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