C macros: advantage/intent of apparently useless macro

Question

I have some experience in programming in C but I would not dare to call myself proficient. Recently, I encountered the following macro:

#define CONST(x) (x)

I find it typically used in expressions like for instance:

double x, y;
x = CONST(2.0)*y;

Completely baffled by the point of this macro, I extensively researched the advantages/disadvantages and properties of macros but still I can not figure out what the use of this particular macro would be. Am I missing something?

This macro achieves nothing, other than perhaps as a visual reminder that something is a constant... — Oliver Charlesworth, Jan 31 '13 at 15:53
The only reason I can think of to do this would be to allow an external tool to more easily find specific constants in the program. It's used for marking text to be translated sometimes although I'm not sure how it would help here — jcoder, Jan 31 '13 at 15:56

WildCrustacean · Accepted Answer · 2013-01-31T16:20:26.650

11

As presented in the question, you are right that the macro does nothing.

This looks like some artificial structure imposed by whoever wrote that code, maybe to make it abundantly clear where the constants are, and be able to search for them? I could see the advantage in having searchable constants, but this is not the best way to achieve that goal.

It's also possible that this was part of some other macro scheme that either never got implemented or was only partially removed.

edited Jan 31 '13 at 16:20

answered Jan 31 '13 at 15:54

WildCrustacean

5,896
2
31
42

1

+1, that's the only sane thing I can think of for this as well. – Theodoros Chatzigiannakis Jan 31 '13 at 15:55
5

That is not sane, a sane programmer wouldn't use "magic numbers" but instead `#define SOME_FLOAT 2.0`, followed by `x = SOME_FLOAT*y;`. Or even better, use a `const` variable, if applicable. – Lundin Jan 31 '13 at 15:57
2

A const variable is not 'even better'. It's Considered Harmful. – R.. GitHub STOP HELPING ICE Jan 31 '13 at 16:17
@R.. what? would you care to explain? I don't want to be laboring under the false impression that const ints are preferred over defines (in C++. is it different in C?), so any light you can shed on this would be wonderful, thanks! – im so confused Jan 31 '13 at 16:54
It's **very** different in C. `const` variables are not constant expressions. In C you should always use macros or enums for symbolic constants. – R.. GitHub STOP HELPING ICE Jan 31 '13 at 17:04

npclaudiu · Answer 2 · 2013-01-31T16:34:27.233

Some (old) C compilers do not support the const keyword and this macro is most probably a reminiscence of a more elaborate sequence of macros that handled different compilers. Used like in x = CONST(2.0)*y; though makes no sense.

You can check this section from the Autoconf documentation for more details.

EDIT: Another purpose of this macro might be custom preprocessing (for extracting and/or replacing certain constants for example), like Qt Framework's Meta Object Compiler does.

score 2 · Answer 3 · answered Jan 31 '13 at 15:54

2

There is absolutely no benefit of that macro and whoever wrote it must be confused. The code is completely equivalent to x = 2.0*y;.

answered Jan 31 '13 at 15:54

Lundin

195,001
40
254
396

score 1 · Answer 4 · edited May 23 '17 at 11:49

1

Well this kind of macro could actually be usefull when there is a need to workaround the macro expansion.
A typical example of such need is the stringification macro. Refer to the following question for an example : C Preprocessor, Stringify the result of a macro

Now in your specific case, I don't see the benefit appart from extreme documention or code parsing purposes.

edited May 23 '17 at 11:49

Community

1
1

answered Jan 31 '13 at 16:05

greydet

5,509
3
31
51

Good point about the stringification macro which I also encountered in my research in an even earlier SO post http://stackoverflow.com/questions/1597007/creating-c-macro-with-and-line-token-concatenation-with-positioning-macr However in the code I found , it is not used in that manner – Pankrates Jan 31 '13 at 16:12

score 0 · Answer 5 · answered Jan 31 '13 at 16:06

0

Another use could be to reserve those values as future function invocations, something like this:

 /* #define CONST(x) (x) */
 #define CONST(x) some_function(x)

 // ...

 double x, y;
 x = CONST(2.0)*y; // x = some_function(2.0)*y;

answered Jan 31 '13 at 16:06

higuaro

15,730
4
36
43

score 0 · Answer 6 · answered Jan 31 '13 at 20:14

Another good thing about this macro would be something like this

result=CONST(number+number)*2;

or something related to comparisons

result=CONST(number>0)*2;

If there is some problem with this macro, it is probably the name. This "CONST" thing isn't related with constants but with some other thing. It would be nice to look for the rest of the code to know why the author called it CONST.

score 0 · Answer 7 · answered Feb 07 '13 at 19:08

This macro does have the effect of wrapping parenthesis around x during the macro expansion.

I'm guessing someone is trying to allow for something along the lines of

CONST(3+2)*y

which, without the parens, would become

3+2*y

but with the parens becomes

(3+2)*y

I seem to recall that we had the need for something like this in a previous development lifetime.

C macros: advantage/intent of apparently useless macro

7 Answers7