总结回顾html
昨天在看一道笔试题的时候本觉得很简单,可是结果不是我想象的那样,直接上笔试题。前端
const array = new Array(5).map((item) => {
return item = {
name: '1'
}
});
console.log(array);
// 请写出输出结果
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我想象的答案:[{name: '1'}, {name: '1'}, {name: '1'}, {name: '1'}, {name: '1'}]; 面试
实际的答案:[empty × 5] 数组
为何会这样了?bash
我第一个想到的是new Array(5)生成的数组是[undefined, undefined, undefined, undefined, undefined]。网络
const array = [undefined, undefined, undefined, undefined, undefined];
const newArr = array.map((item) => {
return item = {
name: '1'
}
});
console.log(newArr);
// 结果是[{name: '1'}, {name: '1'}, {name: '1'}, {name: '1'}, {name: '1'}];
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猜测1错误闭包
new Array(5)生成的数组在每一项都没有值,意思就是生成了[,,,,,]一个这样的数组。app
const array = [,,,,,];
const newArr = array.map((item) => {
return item = {
name: '1'
}
});
console.log(newArr);
// 结果是[empty × 5];
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猜测2正确(这里大喊本身牛逼)ide
map calls a provided callback function once for each element in an array, in order, and constructs a new array from the results. callback is invoked only for indexes of the array which have assigned values (including undefined).函数
It is not called for missing elements of the array; that is:
map依次为数组中的每一个元素调用一次提供的callback函数,而后根据结果构造一个新的数组。-----仅对已分配值(包括)的数组索引进行调用----。 map函数的回调函数只会被赋过值的项调用。new Array(1) 和 [undefined]不同。new Array(1)没有为数组中的项赋过值,而[undefined]为数组中的项赋了一个undefined值。
// 处理1
const array = new Array(5).fill().map((item) => {
return item = {
name: '1'
}
});
console.log(array);
// [{name: '1'}, {name: '1'}, {name: '1'}, {name: '1'}, {name: '1'}];
// 处理2
var array = Array.apply(0,{length:5}).map((item) => {
return item = {
name: '1'
}
});
console.log(array);
// [{name: '1'}, {name: '1'}, {name: '1'}, {name: '1'}, {name: '1'}];
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