Calculating using functions - can someone explain?

JavaScript
#1

Hi there, I have a solution to this but i’m not sure i understand it 100%

Taking (seven(times(five())) as an example:

  1. firstly five() doesn’t take in anything so return 5:
    seven(times(5))

  2. (times(5)): times takes in 5 as “number”. we pass 5 into the times function

  3. ?? now i get confused.

I wonder if anyone has a good way of explaining the step by step of this?

thanks as always

const zero = (operation) => !operation ? 0 : operation(0);
const one = (operation) => !operation ? 1 : operation(1);
const two = (operation) => !operation ? 2 : operation(2);
const three = (operation) => !operation ? 3 : operation(3);
const four = (operation) => !operation ? 4 : operation(4);
const five = (operation) => !operation ? 5 : operation(5);
const six = (operation) => !operation ? 6 : operation(6);
const seven = (operation) => !operation ? 7 : operation(7);
const eight = (operation) => !operation ? 8 : operation(8);
const nine = (operation) => !operation ? 9 : operation(9);

const plus = (number) => (other_number) => other_number + number;
const minus = (number) => (other_number) => other_number - number;
const times = (number) => (other_number) => other_number * number;
const dividedBy = (number) => (other_number) => other_number / number;

console.log(seven(times(five())), 35); // must return 35
console.log(four(plus(nine())), 13); // must return 13
console.log(eight(minus(three())), 5); // must return 5
console.log(six(dividedBy(two())), 3); // must return 3
#2

So like in math, things in brackets go first, right? So:

  1. Correct
  2. times(5) returns a function that will multiply 5 to a number that will be passed as argument
  3. seven(fn) has function from step 2 as argument, so it cannot just return 7, but runs fn passing 7 as argument and returns the result.

Hope this will help

#3

Step through it. Just as iiiked described above:

  1. five() returns 5
  2. times(5) returns the function (other_number) => other_number * 5;
  3. seven( (other_number) => other_number * 5 ) executes 7 * 5 and returns 35
#4

Hi,

Step 2: If there are two arrow functions, the argument is passed into the first function and then the second function with the logic is executed? I didn’t know this.

so another example would be:
four(plus(nine()

  1. four(plus(9)
    2)(other_number) => other_number + 9)

step 3: is this where operation(4) is called?

const four = (operation) => !operation ? 4 : operation(4);`

such that:

const four = (other_number) => other_number + 9) => 4 + 9

whereby other_number is the argument passed into the operation parameter?

is this the right thinking?

#5

Yes, you seem to be on the right track.

The function you list in 2) is used as the argument operation in the call to four().

#6

I am getting the value of each number and operation into one common function then processing it. Why it is not correct?

let i=0;
let a=[];
function zero() {
process(0)}
function one() {
process(1)}
function two() {
process(2)}
function three() {
process(3)}
function four() {
process(4)}
function five(a) {

//console.log(" Hii Five")
process(5)
}
function six() {
process(6)}
function seven() {

//console.log("hii seven")
process(7)
}
function eight() {
process(8)}
function nine() {
process(9)}


function plus() {
process('+')}
function minus() {
process('-')}
function times() {
//console.log("times hiii")
process('*')
}
function dividedBy() {
process('/')}

function process(args)
{
i=i+1

a[i]=args
//console.log("a is "+ a[i])
//console.log(a[1],a[2],a[3])
if(a[2]=='*' && a[3]!==undefined)
{i=0 
r=(a[1]*a[3])
a=[]
console.log(r)
}

if(a[2]=='+' && a[3]!==undefined)
{ i=0
r=(a[1]+a[3])

a=[]
console.log(r)
}

if(a[2]=='-' && a[3]!==undefined)
{ i=0
r=(a[3]-a[1])

a=[]
console.log(r)
}

if(a[2]=='/' && a[3]!==undefined)
{ i=0
r=(a[3]/a[1])

a=[]
console.log(r)
}


}
#7