var add = (x, y) => x + y

var curry = function(fn) { var limit = fn.length ... }

curry函数要返回一个函数, 这个函数是要执行的，那么问题就是，我们要判断这个函数的执行是否激活了原始函数的执行，问题就出现在传入的参数上面。返回函数还是结果？这的确是一个问题

var curry = function(fn) { var limit = fn.length return function (...args) { if (args.length >= limit) { return fn.apply(null, args) } } }

var curry = function(fn) { var limit = fn.length return function (...args) { if (args.length >= limit) { return fn.apply(null, args) } else { return function(...args2) { } } } }

var curry = function(fn) { var limit = fn.length return function judgeCurry (...args) { if (args.length >= limit) { return fn.apply(null, args) } else { return function(...args2) { return judgeCurry.apply(null, args.concat(args2)) } } } }

var currySingle = fn => judgeCurry = (...args) => args.length >= fn.length ? fn.apply(null, args) : (...args2) => judgeCurry.apply(null, args.concat(args2))

var currySingle = fn => { var limit = fn.length var judgeCurry = null return judgeCurry = (...args) => args.length >= limit ? fn.apply(null, args) : (...args2) => judgeCurry.apply(null, args.concat(args2)) }

var currySingle = fn => ((limit) => judgeCurry = (...args) => args.length >= limit ? fn.apply(null, args) : (...args2) => judgeCurry.apply(null, args.concat(args2)))(fn.length)

tags: args,fn,函数,var,gt,length,limit,curry,参数,null,function,args2,return,judgeCurry,apply

1.凡CodeSecTeam转载的文章,均出自其它媒体或其他官网介绍,目的在于传递更多的信息,并不代表本站赞同其观点和其真实性负责；
2.转载的文章仅代表原创作者观点,与本站无关。其原创性以及文中陈述文字和内容未经本站证实,本站对该文以及其中全部或者部分内容、文字的真实性、完整性、及时性，不作出任何保证或承若；
3.如本站转载稿涉及版权等问题,请作者及时联系本站,我们会及时处理。