The logical and (&&) and or (||) are logical operators in JavaScript. Normally, you're using these operators on booleans:
true && true // => true
true && false // => false
true || true // => true
true || false // => true
However, can you use && and || with operands other than booleans? Turns out, you can!
This post explains in detail how && and || operators work in JavaScript.
Before jumping into how the operators work, let's start with the basic concepts of truthy and falsy.
Because JavaScript is a loosely typed language, logical operations can be performed on any type. The expressions like 1 && 2, null || undefined, 'hello' && true are weird, but still valid in JavaScript.
To perform logical operations on any type, JavaScript decides whether a particular value can be considered falsy (an equivalent of false) or truthy (an equivalent of true).
Falsy is a value for which Boolean(value) returns false. Falsy values in JavaScript are only false, 0, '', null, undefined and NaN.
Boolean(false); // => false
Boolean(0); // => false
Boolean(''); // => false
Boolean(null); // => false
Boolean(undefined); // => false
Boolean(NaN); // => false
Truthy is a value for which Boolean(value) returns true. Saying it differently, truthy are the non-falsy values.
Examples of truthy values are true, 4, 'Hello', { name: 'John' } and everything else that's not falsy.
Boolean(true); // => true
Boolean(4); // => true
Boolean('Hello'); // => true
Boolean({ name: 'John' }); // => true
Now let's continue with proper learning of how && operator works. Note that the operator works in terms of truthy and fasly, rather than true and false.
Here's the syntax of the && operator in a chain:
operand1 && operand2 && ... && operandN
The expression is evaluated as follows:
Starting from left and moving to the right, return the first operand that is falsy. If no falsy operand was found, return the latest operand.
Let's see how the algorithm works in a few examples.
When the operands are booleans, it's simple:
true && false && true; // => false
The evaluation starts from left and moves to the right. The first true operand is passed. However, the second operand false is a falsy value, and evaluation stops. false becomes the result of the entire expression. The third operand true is not evaluated.
When operands are numbers:
3 && 1 && 0 && 10; // => 0
The evaluation is performed from left to right. 3 and 1 are passed because they are truthy. But the evaluation stops at the third operand 0 since it's falsy. 0 becomes the result of the entire expression. The fourth operand 10 is not evaluated.
A slighly more complex example with different types:
true && 1 && { name: 'John' }
Again, from left to right, the operands are checked for falsy. No operand is falsy, so the last operand is returned. The evaluation result is { name: 'John' }.
The && evaluation algorithm is optimal because the evaluation of operands stops as soon as a falsy value is encountered.
Let's consider an example:
const person = null;
person && person.address && person.address.street; // => null
The evaluation of the long logical expression person && person.address && person.address.street stops right at the first operand person (which is null - i.e. falsy). person.address and person.address.street operands are skipped.
Here's a generalized syntax of || operator in chain:
operand1 || operand2 || ... || operandN
The evaluation of || happens this way:
Starting from left and moving to the right, return the first operand that is truthy. If no truthy operand was found, return the latest operand.
|| works the same way as &&, with the only difference that || stops evaluation when encounters a truthy operand.
Let's study some || examples.
A simple expression having 2 booleans:
true || false; // => true
The evaluation starts from left and moves to the right. Luckily, the first operand true is a truthy value, so the whole expression evaluates to true. The second operand false is not checked.
Having some numbers as operands:
The first operand 0 is falsy, so the evaluation continues. The second argument -1 is already truthy, so the evaluation stops, and the result is -1.
You can use a side-effect of the || evaluation to access an object property providing a default value when the property is missing.
For example, let's access the properties name and job of the person object. When the property is missing, simply default to a string 'Unknown'. Here's how you could use || operator to achieve it:
const person = {
name: 'John'
};
person.name || 'Unknown'; // => 'John'
person.job || 'Unknown'; // => 'Unknown'
Let's look at the expression person.name || 'Unknown'. Because the first operand person.name is 'John' (a truthy value), the evaluation early exists with the truthy value as a result. The expression evaluates to 'John'.
person.job || 'Unknown' operates differently. person.job is undefined, so the expression is the same as undefined || 'Unknown'. The || evaluation algorithm says that the expression should return the first operand that is truthy, which in this case is 'Unknown'.
Because JavaScript is a loosely typed language, the operands of && and || can be of any type.
The concepts of falsy and truthy are handy to deal with types conversion within logical operators. Falsy values are false, 0, '', null, undefined and NaN, while the rest of values are truthy.
&& operator evaluates the operands from left to right and returns the first falsy value encountered. If no operand is falsy, the latest operand is returned.
The same way || operator evaluates the operands from left to right but returns the first truthy value encountered. If no truthy value was found, the latest operand is returned.
While && and || evaluation algorithms seem weird at first, in my opinion, they're quite efficient. The algorithms perform early exit, which is a good performance optimization.
In terms of usage, I recommend to stick to booleans as operands for both && and ||, and avoid other types if possible. Logical expressions that operate only on booleans are easier to understand.
Can you explain how 0 || 1 && 2 is evaluated?
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