freeCodeCamp/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-the-rest-parameter-with-function-parameters.md
Oliver Eyton-Williams ee1e8abd87
feat(curriculum): restore seed + solution to Chinese (#40683)
* feat(tools): add seed/solution restore script

* chore(curriculum): remove empty sections' markers

* chore(curriculum): add seed + solution to Chinese

* chore: remove old formatter

* fix: update getChallenges

parse translated challenges separately, without reference to the source

* chore(curriculum): add dashedName to English

* chore(curriculum): add dashedName to Chinese

* refactor: remove unused challenge property 'name'

* fix: relax dashedName requirement

* fix: stray tag

Remove stray `pre` tag from challenge file.

Signed-off-by: nhcarrigan <nhcarrigan@gmail.com>

Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
2021-01-12 19:31:00 -07:00

1.7 KiB

id, title, challengeType, forumTopicId, dashedName
id title challengeType forumTopicId dashedName
587d7b88367417b2b2512b47 将 rest 操作符与函数参数一起使用 1 301221 use-the-rest-parameter-with-function-parameters

--description--

ES6 推出了用于函数参数的rest 操作符帮助我们创建更加灵活的函数。在rest操作符的帮助下,你可以创建有一个变量来接受多个参数的函数。这些参数被储存在一个可以在函数内部读取的数组中。

请看以下代码:

function howMany(...args) {
  return "You have passed " + args.length + " arguments.";
}
console.log(howMany(0, 1, 2)); // You have passed 3 arguments.
console.log(howMany("string", null, [1, 2, 3], { })); // You have passed 4 arguments.

rest操作符可以避免查看args数组的需求,并且允许我们在参数数组上使用map()fiter()reduce()

--instructions--

修改sum函数,来让它使用rest操作符,并且它可以在有任何数量的参数时以相同的形式工作。

--hints--

sum(0,1,2)的返回结果应该为3。

assert(sum(0, 1, 2) === 3);

sum(1,2,3,4)的返回结果应该为10。

assert(sum(1, 2, 3, 4) === 10);

sum(5)的返回结果应该为5。

assert(sum(5) === 5);

sum()的返回结果应该为 0。

assert(sum() === 0);

sum函数的args参数使用了...展开操作符。

assert(code.replace(/\s/g, '').match(/sum=\(\.\.\.args\)=>/));

--seed--

--seed-contents--

const sum = (x, y, z) => {
  const args = [x, y, z];
  return args.reduce((a, b) => a + b, 0);
}

--solutions--

const sum = (...args) => {
  return args.reduce((a, b) => a + b, 0);
}