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Is the C preprocessor Turing complete?

pfultz2 edited this page Nov 22, 2012 · 16 revisions

One of the requirements for a language to be turing complete, is that it must be able to implement recursion. As we all know macros don't expand recursively, but there are ways we can work around this.

The easiest way of doing recursion in the preprocessor is to use a deferred expression. A deferred expression is an expression that requires more scans to fully expand:

#define EMPTY()
#define DEFER(id) id EMPTY()
#define EXPAND(...) __VA_ARGS__

#define A() 123
A() // Expands to 123
DEFER(A)() // Expands to A () because it requires one more scan to fully expand
EXPAND(DEFER(A)()) // Expands to 123, because the EXPAND macro forces another scan

Why is this important? Well when a macro is scanned and expanding, it creates a disabling context. This disabling context will cause a token, that refers to the currently expanding macro, to be painted blue. Thus, once its painted blue, the macro will no longer expand. This is why macros don't expand recursively. However, a disabling context only exists during one scan, so by deferring an expansion we can prevent our macros from becoming painted blue. We will just need to apply more scans to the expression. We can do that using this EVAL macro:

#define EVAL(...)  EVAL1(EVAL1(EVAL1(__VA_ARGS__)))
#define EVAL1(...) EVAL2(EVAL2(EVAL2(__VA_ARGS__)))
#define EVAL2(...) EVAL3(EVAL3(EVAL3(__VA_ARGS__)))
#define EVAL3(...) EVAL4(EVAL4(EVAL4(__VA_ARGS__)))
#define EVAL4(...) EVAL5(EVAL5(EVAL5(__VA_ARGS__)))
#define EVAL5(...) __VA_ARGS__

Now if we want to implement a REPEAT macro using recursion, first we need some increment and decrement operators to handle state:

#define CAT(a, ...) PRIMITIVE_CAT(a, __VA_ARGS__)
#define PRIMITIVE_CAT(a, ...) a ## __VA_ARGS__

#define INC(x) PRIMITIVE_CAT(INC_, x)
#define INC_0 1
#define INC_1 2
#define INC_2 3
#define INC_3 4
#define INC_4 5
#define INC_5 6
#define INC_6 7
#define INC_7 8
#define INC_8 9
#define INC_9 9

#define DEC(x) PRIMITIVE_CAT(DEC_, x)
#define DEC_0 0
#define DEC_1 0
#define DEC_2 1
#define DEC_3 2
#define DEC_4 3
#define DEC_5 4
#define DEC_6 5
#define DEC_7 6
#define DEC_8 7
#define DEC_9 8

Next we need a few more macros to do logic:

#define CHECK_N(x, n, ...) n
#define CHECK(...) CHECK_N(__VA_ARGS__, 0,)

#define NOT(x) CHECK(PRIMITIVE_CAT(NOT_, x))
#define NOT_0 ~, 1,

#define COMPL(b) PRIMITIVE_CAT(COMPL_, b)
#define COMPL_0 1
#define COMPL_1 0

#define BOOL(x) COMPL(NOT(x))

#define IIF(c) PRIMITIVE_CAT(IIF_, c)
#define IIF_0(t, ...) __VA_ARGS__
#define IIF_1(t, ...) t

#define IF(c) IIF(BOOL(c))

#define EAT(...)
#define EXPAND(...) __VA_ARGS__
#define WHEN(c) IF(c)(EXPAND, EAT)

Now with all these macros we can write a recursive REPEAT macro. We use a REPEAT_INDIRECT macro to refer back to itself recursively. This prevents the macro from being painted blue, since it will expand on a different scan(and using a different disabling context).

#define REPEAT(count, macro, ...) \
    WHEN(count) \
    ( \
        DEFER(REPEAT_INDIRECT) () \
        ( \
            DEC(count), macro, __VA_ARGS__ \
        ) \
        DEFER(macro) \
        ( \
            DEC(count), __VA_ARGS__ \
        ) \
    )
#define REPEAT_INDIRECT() REPEAT

//An example of using this macro
#define M(s, i, _) i
EVAL(REPEAT(8, M, ~)) // 0 1 2 3 4 5 6 7

Now this example is limited to 10 repeats, because of limitations of the counter. Just like a repeat counter in a computer would be limited by the finite memory. Multiple repeat counters could be combined together to workaround this limitation, just like in a computer. Furthermore, we could define a FOREVER macro:

#define FOREVER() \
    ? \
    DEFER(FOREVER_INDIRECT) () ()
#define FOREVER_INDIRECT() FOREVER
// Outputs question marks forever
EVAL(FOREVER())

This will try to output ? forever, but will eventually stop because there are no more scans being applied. Now the question is, if we gave it an infinite number of scans would this algorithm complete? This is known as the halting problem, and Turing completeness is necessary to prove the undecidability of the halting problem. So as you can see, the preprocessor can act as a Turing complete language, but instead of being limited to the finite memory of a computer it is instead limited by the finite number of scans applied.

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