Here are three functions that can be combined to produce an output of 30:
const add = (a) => a + 10;
const mul = (a) => a * 10;
const divide = (a) => a / 5;
// How would you write the `compositionFunction` to achieve this?
compositionFunction(add, mul, divide)(5);
//=> 30
The expected result is 30 because:
5 + 10 = 15
15 * 10 = 150
150 / 5 = 30
Here is an example
const sum = (a) => a + 10 ;
const product = (a) => a * 10 ;
const divideByFive = (a) => a / 5 ;
// How to utilize these functions -----
const customFunction = (...f) => v => f.reduce((res, f) => f(res), v)
console.log(customComposeFn(sum, product, divideByFive)(5));There are two types of function composition:
Here is a fun recursive approach to implementation:
This assumes there are at least two functions to combine
const compose = (...fn) => {
const [[f, g], x] = [fn.slice(-2), fn.slice(0, -2)];
const h = a => f(g(a));
return x.length ? compose(...x, h) : h;
}
const pipe = (...fn) => {
const [f, g, ...x] = fn;
const h = a => g(f(a));
return x.length ? pipe(h, ...x) : h;
}
Let's test it out:
const foo = x => x + 'foo';
const bar = x => x + 'bar';
const baz = x => x + 'baz';
pipe(foo, bar, baz)('');
//=> 'foobarbaz'
compose(foo, bar, baz)('');
//=> 'bazbarfoo'
const add = (num) => num + 10;
const multiply = (num) => num * 10;
const divideByFive = (num) => num / 5;
const customFunction = (...functions) => {
return function (argument) {
if (functions.length > 0) {
const output = functions[0](argument);
return customFunction(...functions.splice(1))(output);
} else {
return argument;
}
};
};
const result = customFunction(add, multiply, divideByFive)(5);
console.log(`result`, result);My approach would be the following:
function combineFunctions(...functions) {
return function(input) {
let func, result = input;
while (func = functions.shift()) {
result = func(result);
}
return result;
}
}
const reduceByOne = num => num - 1;
const multiplyByTwo = num => num * 2;
const divideByThree = num => num / 3;
// Usage example -----
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