very often. For example, if we are getting values from HTML form fields, then they are usually strings.
What if we want to sum them?
The binary plus would add them as strings:
let apples = "2";
let oranges = "3";
alert( apples + oranges ); // "23", the binary plus concatenates strings
If we want to treat them as numbers, then we can convert and then sum:
let apples = "2";
let oranges = "3";
// both values converted to numbers before the binary plus
alert( +apples + +oranges ); // 5
// the longer variant
// alert( Number(apples) + Number(oranges) ); // 5
From a mathematician’s standpoint the abundance of pluses may seem strange. But from a programmer’s standpoint, there’s nothing special: unary pluses are applied first, they convert strings to numbers, and then the binary plus sums them up.
Why are unary pluses applied to values before the binary one? As we’re going to see, that’s because of their higher precedence.
If an expression has more than one operator, the execution order is defined by their precedence, or, in other words, there’s an implicit priority order among the operators.
From school we all know that the multiplication in the expression 1 + 2 * 2 should be calculated before the addition. That’s exactly the precedence thing. The multiplication is said to have a higher precedence than the addition.
Parentheses override any precedence, so if we’re not satisfied with the order, we can use them, like: (1 + 2) * 2.
There are many operators in JavaScript. Every operator has a corresponding precedence number. The one with the bigger number executes first. If the precedence is the same, the execution order is from left to right.
An extract from the precedence table (you don’t need to remember this, but note that unary operators are higher than corresponding binary ones):
| Precedence | Name | Sign |
|---|---|---|
| ... | ... | ... |
| 16 | unary plus | + |
| 16 | unary negation | - |
| 14 | multiplication | * |
| 14 | division | / |
| 13 | addition | + |
| 13 | subtraction | - |
| ... | ... | ... |
| 3 | assignment | = |
| ... | ... | ... |
As we can see, the “unary plus” has a priority of 16, which is higher than 13 for the “addition” (binary plus). That’s why in the expression "+apples + +oranges" unary pluses work first, and then the addition.
Let’s note that an assignment = is also an operator. It is listed in the precedence table with the very low priority of 3.
That’s why when we assign a variable, like x = 2 * 2 + 1, then the calculations are done first, and afterwards the = is evaluated, storing the result in x.
let x = 2 * 2 + 1;
alert( x ); // 5
It is possible to chain assignments:
let a, b, c;
a = b = c = 2 + 2;
alert( a ); // 4
alert( b ); // 4
alert( c ); // 4
Chained assignments evaluate from right to left. First the rightmost expression 2 + 2 is evaluated then assigned to the variables on the left: c, b and a. At the end, all variables share a single value.
An operator always returns a value. That’s obvious for most of them like an addition + or a multiplication *. But the assignment operator follows that rule too.
The call x = value writes the value into x and then returns it.
Here’s the demo that uses an assignment as part of a more complex expression:
let a = 1;
let b = 2;
let c = 3 - (a = b + 1);
alert( a ); // 3
alert( c ); // 0
In the example above, the result of (a = b + 1) is the value which is assigned to a (that is 3). It is then used to subtract from 3.
Funny code, isn’t it? We should understand how it works, because sometimes we can see it in 3rd-party libraries, but shouldn’t write anything like that ourselves. Such tricks definitely don’t make the code clearer and readable.
The remainder operator % despite its look does not have a relation to percents.
The result of a % b is the remainder of the integer division of a by b.
For instance:
alert( 5 % 2 ); // 1 is a remainder of 5 divided by 2
alert( 8 % 3 ); // 2 is a remainder of 8 divided by 3
alert( 6 % 3 ); // 0 is a remainder of 6 divided by 3
The exponentiation operator ** is a recent addition to the language.
For a natural number b, the result of a ** b is a multiplied by itself b times.
For instance:
alert( 2 ** 2 ); // 4 (2 * 2)
alert( 2 ** 3 ); // 8 (2 * 2 * 2)
alert( 2 ** 4 ); // 16 (2 * 2 * 2 * 2)
The operator works for non-integer numbers of a and b as well, for instance:
alert( 4 ** (1/2) ); // 2 (power of 1/2 is the same as a square root, that's maths)
alert( 8 ** (1/3) ); // 2 (power of 1/3 is the same as a cubic root)
Increasing or decreasing a number by one is among the most common numerical operations.
So, there are special operators for that:
-
Increment
++increases a variable by 1:let counter = 2; counter++; // works the same as counter = counter + 1, but is shorter alert( counter ); // 3 -
Decrement
--decreases a variable by 1:let counter = 2; counter--; // works the same as counter = counter - 1, but is shorter alert( counter ); // 1
Increment/decrement can be applied only to a variable. An attempt to use it on a value like 5++ will give an error.
