“foo”下方的功能不起作用。我想要一种方法来获得下面显示的输出,解决“foo”中显示的原理。
function foo() {
var i;
i = "hello world";
i.a = "hello kitten";
i.b = "hello... Is there anybody out there?"
return i; // This doesn't work
}
这就是我想要的:
alert(foo()); // "hello world"
var bar = foo();
alert(bar.a); // "hello kitten"
alert(bar.b); // "hello... Is there anybody out there?"
感谢。
答案 0 :(得分:4)
使用字符串对象而不是字符串值。
i = new String("hello world");