Is the C99 preprocessor Turing complete?

Question

After discovering the Boost preprocessor\'s capabilities I found myself wondering: Is the C99 preprocessor Turing complete?

If not, what does it lack to not qualify?

Answer 1

To be Turing complete, one needs to define recursion that may never finish -- one calls them mu-recursive operator.

To define such an operator one needs an infinite space of defined identifiers (in case that each identifier is evaluated a finite number of times), as one cannot know a priori an upper limit of time in which the result is found. With a finite number of operators inside the code one needs to be able to check an unlimited number of possibilities.

So this class of functions cannot be computed by the C preprocessor because in C preprocessor there is a limited number of defined macros and each one is expanded only once.

The C preprocessor uses the Dave Prosser's algorithm (written by Dave Prosser for the WG14 team in 1984). In this algorithm a macro is painted blue in the moment of the first expansion; a recursive call (or mutual recursive call) does not expand it, as it has already been painted blue in the moment when the first expansion starts. So with a finite number of preprocessing lines it is impossible to make infinite calls of functions(macros), which characterizes the mu-recursive operators.

The C preprocessor can compute only sigma-recursive operators .

For details see the course of computation of Marvin L. Minsky (1967) -- Computation: Finite and Infinite Machines, Prentice-Hall, Inc. Englewood Cliffs, N.J. etc.