TITLE: Intuitive meaning of the concept “computable”
QUESTION [7 upvotes]: My question is a follow-up question to this one: How to show that a function is computable?
The original question was:

Is the following function $$g(x) = \begin{cases} 1 & \mbox{if } \phi_x(x) \downarrow \mbox{or } x \geq 1 \\ 0 & \mbox{otherwise } \end{cases}$$ computable?

The accepted answer was that this function is computable because it is either the constant 1 function or the function that is 1 everywhere except on input 0.
According to my professor at university, this answer is in fact correct. But I don't understand the intuition behind it.
As far as I know, “computable” means (amongst other things) that there is an algorithm that actually can compute the function. I don't really get how such an algorithm would look in this case. Obviously, if $x\geq 1$, the algorithm can immediately return 1. But if the input is 0, the algorithm would have to simulate $\phi_0(0)$. If $\phi_0$ terminates on input 0, then the algorithm can return 1. But if $\phi_0$ doesn't terminates on input 0, the algorithm will run forever and never return 0.
So my understanding of the concept “computable” is in conflict with its actual meaning. Can someone explain where the error in my reasoning lies?

REPLY [5 votes]: A function is not the same thing as the words used to describe it. The function is the just the mapping of inputs to outputs. That mapping can be described in infinitely many different ways. For example, the identity function could be described both as "Given $x$, output $x$", or as "Given $x$, add $5$ to $x$. Now multiply $x$ by $3$, and subtract $15$. Now divide by $3$. Output the result."
The question "is this function computable" means "does there exist an algorithm which happens to produce the same outputs as this function for all inputs", and not "does there exist an algorithm which resembles this description of the function", because you're asking about the function, not the particular description of it.