the problem of "#define hash_hash # ## #"

the problem of "#define hash_hash # ## #"

[f z]

Hi , 

Anybody could help to explain the following code which is the example in C99 std!

/***********************************************************/
4 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.
/**********************************************************/

Where define this syntax "#define hash_hash # ## #" in std?

In C99 std, we all know that "'##' cannot appear at either end of a macro expansion". 

In this definiation of macro , the "hash_hash" are defined with "# ## #",but I don't know what is the mean of the first and last sharp(#).



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Re: the problem of "#define hash_hash # ## #"

[Matthew Woehlke]

f z wrote:
> Hi , 
> 
> Anybody could help to explain the following code which is the example in C99 std!
> 
> /***********************************************************/
> 4 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.
> /**********************************************************/
> 
> Where define this syntax "#define hash_hash # ## #" in std?
> In this definiation of macro , the "hash_hash" are defined with "# ##
> #",but I don't know what is the mean of the first and last sharp(#).

Re-read your preprocessor specification. This is the symbol '#' 
concatenated ('##' preprocessor operator) with the symbol '#'. Which 
results in the string '##' being output.

This is apparently being used to get '##' into a literal string via a macro.

-- 
Matthew
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