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If $M$ halts on every input, simulate $M$ on input $n$ while counting its steps, and output that count when it halts. The resulting function $t_M(n)$ is a total computable function. Consequently yes: one may take the exact running time,This construction does not promise a simple closed expression or a bound of any particular complexity class; it uses the promised totality of this particular machine.

For a machine computing a partial computable function, its exact halting-time function is still a partial computable function, with the same domain. A finite bound cannot cover a genuinely infinite computation. Even restricting attention to inputs on which $M$ halts, a total computable function bounding all halting times need not exist. In fact the precise characterization is the computable bound on halting time criterion:For the forward implication, compute $f(n)$ and simulate $M(n)$ for that many steps. If it has not halted, the proposed bound ensures that it never will. This decides the domain. Conversely, first decide whether $n$ is in the domain; return zero outside it and the simulated halting time inside it. This defines a total computable function satisfying the bound. The time needed to compute the bound itself is irrelevant to this argument.

A universal halting recognizer has undecidable domain by the halting problem, so it has no such total bound. On the other hand, a machine that immediately halts on even inputs and loops on odd inputs computes a nontotal partial computable function but has a constant bound on its halting times. Thus nontotality alone does not decide whether a total bound exists.