How to make a function that return an index with a limit-CodePudding

I've wondered how can I optimize the conditions to function that is flexible enough to return an index that has a limitation of 3 index

if(val === 0) return 0;
if (val === -90) return 3;
if (val === -180) return 2;
if (val === -270) return 1;
if (val === -360) return 0;
if (val === -450) return 3;

if(val === 0) return 0;
if (val === 90) return 1;
if (val === 180) return 2;
if (val === 270) return 3;
if (val === 360) return 0;
if (val === 450) return 1;

and so on

I wonder how to make a function that if a user increments by 90 the returns should get incremented with a limit of 3 then after 3 it will return 0 index

CodePudding user response：

function f(x) {
 let z = x / 90;
 let y = z % 4;
 return y < 0 ? 4   y : y;   
}



for(let i =  -90 * 10; i <= 90 * 10; i = i   90) {
 console.log(`x = ${i} => result => ${f(i)}`);
}

function f(x) {
  return ((x / 90) % 4   4) % 4;
}


 for(let i =  -90 * 10; i <= 90 * 10; i = i   90) {
  console.log(`x = ${i} => result => ${f(i)}`);
 }

CodePudding user response：

you should get the minimum input first, then add 360 to it.

val = 360 val;

if(val % 360 === 90){
    return 1
}
if(val % 360 === 180){
    return 2
}
if(val % 360 === 270){
    return 3
}
if(val % 360 === 0){
    return 0
}

CodePudding user response：

You can try dividing by 90 for the positive numbers and for the negative numbers add 360 first then divide by 90.

CodePudding user response：

You can do (val % 360) / 90

if (val % 90 !== 0) return null.

return (val % 360) / 90;

CodePudding user response：

Try

const f = v => (v = v / 90 % 4, v < 0 ? 4   v : v | 0);

for (let i = -450; i <= 450; i  = 90) {
 console.log(`f(${i}) = ${f(i)}`);
}