JavaScript函数柯里化实现原理及过程

// 如何实现 add(1,2) === add(1)(2)
let add = (...args) => {
  return args.length === 1 ? a => a + args[0] : args[0] + args[1]
}

let add = (...args) => args.reduce((a,b)=>a+b)
 
let currying = (fn)=>{
  return function(...args){
    return fn.apply(this, args)
  }
}
// add(1,2,3,,,) === currying(add)(1,2,3,,,)

let currying = (fn) => {
  var args = [].slice.call(arguments, 1)
  return function() {
    var newArgs = args.concat([].slice.call(arguments))
    return fn.apply(this, newArgs)
  }
}
 
let addCurry = currying(add, 1, 2);
addCurry() // 3
 
addCurry = currying(add, 1);
addCurry(2) // 3
 
addCurry = currying(add);
addCurry(1, 2) // 3