ES6 块是否只能防止功能提升？

Question

有人请帮我理解以下场景：

//Outer funtion 
function foo() {
  console.log('outer foo');
}

{
  //Inner function
  function foo() {
    console.log('inner foo');
  }
}

foo(); //Says "inner foo"

我假设在上述情况下，内部函数的显式减速会在块执行后替换提升的外部函数。

这是否意味着 ES6 块仅在内部声明时阻止函数提升？

更新

许多人认为这些块没有影响功能。但是，请根据 MDN 查看以下场景 -

foo('outside');  // TypeError: foo is not a function
{
  function foo(location) {
   console.log('foo is called ' + location);
  }
  foo('inside'); // works correctly and logs 'foo is called inside'
}
To be more precise, the block statement is preventing the function declaration from being hoisted to the top of the scope. The function is behaving as if it were defined as a function expression and, as such, it is only the implicit variable declaration that gets hoisted to the top of the scope

另一个更新

文档有误，刚刚由提供所选答案的专家修复。

Answer 1

I assume that in the above case the Inner function's explicit deceleration is replacing the hoisted outer function after the block is executed.

函数声明，如 var 语句，在执行函数中的代码之前的函数初始扫描期间被提升到函数的顶部。

块与提升 var 语句或函数声明（两者都具有函数范围）无关。它们仅对具有块作用域的 let 和 const 重要。

ES6 块是否只能防止功能提升？

Are ES6 blocks only prevent function hoisting?

javascript

scope

hoisting