CODEWITHSUNDEEP
ccompile-time-constant
phs
phs

Reputation: 563

Define (and compute) a C constant at compile time

Say that my C code uses a constant RANGEMAX that is the largest power of 10 fitting in an int. I can define it with:

#include <limits.h>
#if (INT_MAX < 1)
#define RANGEMAX ERROR1
#elif (INT_MAX >= 1) && (INT_MAX < 10)
#define RANGEMAX 1
#elif (INT_MAX >= 10) && (INT_MAX < 100)
#define RANGEMAX 10
#elif (INT_MAX >= 100) && (INT_MAX < 1000)
#define RANGEMAX 100
...             /* well, you get the picture */
#elif (INT_MAX >= 10000000000000000)
#define RANGEMAX ERROR2
#endif

Is there a smarter way of doing such simple computations in the macro preprocessing phase?

And since these are "simple" computations, I prefer solutions that an average guy like me will understand by just reading the code.

Upvotes: 0

Views: 143

Answers (2)

chux
chux

Reputation: 154242

With a series of test confirming the approximate range of INT_MAX, multiplication by 1, 10, 100, 10000, 100000000, 10000000000000000, etc. can be done.

The following should work for 10000 <= INT_MAX < power(10,32)

#include <stdio.h>
#include <limits.h>

#define F1(x) (1)
#define F2(x) (((x) >= 10) ? 10 * F1((x)/10): F1(x))
#define F3(x) (((x) >= 100) ? 100 * F2((x)/100): F2(x))
#define F4(x) (((x) >= 10000) ? 10000 * F3((x)/10000): F3(x))
#define F5(x) (((x) >= 100000000) ? 100000000 * F4((x)/100000000): F4(x))

#define ITEST (INT_MAX)

#if ITEST/10000 >= 10000
  #if  ITEST/100000000 >= 100000000
    #define INT10_MAX (10000000000000000 * F6(ITEST/10000000000000000))
  #else
    #define INT10_MAX (100000000 * F5(ITEST/100000000))
  #endif
#else
  #define INT10_MAX (10000 * F4(ITEST/10000))
#endif

int main(void) {
  printf("%d\n", ITEST);
  printf("%d\n", INT10_MAX);
  return 0;
}

A pragmatic approach might assume INT_MAX to be power(2,15)-1, or power(2,31)-1, power(2,63)-1. Macro arithmetic needs to be careful about using values outside 32-bit range.

#if INT_MAX/10000 == 3
  #define INT10_MAX 10000
#elif INT_MAX/1000000000 == 2
  #define INT10_MAX 1000000000
#elif INT_MAX/1000000000/1000000000 == 9
  #define INT10_MAX 1000000000000000000
#else
  #error Unusual INT_MAX
#endif

Upvotes: 1

abligh
abligh

Reputation: 25189

INT_MAX is defined to be at least 32767 (i.e. 2^15-1). As int is signed, in practice you need only check for 2^15-1, 2^31-1 and (theoretically) 2^63-1. This will reduce the size of your #if block.

Upvotes: 3

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