Call a C preprocessor macro multiple times (through a variable)

Question

I'd like to clean up my assembly code and povide a way to call "NOP" multiple times through a macro:

#define NOP() asm(" nop")

#define NOP_N( N ) \
    NOP(); \
    NOP(); \
    .... call NOP() N times

I can't figure if this is possible in a macro.

Obviously, for performance reasons, I don't want something like this:

#define NOP_N( n ) { register int i; for(i=0;i<n;i++) asm(" nop"); }

Which defeats the purpose of NOP:

L17:                                    ; NOP_N(3);
        nop
        addi      1,r0                  ; Unsigned
        cmpi      3,r0
        blo       L17

The code is in C and assembly, so no C++ can be involved in here. Also, the compiler is fairly old and doesn't support variadic macros...

how do you write it as macro for N, call NOP() N times, N has to be know at coding time — Grijesh Chauhan
– Grijesh Chauhan, Commented May 17, 2013 at 8:59
@Mogria: Well if xgbi can use C++, then this could be done fairly simply via template metaprogramming, and no preprocessor trickery would be necessary at all. — jamesdlin
– jamesdlin, Commented May 17, 2013 at 9:08
The code is in C and assembly, so no C++ can be involved in here. Also, the compiler is fairly old and doesn't support variadic macros... — Gui13
– Gui13, Commented May 17, 2013 at 9:26

krlmlr · Accepted Answer · 2013-05-17 09:07:54Z

7

I don't think a solution for unbounded N is possible. For bounded N you could do something along the following lines:

#define REPEAT_0(WHAT)
#define REPEAT_1(WHAT) WHAT REPEAT_0(WHAT)
#define REPEAT_2(WHAT) WHAT REPEAT_1(WHAT)
#define REPEAT_3(WHAT) WHAT REPEAT_2(WHAT)

#define NOP_N(N) REPEAT_##N(asm("nop");)

The first part can be autogenerated easily. The technique employed for the second part is sometimes called token pasting.

edited May 17, 2013 at 9:07

answered May 17, 2013 at 9:01

krlmlr

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1 Comment

Gui13 Over a year ago

That's the closest that I can foresee as possible without variadic macros.

Community · Accepted Answer · 2017-05-23 12:00:07Z

2

what about this and Is the C preprocessor Turing complete?:

#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__

AND REPEAT_INDIRECT

#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

AND FOREVER:

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

edited May 23, 2017 at 12:00

CommunityBot

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answered May 17, 2013 at 9:07

0x90

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1 Comment

krlmlr Over a year ago

Care to elaborate what this code does? Otherwise the link is great in a comment but does not qualify for an answer.

Jens Gustedt · Accepted Answer · 2013-05-17 10:04:47Z

0

If you really want to do this in the preprocessor (and have a C99 compliant compiler) you could use P99_UNROLL from P99.

But you are completely underestimating the optimization that modern compilers are capable. As long as the bounds are compile time constants, a good compiler with optimization should just unroll the code for you. Look into the assembler to be sure (gcc has -S for that).

But you probably could help the compiler a bit by coding this "correctly":

#define NOP_N(N) for(register unsigned i=0; i < (N); i++) asm(" nop")

That is, have the loop counter local to the for and use an unsigned type such that there can't be theoretical issues with overflow.

answered May 17, 2013 at 10:04

Jens Gustedt

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1 Comment

Gui13 Over a year ago

The thing is that the compiler is old. I checked that using a for loop will actually initialize the integer and not unroll the loop.

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