Saman - 7 months ago 23

Javascript Question

What is the interpretation of this line in Javascript.

`var x[],matrix[],n;`

...

n = (matrix = x) && matrix.length;

Despite I searched for it, I couldn't find any tips.

Thank you

Answer

It does this:

- Assigns the value of
`x`

to`matrix`

; the result of the`matrix = x`

expression is the value that was assigned (this is true of all assignment expressions). Let's call that value "x-value". I don't want to call it`x`

from here on out, because`x`

is only evaluated once. - If x-value is
*truthy*^{1}(coerces to true), it assigns`matrix.length`

to`n`

; otherwise, assigns x-value to`n`

.

So for instance, if `x`

is `[]`

, the code sets `matrix`

to point to the same empty array `x`

does and sets `n`

to `0`

(`matrix.length`

after the assignment). Other examples *(I'd written these before you edited your question)*: If `x`

is `"foo"`

, it sets `matrix`

to `"foo"`

and sets `n`

to `3`

(the `length`

of `matrix`

). If `x`

is `""`

(a falsy value), it sets `matrix`

to `""`

*and* sets `n`

to `""`

. If `x`

is `{foo:"bar"}`

, it sets `matrix`

to refer to that same object and sets `n`

to `undefined`

(since the object has no `length`

property). You get the idea.

#2 above comes about because `&&`

is not just a simple logical AND operator. `a && b`

works like this:

- Evaluate
`a`

to get its value; let's call that a-value - If a-value is
*falsy*, the result of the`&&`

operator is a-value - Otherwise, evaluate
`b`

and make that the result of the`&&`

operator

^{1} "Truthy" values are any values that aren't "falsy." The falsy values are `0`

, `null`

, `undefined`

, `""`

, `NaN`

, and of course, `false`

.