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@neftaly
Last active April 10, 2022 18:47
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ES6 Auto-curry
const curry = (fn, ...oldArgs) => (...newArgs) => {
const args = [...oldArgs, ...newArgs];
return (args.length < fn.length) ? curry(fn, ...args) : fn(...args);
};
const curryN = (arity, fn) => {
const receiver = preArgs => (...postArgs) => {
const args = [...preArgs, ...postArgs];
return (args.length < arity) ? receiver(args) : fn(...args);
}
return receiver([]);
}
const curry = fn => curryN(fn.length, fn);
const curryN = (arity, fn) => {
// Take old arguments, and an array of new arguments
const receiver = oldArgs => (...newArgs) => {
const args = [...oldArgs, ...newArgs]; // Combine old and new arguments
if (args.length >= arity) { // Have enough arguments have been received?
return fn(...args); // Run fn with argument list
}
return receiver(args); // Recurse, and await further arguments
};
return receiver([]); // Start with an empty array of prior arguments
};
const curry = fn => curryN(fn.length, fn);
const add3 = function (a, b, c) {
return a + b + c;
};
const x = curry(add3);
x(1)(1)(1); //=> 3
x(1, 1)(1); //=> 3
x(1, 1, 1); //=> 3
@meChrisReed
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I like what you have here.
I think you can reduce this to:

const curry = (fn, ...args) => (
  args.length === fn.length ?
    fn(...args) :
    curry.bind(null, fn, ...args)
);

const add3 = curry((a,b,c) => a+b+c);
add3(1)(1)(1); //=> 3
add3(1, 1)(1); //=> 3
add3(1, 1, 1); //=> 3

Please let me know your thoughts or if there is some functionality lost.

@neftaly
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neftaly commented Jun 21, 2016

Looks good! This was actually for explaining how auto-currying worked in a talk, so I wouldn't have used bind (too much cognitive load for audience), but it'd be great otherwise.

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