ohk
i try to xplain ,
first u understand abt two macros operator
# operator makes string
like
#define g(a) *#*a
WTEVER value of a will come ,this operator wil convert that into string
noe 2nd operator
is concatenation one
like
#define f(a,b) a*##*b
here a and b wil get concatenated to form string
but ths question test also the order of evalaution of fucntion
like
h(f(1,2))
first h macro wil get called and then it wil become
g(f(1,2))
but now it wil call
f(1,2)
macro
so tht makes
g(12)
n now
g macro
makes it
12
fr 2nd part
g(f(1,2))
it calls
g macro n converts it into string
"f(1,2)"
so here f macor wil nt get called
as it has convereted into string :)
cheers
On Sat, Feb 5, 2011 at 12:50 AM, priya mehta <priya.mehta...@gmail.com>wrote:
> thats amazing :) :)
> but i wanted to know the explanation dear.
>
>
> On Sat, Feb 5, 2011 at 12:49 AM, rajat ahuja
> <catch.rajatah...@gmail.com>wrote:
>
>> 12
>> and
>> f(1,2)
>>
>> On Sat, Feb 5, 2011 at 12:46 AM, priya mehta <priya.mehta...@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\n",h(f(1,2)));
>>>
>>>
>>>
>>> printf("%s\n",g(f(1,2)));
>>>
>>>
>>>
>>> *return* 0;
>>> }
>>>
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