javascript - 参数传递: In and Out

Question

考虑:

Primitive data types are passed by value in JavaScript. This means that a copy is effectively made of a variable when it is passed to a function, so any manipulation local to the function leaves the original variables untouched.

function fiddle(arg1) {
      arg1 = "Fiddled with";
      console.log("In function fiddle str = "+arg1+"<br>");
}
var str = "Original Value";
console.log("Before function call str = "+str+"<br>");
fiddle(str);
console.log("After function call str ="+str+"<br>");

简单

composite types such as arrays and objects if used, they are passed by reference rather than value

考虑以下修改之前的 fiddle() 函数:

function fiddle(arg1) {
       arg1[0] = "Fiddled with";
       console.log("In function fiddle arg1 = "+arg1+"<br>");
}
var arr = ["Original", " Original ", " Original "];
console.log("Before function call arr = "+arr+"<br>");
fiddle(arr);
console.log("After function call arr ="+arr+"<br>");

到这里为止还好

从这里我发现自己真的很困惑，

function fiddle(arg1) {
      arg1 = ["Blasted!","Blasted!"];
      console.log("In function fiddle arg1 = "+arg1+"<br>");
}
var arr = ["Original", " Original ", " Original "];
console.log("Before function call arr = "+arr+"<br>");
fiddle(arr);
console.log("After function call arr ="+arr+"<br>"); // Why this hasn't changed?

有什么建议吗？可以说吗

composite types such as arrays and objects if used, they are passed by reference rather than value

在这种情况下也是如此吗？

Answer 1

看起来很清楚，在 fiddle 中，您正在用您当场创建的数组覆盖对 arr 的引用。您从未通过其引用触及 arr 本身，而只是引用，因此一旦您再次外出，arr 仍然是原始数组。