javascript - 为什么闭包在这些函数中不起作用？

Question

为什么闭包在这里不起作用？ createTreeText() 函数不是应该从调用它的函数中获取文本变量吗？我知道我可以在参数中传递它，但为什么我不能通过闭包来做到这一点？

function createTree(){
    var text = "example";
    text = createTreeText();
    console.log(text);
}

function createTreeText(){
    var newText = text.toUpperCase(); // error happens here
    return newText;
}

createTree();

Answer 1

Isn't the createTreeText() function supposed to take the text variable from function where it's called

不，一点也不。函数在创建变量的范围内关闭变量，而不是在调用变量的范围内。所有函数从调用它们的地方获取的是作为参数传递给它们的信息(有时是 this，具体取决于它们的调用方式以及它们是什么类型的函数)。

这个例子可能会澄清，请参阅评论:

function wrapper() {
  var a = Math.random();
  
  function foo() {
    // `foo` closes over `a`, because `a` is in scope
    // where `foo` was created
    console.log("a = " + a);

    // `foo` does not close over `b`, because `b` is not in scope
    // where `foo` was created
    try {
      console.log("b = " + b); // throws ReferenceError
    } catch (e) {
      console.log("error: " + String(e));
    }
  }
  
  function bar() {
    var b = Math.random();
    // Calling `foo` here does nothing to grant it access to `b`
    foo();
  }
  
  bar();
}

wrapper();