Defining constants in C

Question

I was working with C++ for a long time and now I am on a C project.
I am in the process of converting a C++ program to C.

I am having difficulty with the constants used in the program.
In the C++ code we have constants defined like

static const int X = 5 + 3;
static const int Y = (X + 10) * 5
static const int Z = ((Y + 8) + 0xfff) & ~0xfff

In C, these definitions throw error. When I use #defines instead of the constants like

#define X (5+3);
#define Y (((X) + 10) * 5)
#define Z ((((Y) + 8) + 0xfff) & ~0xfff)

the C compiler complains about the definitions of "Y" and "Z".

Could anyone please help me to find a solution for this.

People have already given the solution - but I just want to point out the reason why you need to remove the semicolon. #define in c causes the preprocessor to do a textual substitution before compilation, so imagine taking the contents of X and plugging it into Y and you will likely see what the problem is. — Gavin H
– Gavin H, Commented Apr 19, 2011 at 19:44
Some people also use enums to declare constants in C, e.g. enum { N = 100 };. — Pascal Cuoq
– Pascal Cuoq, Commented Apr 19, 2011 at 19:53
static is redundant in the C++ example. const objects that are not explicitly declared extern have internal linkage anyway. — Lara Bailey
– Lara Bailey, Commented Apr 19, 2011 at 20:06

Kyle Sletten · Accepted Answer · 2011-04-19 19:41:42Z

5

You need to remove the semi-colon from the #define X line

#define X (5+3)
#define Y (((X) + 10) * 5)
#define Z ((((Y) + 8) + 0xfff) & ~0xfff)

answered Apr 19, 2011 at 19:41

Kyle Sletten

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amit · Accepted Answer · 2011-04-19 19:41:20Z

3

#define X (5+3); is wrong, it needs to be #define X (5+3) (without ';')
also be aware of the difference between using static const and #define: in static const, the value is actually evaluated, in #define, it's pre-processor command, so

#define n very_heavy_calc()
...
n*n;

will result in evaluating very_heavy_calc() twice

answered Apr 19, 2011 at 19:41

amit

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Neil · Accepted Answer · 2011-04-19 19:59:04Z

1

Another option is to use an enum:

enum {
  X = 5 + 3,
  Y = (X + 10) * 5,
  Z = ((Y + 8) + 0xfff) & ~0xfff
};

answered Apr 19, 2011 at 19:59

Neil

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