TypeScript write "if" in short

Question

I have the code that sets a property only if the argument is truthy:

export class Vector {
    protected x: number = 0
    protected y: number = 0
    protected z: number = 0

    set(x?: number, y?: number, z?: number) {
        this.x = x ? x : this.x
        this.y = y ? y : this.y
        this.z = z ? z : this.z
    }
}

However, I don't want to provide a fallback value of the current value. Ideally, I want to just do nothing if the value is falsy, like this:

if (x) {
  this.x = x
}

if (x) {
  this.y = y
}

if (z) {
  this.z = z
}

… but I don't want to write it like this, it is not cool. I'm looking for something like this:

this.x = x ? x
this.y = y ? y
this.z = z ? z

Is there any similar syntax that does what I want?

a ? a : b is equivalent to a || b (unless a expression has side effects). — raina77ow
– raina77ow, Commented Aug 30, 2018 at 11:10
You could also remove the set method and make your properties public and set them when they need to be different from default — Mike Bovenlander
– Mike Bovenlander, Commented Aug 30, 2018 at 11:11
Thank you for a comment!!! Is it possible to erase ":b"? I want to do nothing if there wan no a. I don't want to do b. — Kuru
– Kuru, Commented Aug 30, 2018 at 11:12
What @raina77ow means is usage like this: this.x = x || this.x — Mike Bovenlander
– Mike Bovenlander, Commented Aug 30, 2018 at 11:13

Lars Holdaas · Accepted Answer · 2018-08-30 12:25:41Z

4

Usually x && this.x=x would be the shortest syntax to get this done.

However x, y and z are all numbers. It's a bit dangerous to use this short-hand syntax for numbers. Consider the case where set is called with x=0. 0 && this.x=x would not execute this.x as 0 is falsey in Javascript. From reading your code this does not seem like what you want to achieve, instead you want to skip setting this.x in the case x is undefined.

In that case, I suggest the following code:

set(x?: number , y? : number , z? : number){
    typeof x === 'number' && (this.x = x);
    typeof y === 'number' && (this.y = y);
    typeof z === 'number' && (this.z = z);
}

That way your set-function will support sending 0 as an argument, which it currently does not.

edited Aug 30, 2018 at 12:25

answered Aug 30, 2018 at 11:24

Lars Holdaas

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7 Comments

Kuru Over a year ago

OH!!! You are so smart to notice that 0 is false!!! Very good study for me.But VSCODE give me error.";" is required(T_T)

Lars Holdaas Over a year ago

Can you see which line you're required to add ;?

Kuru Over a year ago

I add message to you on the question!

Kuru Over a year ago

Oh!!! No error!!!! I think this is the coolest way!!!!! Thank you so much , handsome man!!!

Parzh Over a year ago

The OP's message was edited out of the question. The code that produced the error was: typeof x === 'number' && this.x = x; (screenshot of the error: i.sstatic.net/Z2ryk.png)

|

webNeat · Accepted Answer · 2018-08-30 11:36:34Z

A clean way to write your method is

set(x?: number , y? : number , z? : number){
  this.x = x || this.x;
  this.y = y || this.y;
  this.z = z || this.z;
}

An other way is

set(x?: number , y? : number , z? : number){
  x && this.x = x;
  y && this.y = y;
  z && this.z = z;
}

But, as @Lars Holdaas already mentioned, this will not support the falsy values (like 0 or ""). The generic way to solve this is by writing a validation or filter function to tell if the value is actually truly for that parameter or not.

// returns `true` if x is a number
const n = x => typeof n === 'number';

set(x?: number , y? : number , z? : number){
  n(x) && this.x = x;
  n(y) && this.y = y;
  n(z) && this.z = z;
}

Hope this helps :)

Parzh · Accepted Answer · 2024-02-06 14:41:57Z

0

Aside from other answers, you have three more options:

Use if one-liner:

if (x) this.x = x
if (y) this.y = y
if (z) this.z = z

Use ?? operator (it wasn't available in 2018, when the question was asked):
```
this.x = x ?? this.x
this.y = y ?? this.y
this.z = z ?? this.z
```

Use default values for optional parameters:

export class Vector {
    protected x: number = 0
    protected y: number = 0
    protected z: number = 0

    set(x = this.x, y = this.y, z = this.z) {
        this.x = x
        this.y = y
        this.z = z
    }
}

answered Feb 6, 2024 at 14:41

Parzh

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