Question

我正在编写一个函数来替换所有出现的变量p和q及其各自的值，而不使用eval()，但是，我遇到了一些意想不到的行为。顺便说一句，我正在使用phpjs作为str_replace

function table(str){
    str=str_replace(["nand","nor","implies","equals","not","xor","and","or","(",")"],[" 3 "," 4 "," 5 "," 6 "," 7 "," 8 ", " 9 ", " 10 ", " ( "," ) "],str).replace(/\s{2,}/g, ' ').trim();
    str=str_replace(["3","4","5","6","7","8", "9", "10", "(",")"],["nand","nor","implies","equals","not","xor","and","or","(",")"],str).split(" ");

    var vars={p:1,q:1};
    for(vars['p']=1;vars['p']>=0;vars['p']--){
        for(vars['q']=1;vars['q']>=0;vars['q']--){
            alert(str);
            newinput=str;

            for(var i=0;i<newinput.length;i++){
                var token=newinput[i];
                if(token.length===1){
                    console.log(newinput[i]);
                    newinput[i]=vars[newinput[i]];


                }
            }
//          console.log(n.join(" "));

        }
    }
}

我有这个代码来替换所有的事件，但它不起作用。我警告每次输入的原始字符串，但字符串会发生变化。该函数的预期输出为p,and,q重复4次，相反，我有p,and,q，然后1,and,1重复3次。但是，我似乎没有对str的任何任务。有谁知道为什么会这样？

Answer 1

当您将newinput设置为str时，您仍然会引用该原始对象。当您稍后在newinput中更改该值时，会影响str变量。

如果要克隆对象，可以迭代str的属性，如下所示：

var newinput = {};
for(var key in str) {
    newinput[key] = str[key];
}

因此，复制原始对象并且不会影响它的值。假设您没有要在str对象内克隆的对象。如果这样做，只需递归运行此函数。

Updated Fiddle

Javascript意外的分配行为

1 个答案: