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freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/i-before-e-except-after-c.md
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---
id: 5a23c84252665b21eecc7eb0
title: I 在 E 之前,除了 C 之后
challengeType: 1
forumTopicId: 302288
dashedName: i-before-e-except-after-c
---
# --description--
**<a href="http://www.rosettacode.org/wiki/I_before_E_except_after_C" target="_blank" rel="noopener noreferrer nofollow">"I before E, except after C"</a>** is a general rule for English language spelling. If one is unsure whether a word is spelled with the digraph `ei` or `ie`, the rhyme suggests that the correct order is `ie` unless the preceding letter is `c`, in which case it may be `ei`.
Using the words provided, check if the two sub-clauses of the phrase are plausible individually:
<ol>
<li>
<i>"I before E when not preceded by C".</i>
</li>
<li>
<i>"E before I when preceded by C".</i>
</li>
</ol>
如果两个子短语都是合理的,则原始短语可以说是合理的。
# --instructions--
Write a function that accepts a word and check if the word follows this rule. The function should return true if the word follows the rule and false if it does not.
# --hints--
`IBeforeExceptC` 应该是一个函数。
```js
assert(typeof IBeforeExceptC == 'function');
```
`IBeforeExceptC("receive")` 应该返回一个布尔值。
```js
assert(typeof IBeforeExceptC('receive') == 'boolean');
```
`IBeforeExceptC("receive")` 应该返回 `true`
```js
assert.equal(IBeforeExceptC('receive'), true);
```
`IBeforeExceptC("science")` 应该返回 `false`
```js
assert.equal(IBeforeExceptC('science'), false);
```
`IBeforeExceptC("imperceivable")` 应该返回 `true`
```js
assert.equal(IBeforeExceptC('imperceivable'), true);
```
`IBeforeExceptC("inconceivable")` 应该返回 `true`
```js
assert.equal(IBeforeExceptC('inconceivable'), true);
```
`IBeforeExceptC("insufficient")` 应该返回 `false`.
```js
assert.equal(IBeforeExceptC('insufficient'), false);
```
`IBeforeExceptC("omniscient")` 应该返回 `false`.
```js
assert.equal(IBeforeExceptC('omniscient'), false);
```
# --seed--
## --seed-contents--
```js
function IBeforeExceptC(word) {
}
```
# --solutions--
```js
function IBeforeExceptC(word)
{
if(word.indexOf("c")==-1 && word.indexOf("ie")!=-1)
return true;
else if(word.indexOf("cei")!=-1)
return true;
return false;
}
```