currying function instead of if condition in javascript

Question

Tried to use curry function to write if condition (without "if" "true" and "false"). Have no idea how to do it right! Can anyone help and show me how to do it?

const True = () => {};
const False = () => {};
const If = True => False => ifFunction(True)(False);

const ifFunction = If(True);
ifFunction('1')('2'); // 1

console.log(If(True)('1')('2'));  // 1
console.log(If(False)('1')('2')); // 2

It should return 1 or 2 depends on which function is getting pass to if condition. But this code does not work at all.

For the record, here is Ramda's implementation of ifElse which implements ifElse(conditionFn, trueBranchFn, falseBranchFn). The library uses a simple conditional operator as the implementation (simplified): conditionFn() ? trueBranchFn() : falseBranchFn() — VLAZ
– VLAZ, Commented Jan 6, 2020 at 11:05

Nina Scholz · Accepted Answer · 2020-01-06 15:28:52Z

7

By using Church_Booleans, you could use the following functions.

const
    TRUE = a => b => a,
    FALSE = a => b => b,
    IF = p => a => b => p(a)(b);

console.log(IF(TRUE)('1')('2'));  // 1
console.log(IF(FALSE)('1')('2')); // 2

edited Jan 6, 2020 at 15:28

answered Jan 6, 2020 at 11:00

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Aadit M Shah · Accepted Answer · 2020-01-08 02:34:16Z

2

Here's a way to implement True, False, and If without using any control flow statements like if...else or switch..case, control flow expressions like ?:, or higher-order functions.

const True  = 1;
const False = 0;

const If = Bool => Then => Else => [Else, Then][Bool];

console.log(If(True)('1')('2'));  // 1
console.log(If(False)('1')('2')); // 2

This is a defunctionalization of the Church encoding solution. Hence, it's more efficient.

edited Jan 8, 2020 at 2:34

answered Jan 8, 2020 at 2:29

Aadit M Shah

74.5k31 gold badges177 silver badges311 bronze badges

1 Comment

A1rPun Over a year ago

You can accept truthy/falsy values if you modify [Bool] to [Number(!!Bool)].

A1rPun · Accepted Answer · 2020-01-06 10:58:47Z

1

What about this?

const True = () => true;
const False = () => false;
const If = exp => exp() ? a => b => a : a => b => b;

console.log(If(True)('1')('2'));  // 1
console.log(If(False)('1')('2')); // 2

answered Jan 6, 2020 at 10:58

A1rPun

17k8 gold badges60 silver badges93 bronze badges

1 Comment

Lex27 Over a year ago

thank you. but I trying to do it without "if" (even ternary)

Collectives™ on Stack Overflow

currying function instead of if condition in javascript

3 Answers 3

Comments

1 Comment

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related