Why does the absence of the assignment operator permit me to modify a Ruby constant with no compiler warning?

问题

In the following two examples I do the same thing, creating a constant String and using the concat method to modify it. Because it's a constant, I expect a compiler warning but only receive one in the second example when I use the assignment operator. Why is this?

X = "hello"
X.concat(" world")
puts X # no warning

X = "hello"
X = X.concat(" world")
puts X # warning: already initialized

Since the concat method modifies the string in place, that's normally what I would do, since there's no need to use an assigment operator. So, why does the presence of the assignment operator cause the compiler to identify these two operations as being different?

回答1:

This is because you're re-defining a new X. When you redefine a constant it gives you the "already initialized" error. The first example does not give this error because you're not redefining X, you're modifying it.

回答2:

In Ruby, variables are essentially pointers to a place in a memory containing an object -- not the object itself. In the second example, you are initializing a constant X to point to an object in the first line (X = "hello"), and in the second line, you are again initializing the constant -- but it already points to an object, so you get the error.

A constant's immutability doesn't mean you can't alter the object -- it just means you can't change the constant to point to another object.

回答3:

If you want to make your string "real" constant, try 'freeze':

X = "foo".freeze        # => "foo" 
X.concat("bar")

TypeError: can't modify frozen string
    from (irb):2:in `concat'
    from (irb):2

I really encourage you to read The Ruby Programming Languge.

回答4:

This is because the constant X is storing a reference to a String object. In your first example, you are modifying the internal state of the String object, but not the reference stored by the constant. In the second example, you are changing the reference stored by the constant to a new String object which is returned from the concat method.

The PickAxe book explains this here.

来源：https://stackoverflow.com/questions/670094/why-does-the-absence-of-the-assignment-operator-permit-me-to-modify-a-ruby-const