How to use #if inside #define in the C preprocessor

cc-preprocessor

I want to write a macro that spits out code based on the boolean value of its parameter. So say DEF_CONST(true) should be expanded into const, and DEF_CONST(false) should be expanded into nothing.

Clearly the following doesn't work because we can't use another preprocessor inside #defines:

#define DEF_CONST(b_const) \
#if (b_const) \
  const \
#endif

Best Answer

You can simulate conditionals using macro token concatenation as follows:

#define DEF_CONST(b_const) DEF_CONST_##b_const
#define DEF_CONST_true const
#define DEF_CONST_false

Then,

/* OK */
DEF_CONST(true)  int x;  /* expands to const int x */
DEF_CONST(false) int y;  /* expands to int y */

/* NOT OK */
bool bSomeBool = true;       // technically not C :)
DEF_CONST(bSomeBool) int z;  /* error: preprocessor does not know the value
                                of bSomeBool */

Also, allowing for passing macro parameters to DEF_CONST itself (as correctly pointed out by GMan and others):

#define DEF_CONST2(b_const) DEF_CONST_##b_const
#define DEF_CONST(b_const) DEF_CONST2(b_const)
#define DEF_CONST_true const
#define DEF_CONST_false

#define b true
#define c false

/* OK */
DEF_CONST(b) int x;     /* expands to const int x */
DEF_CONST(c) int y;     /* expands to int y */
DEF_CONST(true) int z;  /* expands to const int z */

You may also consider the much simpler (though potentially less flexible):

#if b_const
# define DEF_CONST const
#else /*b_const*/
# define DEF_CONST
#endif /*b_const*/

Setting a bit

Use the bitwise OR operator (|) to set a bit.

number |= 1UL << n;

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

Clearing a bit

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

That will toggle the nth bit of number.

Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

That will put the value of the nth bit of number into the variable bit.

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

It's generally a good idea to use unsigned types for portable bit manipulation.

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.

Best Answer

Related Solutions

C++ – the difference between #include and #include “filename”

C++ – How to set, clear, and toggle a single bit

Setting a bit

Clearing a bit

Toggling a bit

Checking a bit

Changing the nth bit to x

Related Topic