Operators ++ and -- can be placed both after and before the variable.
-
When the operator goes after the variable, it is called a “postfix form”:
counter++. -
The “prefix form” is when the operator stands before the variable:
++counter.
Operators ++ and -- can be placed both after and before the variable.
-
When the operator goes after the variable, it is called a “postfix form”: counter++.
-
The “prefix form” is when the operator stands before the variable: ++counter.
Both of these records do the same: increase counter by 1.
Is there any difference? Yes, but we can only see it if we use the returned value of ++/--.
Let’s clarify. As we know, all operators return a value. Increment/decrement is not an exception here. The prefix form returns the new value, while the postfix form returns the old value (prior to increment/decrement).
To see the difference, here’s the example:
let counter = 1;
let a = ++counter; // (*)
alert(a); // 2
Here in the line (*) the prefix call ++counter increments counter and returns the new value that is 2. So the alert shows 2.
Now let’s use the postfix form:
let counter = 1;
let a = counter++; // (*) changed ++counter to counter++
alert(a); // 1
In the line (*) the postfix form counter++ also increments counter, but returns the old value (prior to increment). So the alert shows 1.
To summarize:
-
If the result of increment/decrement is not used, then there is no difference in which form to use:
let counter = 0; counter++; ++counter; alert( counter ); // 2, the lines above did the same -
If we’d like to increase the value and use the result of the operator right now, then we need the prefix form:
let counter = 0; alert( ++counter ); // 1 -
If we’d like to increment, but use the previous value, then we need the postfix form:
let counter = 0;
alert( counter++ ); // 0
Operators ++/-- can be used inside an expression as well. Their precedence is higher than most other arithmetical operations.
For instance:
let counter = 1;
alert( 2 * ++counter ); // 4
Compare with:
let counter = 1;
alert( 2 * counter++ ); // 2, because counter++ returns the "old" value
Though technically allowable, such notation usually makes the code less readable. One line does multiple things – not good.
While reading the code, a fast “vertical” eye-scan can easily miss such counter++, and it won’t be obvious that the variable increases.
The “one line – one action” style is advised:
let counter = 1;
alert( 2 * counter );
counter++;
Bitwise operators treat arguments as 32-bit integer numbers and work on the level of their binary representation.
These operators are not JavaScript-specific. They are supported in most programming languages.
The list of operators:
-
AND (
&) -
OR (
|) -
XOR (
^) -
NOT (
~) -
LEFT SHIFT (
<<) -
RIGHT SHIFT (
>>) -
ZERO-FILL RIGHT SHIFT (
>>>)
These operators are used very rarely. To understand them, we should delve into low-level number representation, and it would not be optimal to do that right now. Especially because we won’t need them any time soon. If you’re curious, you can read the Bitwise Operators article in MDN. It would be more practical to do that when a real need arises.
We often need to apply an operator to a variable and store the new result in it.
For example:
let n = 2;
n = n + 5;
n = n * 2;
This notation can be shortened using operators += and *=:
let n = 2;
n += 5; // now n = 7 (same as n = n + 5)
n *= 2; // now n = 14 (same as n = n * 2)
alert( n ); // 14
Short “modify-and-assign” operators exist for all arithmetical and bitwise operators: /=, -= etc.
Such operators have the same precedence as a normal assignment, so they run after most other calculations:
let n = 2;
n *= 3 + 5;
alert( n ); // 16 (right part evaluated first, same as n *= 8)
The comma operator , is one of most rare and unusual operators. Sometimes it’s used to write shorter code, so we need to know it in order to understand what’s going on.
The comma operator allows us to evaluate several expressions, dividing them with a comma ,. Each of them is evaluated, but the result of only the last one is returned.
For example:
let a = (1 + 2, 3 + 4);
alert( a ); // 7 (the result of 3 + 4)
Here, the first expression 1 + 2 is evaluated, and its result is thrown away, then 3 + 4 is evaluated and returned as the result.
Please note that the comma operator has very low precedence, lower than =, so parentheses are important in the example above.
Without them: a = 1 + 2, 3 + 4 evaluates + first, summing the numbers into a = 3, 7, then the assignment operator = assigns a = 3, and then the number after the comma 7 is not processed anyhow, so it’s ignored.
Why do we need such an operator which throws away everything except the last part?
Sometimes people use it in more complex constructs to put several actions in one line.
For example:
// three operations in one line
for (a = 1, b = 3, c = a * b; a < 10; a++) {
...
}
Such tricks are used in many JavaScript frameworks, that’s why we mention them. But usually they don’t improve the code readability, so we should think well before writing like that.