Macro arguments are scanned and expanded before the substitution in macro
body unless they are convert to a string representation (using #) or
concatenated with other tokens (using ##).

Xstr(oper)  => Xstr(multiply) =>str(multiply)=>"multiply"    // in Xstr(x)
, x is a simple argument so it will expand fully

str(oper)    =>"oper"    // in str(x) , x is converting to string so it
will not get expanded

check this link :
http://gcc.gnu.org/onlinedocs/cpp/Argument-Prescan.html#Argument-Prescan




On Sun, Oct 28, 2012 at 3:28 PM, rahul sharma <[email protected]>wrote:

> And when char *opername=str(oper);
>
> then o/p is oper....why behaviour is diff. in 2 cases
>
> On Sun, Oct 28, 2012 at 2:53 PM, rahul sharma <[email protected]>wrote:
>
>> #include<stdio.h>
>> #include str(x) #x
>> #define Xstr(x) str(x)
>> #define oper multiply
>>
>>
>> int main()
>> {
>> char *opername=Xstr(oper);
>> printf("%s",opername);
>> }
>> so firstly     Xstr is expanded to str(oper)
>> then str(oper) is expanded to #oper now i have read that
>>
>> If, however, a parameter name is preceded by a # in the replacement text,
>> the
>> combination will be expanded into a quoted string with the parameter
>> replaced by the actual argument...and in this case actual is also oper so
>> it becomes "oper"...so it should print oper .....i thnik i am mistaken
>> anyhre...correct???
>>
>
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-- 
Vikram Pradhan | B.Tech| Computer Science & Engineering | NIT Jalandhar  |
9740186063 |

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