Few Important things about macros, before I explain the output
1. Macros are replaced in passes.
2. Macros are not recursive.
regarding the output remember the rule for expansion
"A parameter in the replacement list, *UNLESS* preceded by a # or ##
preprocessing token or followed by a ## preprocessing token, is replaced by
the corresponding argument after all macros contained therein have been
expanded".
In other words, macros are replaced "inside out" unless # or ## exists
printf("%s",g(f(1,2))); is replaced as #f(1,2) ---> "f(1,2)" according to
the replacement rule.
printf("\t%s",h(f(1,2))); As this does not replace with # or ## directly,
"inside out" expansion leads to h("1,2") --> g('1,2") --> "1,2"
for the first pass
printf("%s", "f(1,2)"); ---> g(a) #a
printf("\t%s", h("1,2"));
second pass
printf("%s", "f(1,2)"); ---->> not processed(exhausted)
printf("\t%s", g("1,2")); --> h(a) g(a)
Third pass
printf("%s", "f(1,2)"); ---->> not processed(exhausted)
printf("\t%s", "1,2"); --> g(a) #a
Hope this answers your question.
-
Azhar.
On Tue, Apr 5, 2011 at 3:22 PM, Vandana Bachani <vandana....@gmail.com>wrote:
> Hi Arvind,
> These are preprocessor specific operators. Check out
> http://msdn.microsoft.com/en-us/library/wy090hkc(v=vs.80).aspx
>
> -Vandana
>
> On Tue, Apr 5, 2011 at 12:45 PM, Arvind <akk5...@gmail.com> wrote:
>
>> #include<stdio.h>
>>
>> #define f(a,b) a##b
>> #define g(a) #a
>> #define h(a) g(a)
>>
>> int main()
>> {
>> printf("%s",g(f(1,2)));
>> printf("\t%s",h(f(1,2)));
>> return 0;
>> }
>>
>>
>>
>> i have run this program in gcc compiler and getting : f(1,2) 12 as
>> output.
>> can anyone explain the reason for getting this output?
>>
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