Maybe has two types: Just / Nothing. Just() will just return the value that passed in. Nothing returns nothing...
Just/ Nothing are both functors, they should have .map() method:
const Just = x => ({ map: f => Just(f(x)), inspect: () => `Just ${x}`, }) const Nothing = x => ({ map: f => Nothing('Nothing'), inspect: () => `${x}`, }) const Maybe = { Just, Nothing }
We added 'inspect' method so that in REPL we can log out understandable message. for example, 'Just 4' instead of '{ map: [Function: map] }'.... or whatever...
Currently, the code below return 'Just 5'
const inc = n => n + 1; const input = Just(4) const result = input.map(inc)
But we don't need 'Just' as a result, we want just 5; in order to achieve that, we add 'option()' method:
const Just = x => ({ map: f => Just(f(x)), inspect: () => `Just ${x}`, option: (_) => x, }) const Nothing = x => ({ map: f => Nothing('Nothing'), inspect: () => `${x}`, option: defVal => defVal })
For Just, it return whatever the current value 'x', ignore the default value we passed in; Nothing it will return the defautl vlaue back.
Now, we got:
const input = Just(4)
const result = input.map(inc).option(0) // 5 const input = Nothing(4) const result = input.map(in) // Nothing const result = input.map(inc).option(0) // 0
Since we don't know it should be Just or Nothing, we can use some predict function as helper:
const safeNum = num => typeof num === 'number' ? Maybe.Just(num): Maybe.Nothing() const input = safeNum(4) const result = input.map(inc).option(0) // 5 const input = safeNum('4') const result = input.map(inc).option(0) // 0
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const Just = x => ({ map: f => Just(f(x)), inspect: () => `Just ${x}`, option: (_) => x, }) const Nothing = x => ({ map: f => Nothing('Nothing'), inspect: () => `${x}`, option: defVal => defVal }) const safeNum = num => typeof num === 'number' ? Maybe.Just(num): Maybe.Nothing() const Maybe = { Just, Nothing } const inc = n => n + 1; const input = safeNum(4) const result = input.map(inc).option(0) console.log(result)