Why does naming a function expression not make the name accessible within the scope, but still possible to delete?

Question

If I assign a named function to a variable, why I can't access to the named function:

var a = function b() {
 console.log("Hello World");
}
a() // hello world;
b() // b is not defined
b // b is not defined

and at this time, I can not delete the a variable, but I can delete b, when I delete the b, the a's function can still there

delete a //false
delete b //true
a() //Hello World

the b function isn't just referenced by the a, not copy, so why the a function is still there?

Answer 1

If you use a named function expression (not a function declaration!), the name of the function is only accessible form inside the function.

From the specification ("Identifier" refers to the name of function, i.e. function Identifier() {}):

The Identifier in a FunctionExpression can be referenced from inside the FunctionExpression's FunctionBody to allow the function to call itself recursively. However, unlike in a FunctionDeclaration, the Identifier in a FunctionExpression cannot be referenced from and does not affect the scope enclosing the FunctionExpression.

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but I can delete b

No, you can't. When you pass anything that is not a reference to delete or the reference cannot be resolved, it will return true.

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This is a great article about all the function definition stuff: http://kangax.github.io/nfe/.