javascript 中 if/else 的简写

Question

I wrote the function to check if a given number is positive, negative, or zero. Is there a shorthanded way?

Is it possible to test the condition with map (using object to index the conditions)? i was inspired by second solution in this question but not sure how it applies here.

const numCheck = (num) => {
  if (num == 0) {
    return "zero";
  } else {
    return num > 0 ? "positive" : "negative";
  }
};

const num1 = 3;
console.log(numCheck(num1));

Answer 1

 const numCheck = (num) => { return (num == 0)? "zero": ((num > 0)? "positive": "negative"); }; const num1 = 3; console.log(numCheck(num1));

Answer 2

you can do that, simply use Math.sign() :

 const numCheck = n => ['negative','zero','positive'][1+Math.sign(n)] console.log( numCheck( 4 ) ) console.log( numCheck( 0 ) ) console.log( numCheck( -3 ) )

Answer 3

A shorter solution:

 var numCheck = (num) => num == 0? "zero": num > 0? "positive": "negative"; const num1 = 3; console.log(numCheck(num1));

Answer 4

Inspired by Mister Jojo answer but without Math.sign

 const sign = x => ['negative','zero','positive'][1 + (x > 0) - (x < 0)] console.log(sign(10)) // positive console.log(sign(0)) // zero console.log(sign(-20)) // negative