## operator

[DL1]

Hello,

I'm slightly puzzled by the following example from Section 6.10.3.3 of the C99 Standard (this one here http://www.open-std.org/JTC1/SC22/WG...cs/n1256.pdf):

"EXAMPLE In the following fragment:

#define hash_hash # ## #
#define mkstr(a) # a
#define in_between(a) mkstr(a)
#define join(c, d) in_between(c hash_hash d)
char p[] = join(x, y); // equivalent to
// char p[] = "x ## y";

The expansion produces, at various stages:

join(x, y)
in_between(x hash_hash y)
in_between(x ## y)
mkstr(x ## y)
"x ## y"

In other words, expanding hash_hash produces a new token, consisting of two adjacent sharp signs, but
this new token is not the ## operator."

I don't really understand how or why the macro "hash_hash" works. I am guessing that the ## in the middle of "# ## #" is concatenating the two outer #s, but why that should work I can't really see. I don't seem to able to find anything in the standard that explains what is going on.

[matsp]

because ## is a symbol used by the preprocessor, you can't use ## in a macro. So hash_hash is a way to produce ## without the preprocessor "noticing", (by concatenating # and #, using the ## symbol).

It's cleverness that you probably never will need.

--
Mats

[DL1]

I thought it probably was (cleverness you'll never need). I was looking for something else yesterday and came across it, and it has been doing my head in ever since.

Anyway I still don't fully understand why it should work - from my understanding, the ## operator normally concatenates two arguments, yet surely the preceding and following #s aren't arguments?

[matsp]

I thought it probably was (cleverness you'll never need). I was looking for something else yesterday and came across it, and it has been doing my head in ever since.

Anyway I still don't fully understand why it should work - from my understanding, the ## operator normally concatenates two arguments, yet surely the preceding and following #s aren't arguments?

You can use it for anything there you want to "paste two bits of stuff together". We could do [I think the first will work, I know the second bit does]:

#define PI 3 ## .141592
#define Symbol(x)  Prefix ## x ## PostFix

--
Mats

[cpjust]

because ## is a symbol used by the preprocessor, you can't use ## in a macro. So hash_hash is a way to produce ## without the preprocessor "noticing", (by concatenating # and #, using the ## symbol).

It's cleverness that you probably never will need.

--
Mats

Actually, it's cleverness that I've needed only once so far in my career.
I would have preferred to find a nice template solution, rather than resorting to macros, but once I worked out all the bugs in the macros, it saved a lot of time and code.
If I remember correctly, the API I was using had a lot of Policy and Result classes with very similar names and identical usage; so I had a macro where I passed the unique part of the class name, and it then dynamically built the names of the Policy and Result classes using the ## pre-processor and did some stuff with them... It's kind of hard to explain in a paragraph, but it did help a LOT with maintainability (maybe not as much with readability), since I only needed to make a code change in one place instead of 10-15 places.

[laserlight]

This is C programming though, so no templates and classes, but we still have the preprocessor

[DL1]

[code]
#define PI 3 ## .141592
#define Symbol(x) Prefix ## x ## PostFix
[code]

Of course. The penny is slowing beginning to drop...

[cpjust]

This is C programming though, so no templates and classes, but we still have the preprocessor

I know this is the C board, but ## is also in C++ and the only place I used it is in C++. The basic idea is pretty much the same, just replace classes with structs...

[Thantos]

Actually I think the cleverness that matsp was referring to was using macro concat to concat # and # to get around the fake that it treats ## as the concat token.

[matsp]

Here is a more complete example showing the use of ## (some of it creative!)

#include <stdio.h>

#define FLOAT(x,y) x ## . ## y
#define Symbol(x)  Prefix_ ## x ## _PostFix
#define __SymbolName(x) #x
#define _SymbolName(x) __SymbolName(x)
#define SymbolName(x) _SymbolName(Symbol(x))

#define PI FLOAT(3, 141592)

int main()
{
  char *Symbol(P) = SymbolName(P);
  printf("PI=%f, Symbol= %s\n", PI, Symbol(P));
  return 0;
}

It compiles and produces what I expected from the executable.

Of course, it's all pretty meaningless.

--
Mats

[matsp]

Actually I think the cleverness that matsp was referring to was using macro concat to concat # and # to get around the fake that it treats ## as the concat token.

Yes. That is what I was referring to. Not being able to use ## in a macro is pretty rare, but there can conceivably be times when it's needed - I can't think of any such time, but I'm sure someone has had the need for it at some point in the past 20 or so years.

--
Mats

[cpjust]

Actually I think the cleverness that matsp was referring to was using macro concat to concat # and # to get around the fake that it treats ## as the concat token.

Oh, I see now.
Yeah, that's one thing I've never had to use.

[slingerland3g]

The use of ##, from what I have read is a token paist operator, and can be used with variadic macros for c_99.

From some obscure online document that I have saved when working with such preprocessor commands:

" Second, the ## token paste operator has a special meaning when placed between a comma and a variable argument. If you write "

#define eprintf(format, …) fprintf (stderr, format, ##__VA_ARGS__)

a google search may turn up more if you are interested in the use of variadic macros. I use them for debugging with the use printf().