Evaluate C macro LINE inside another macro

Question

I'd like to create a macro whose value is the evaluation of __LINE__ at a certain point so that I can do the following:

#define LINE1 __LINE__
int a[] = {
1,
2,
3
};
#define LINE2 __LINE__
printf("There are %d rows", LINE2 - LINE1 - 2);

...but LINE2 and LINE1 end up giving me the line number of the printf statement. Any ideas?

You can't do it that way. I'm not sure you can do it at all. The `#define LINE1` simply says 'when you see `LINE1`, replace it with `__LINE__`. But `__LINE__` evaluates to the line number where `LINE1` is used, not where it is defined. Besides, the correct way to determine the number of elements in an array is `(sizeof(a) / sizeof(a[0]))` — which works correctly even if someone places two entries on a single line, or adds a comment in the middle of the list, etc. The only time that it fails is when the 'array' is actually a pointer parameter to a function. — Jonathan Leffler, Sep 09 '17 at 19:32
https://stackoverflow.com/a/33712984/841108 could inspire you — Basile Starynkevitch, Sep 09 '17 at 19:35
@HolyBlackCat: That creates a variable. C does not have symbolic constants. — too honest for this site, Sep 09 '17 at 19:39
@JonathanLeffler +1 for noticing, what OP is actually trying to do — Erki Aring, Sep 09 '17 at 20:06

army007 · Accepted Answer · 2017-09-09T19:59:04.720

Macro Expansion is a lazy and recursive process. It lazy because when you define a macro with #define it is not evaluated where you have defined it; rather, it will be replaced with its value where you used it. And this replacement will take place recursively until it is replaced with something that is not a macro. So,

printf("There are %d rows", LINE2 - LINE1 - 2);

will be expanded to

printf("There are %d rows", __LINE__ - __LINE__ - 2);

which in turn will be expanded to

printf("There are %d rows", N - N - 2); // N being the current line number.

You can change your code as suggested by HolyBlackCat. Instead of declaring macro using #define use const int.

const int LINE1 = __LINE__;
   .
   .
   .
const int LINE2 = __LINE__;

wildplasser · Answer 2 · 2017-09-10T00:59:46.857

Silly hack:

#include <stdio.h>

int main(void)
{
struct yyy{ char buff[__LINE__ ]; } ;
#define LINE1 ((int)sizeof(struct yyy))
int a[] = {
1,
2,
3,
4
};
struct zzz{ char buff[__LINE__ ]; } ;
#define LINE2 ((int)sizeof(struct zzz))
printf("[%d,%d]There are %d rows\n", LINE1 , LINE2 , LINE2 - LINE1 - 4);

return 0;
}

And it works at file scope,too:

#include <stdio.h>

struct yyy{ char buff[__LINE__ ]; } ;
#define LINE1 ((int)sizeof(struct yyy))
int a[] = {
1,
2,
4
};
struct zzz{ char buff[__LINE__ ]; } ;
#define LINE2 ((int)sizeof(struct zzz))

int main(void)
{
printf("[%d,%d]There are %d rows\n", LINE1 , LINE2 , LINE2 - LINE1 - 4);

return 0;
}

NOTE:this does not create variables, only two (unused) structure types.

Yet another ugly trick: use enums (which are not objects, but constants_in_disguise):

#include <stdio.h>

enum { OMG=__LINE__ , } ;
int a[] = {
#define INDEX(ll) ( (ll)- ((OMG)+3))
INDEX(__LINE__),
INDEX(__LINE__),
INDEX(__LINE__),
INDEX(__LINE__),
INDEX(__LINE__),
INDEX(__LINE__)
};
enum { WTF=__LINE__ , } ;

int main(void)
{
printf("[%d,%d]There are %d rows\n", OMG , WTF , WTF - OMG - 4);

return 0;
}

Evaluate C macro __LINE__ inside another macro

2 Answers2

Evaluate C macro LINE inside another macro