Question

在下面的代码片段中，最后一个控制台日志显示（节点值）空字典，你能告诉我原因吗？

function add(n1, n2) {
    const sum = []
    let carry = 0
    for (let i = 0; i < n1.length || i < n2.length || carry; i++) {
        const added = (n1[i] || 0) + (n2[i] || 0) + carry
        sum[i] = added % 10
        carry = added > 9 //floor(added / 10)
    }
    return sum
}
function times256(n1) {
    for (let i = 8; i; i--) n1 = add(n1, n1)
    return n1
}
function toString(buffer) {
    const isNegative = buffer[0] & 128 //check if high bit is set
    if (isNegative) { //convert to positive, using 2's complement
        buffer = buffer.map(b => ~b) //invert all bits
        let i = buffer.length - 1
        while (buffer[i] === 255) { //add 1 to the number, carrying if necessary
            buffer[i] = 0
            i--
        }
        buffer[i]++
    }
    const result = buffer.reduce((sum, b) =>
        add(
            times256(sum), //multiply sum by 256
            String(b).split('').map(Number).reverse() //then add b
        ),
        []
    )
    const stringResult = result.reverse().join('')
    if (isNegative) return '-' + stringResult
    else return stringResult
}

Answer 1

因为您使用list1[each]创建了一个基本上是1 ... 3的新属性，它是一个对象。此对象已分配给节点。

＆＃13;

var abc = function(t1, key) {
    t1[key] = {};
    return t1;
}

var node = {},
    list1 = [1, 2, 3],
    each;

for (each in list1) {
    console.log(node);
    console.log(abc(node, list1[each])[list1[each]]);
    console.log(list1[each]);
    node = abc(node, list1[each])[list1[each]];
}

console.log(node);

＆＃13;

要获取具有所需属性的对象，可以使用另一个变量并将新对象移动到给定对象内。

＆＃13;

var abc = function(t1, key) {
    t1[key] = {};
    console.log(t1);
    return t1[key]; // return new object
}

var node = {},
    temp = node,
    list1 = [1, 2, 3],
    each;

for (each in list1) {
    temp = abc(temp, list1[each]); // assign just the new object
}

console.log(node);

＆＃13;

Answer 2

这是因为您在每次循环迭代中都覆盖了节点变量。尝试制作备份参考：

node = {};
nodebackup = node;
list1=[1,2,3];
for(each in list1){
    console.log(nodebackup);
    var abc = function(t1, key){
        t1[key] = {};
        return t1;
    }
    node=abc(node, list1[each])[list1[each]]
}
console.log(nodebackup)

为什么从java脚本函数返回空字典

2 个答案: