Decidability or computability in the class P

Question

It is well known that problems in the class P are characterized as those problems that are decidable in polynomial time by means of a deterministic Turing machine. My question is the following: If a decision problem can be decided by a deterministic procedure in polynomial time, i.e. it produces two possible outputs, YES or NOT, then the problem is in P? Is necessary to generate a computational procedure to solve the problem to guarantee that the problem is in P? Here it is clear that computability implies decidability however the inverse is not always true.

Answer 1

Following your comment discussion, this might answer your question:

1. The definition of $P$ (and NP as well) is concerned with decision problems only. So if you find a polynomial algorithm to solve 3-SAT that only gives "yes/no", and no actual solution, that is enough to prove that 3-SAT is in P.

2. For all problems I can think of the distinction between the decision-problem and the computation-problem does not matter actually. Suppose you have a polynomial decision-algorithm for 3-SAT. In order to produce a solution you can simply run this algorithm multiple times on slightly smaller instances in wich you fix one single variable at a time. Depending on the answer of the decision-algorithm (which is only yes/no) you know the value of that single variable. Given that the decision-algorithm is polynomial, calling that algorihtm $n$ times is still polynomial.