Created
November 29, 2019 20:59
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Test demostrating basic memoization for a cached parameter, using previously calculated result
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// our cache | |
let cache = { | |
val: null, | |
res: null | |
} | |
const myLongTask = num => { | |
// if parameter is equal to our cached value, we don't repeat the calculation | |
if (num === cache.val) { | |
const t1 = new Date().valueOf() | |
const t2 = new Date().valueOf() | |
console.log(`Parameter was memoized. Operation time(ms): ${t2-t1}`) | |
return cache.res | |
} | |
// since parameter was not equal to our cached value, it needs to do the calculations | |
const t1 = new Date().valueOf() | |
const res = stress(num) // mock some performance hit | |
cache = {val: num, res: res} // store parameter and response in our cache | |
const t2 = new Date().valueOf() | |
console.log(`Parameter not memoized. Operation time(ms): ${t2-t1}`) | |
return res | |
} | |
// intensive operation | |
const stress = (num) => { | |
let newNum = 0 | |
while (newNum < 2000000000) { | |
newNum += num | |
} | |
return newNum | |
} | |
// use a combination of different and same parameters | |
const myArray = [1,2,3,4,4,4,4,4,4,4,4,4,4,5] | |
// if passing same parameter than one cached = memoized value (instant) | |
// if passing diff parameter than one cached = recalculation (slow) | |
myArray.forEach(number => { | |
myLongTask(number) | |
}) | |
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