我已经测试了我的.indexOf
实现,它似乎适用于我的测试用例,但是我已经看到了其他解决方案,它们添加了if条件array[i] === elem && notFound === -1
。第二个条件的目的是什么?这是我的版本:
var indexOf = function (array, elem) {
var notFound = -1;
for (var i = 0; i < array.length; i++) {
if (array[i] === elem) {return i;}
}
return notFound;
}
你的实现对我来说似乎很好。
我怀疑额外的检查是在找到匹配项时不会脱离循环的函数版本中。当它找到匹配项时,它会设置notFound = i;
,并继续循环(不必要(;循环完成后,它返回notFound
.测试notFound === -1
是否阻止它在找到第二个匹配项时更新变量。
如果不是这样,如果您实际发布使用您显示的检查的实现示例,这将有所帮助。这可能只是一个错误 - 仅仅因为你在互联网上找到了代码,并不意味着它是正确的。
看看下面的 polyfill(来自 mdn(:
// Production steps of ECMA-262, Edition 5, 15.4.4.14
// Reference: http://es5.github.io/#x15.4.4.14
if (!Array.prototype.indexOf) {
Array.prototype.indexOf = function(searchElement, fromIndex) {
var k;
// 1. Let o be the result of calling ToObject passing
// the this value as the argument.
if (this == null) {
throw new TypeError('"this" is null or not defined');
}
var o = Object(this);
// 2. Let lenValue be the result of calling the Get
// internal method of o with the argument "length".
// 3. Let len be ToUint32(lenValue).
var len = o.length >>> 0;
// 4. If len is 0, return -1.
if (len === 0) {
return -1;
}
// 5. If argument fromIndex was passed let n be
// ToInteger(fromIndex); else let n be 0.
var n = +fromIndex || 0;
if (Math.abs(n) === Infinity) {
n = 0;
}
// 6. If n >= len, return -1.
if (n >= len) {
return -1;
}
// 7. If n >= 0, then Let k be n.
// 8. Else, n<0, Let k be len - abs(n).
// If k is less than 0, then let k be 0.
k = Math.max(n >= 0 ? n : len - Math.abs(n), 0);
// 9. Repeat, while k < len
while (k < len) {
// a. Let Pk be ToString(k).
// This is implicit for LHS operands of the in operator
// b. Let kPresent be the result of calling the
// HasProperty internal method of o with argument Pk.
// This step can be combined with c
// c. If kPresent is true, then
// i. Let elementK be the result of calling the Get
// internal method of o with the argument ToString(k).
// ii. Let same be the result of applying the
// Strict Equality Comparison Algorithm to
// searchElement and elementK.
// iii. If same is true, return k.
if (k in o && o[k] === searchElement) {
return k;
}
k++;
}
return -1;
};
}