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));
In the solution below, the numCheck() method is tested by using the Array.prototype.map() method, considering three possible situations.
const numCheck = (num) => {
return (num == 0) ? "zero" : ((num > 0) ? "positive" : "negative");
};
const numbers = [1, -1, 0];
console.log(numbers.map(number => numCheck(number)));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 ) ) A shorter solution:
var numCheck = (num) => num == 0 ? "zero" : num > 0 ? "positive" : "negative";
const num1 = 3;
console.log(numCheck(num1));