Can the C preprocessor perform integer arithmetic?

Question

As the questions says, is the C preprocessor able to do it?

E.g.:

#define PI 3.1416 #define OP PI/100 #define OP2 PI%100 

Is there any way OP and/or OP2 get calculated in the preprocessing phase?

Answer 1

Integer arithmetic? Run the following program to find out:

#include "stdio.h" int main() {     #if 1 + 1 == 2         printf("1+1==2\n");     #endif     #if 1 + 1 == 3         printf("1+1==3\n");     #endif  } 

Answer is "yes", there is a way to make the preprocessor perform integer arithmetic, which is to use it in a preprocessor condition.

Note however that your examples are not integer arithmetic. I just checked, and gcc's preprocessor fails if you try to make it do float comparisons. I haven't checked whether the standard ever allows floating point arithmetic in the preprocessor.

Regular macro expansion does not evaluate integer expressions, it leaves it to the compiler, as can be seen by preprocessing (-E in gcc) the following:

#define ONEPLUSONE (1 + 1) #if ONEPLUSONE == 2     int i = ONEPLUSONE; #endif 

Result is int i = (1 + 1); (plus probably some stuff to indicate source file names and line numbers and such).

Answer 2

The code you wrote doesn't actually make the preprocessor do any calculation. A #define does simple text replacement, so with this defined:

#define PI 3.1416 #define OP PI/100 

This code:

if (OP == x) { ... } 

becomes

if (3.1416/100 == x) { ... } 

and then it gets compiled. The compiler in turn may choose to take such an expression and calculate it at compile time and produce a code equivalent to this:

if (0.031416 == x) { ... } 

But this is the compiler, not the preprocessor.

To answer your question, yes, the preprocessor CAN do some arithmetic. This can be seen when you write something like this:

#if (3.141/100 == 20)    printf("yo"); #elif (3+3 == 6)    printf("hey"); #endif