But this works, where I have both the variable name and function name used:
#include<stdio.h>int charu(int x);int charu(int x)
{
return x;
}
main(){*
int a = charu(45);
int charu = a;*
int scanf=78;
//int printf=45;
int getchar=6;
printf("%d",charu);
//error if this line is uncommented charu=charu(6);
printf("\n%d",getchar);
puts("Wow it ran");
}
http://ideone.com/xN2yi
Local variable names trump function names. You can call the function as long
as compilers doesn't come across a local variable of same name.
Once it does that, call to the function will give compiler error.
(This is not an authoritative answer. This is from what I observe them
behaving as)
--
On Tue, Jul 12, 2011 at 1:14 PM, saurabh singh <[email protected]> wrote:
> For 1st problem learn the concepts of linking.Scanf hasn't been required to
> link yet in the main.
> So during compling the program doesn't give any special treatment to scanf.
> check this,
> #include<stdio.h>
> int charu(int x);
> int charu(int x)
> {
> return x;
> }
> main()
> {
> int charu=67;
> int scanf=78;
> //int printf=45;
> int getchar=6;
> printf("%d",charu);
> //error if this line is uncommented charu=charu(6);
> printf("\n%d",getchar);
> puts("Wow it ran");
> }
>
> On Tue, Jul 12, 2011 at 1:03 PM, Abhi <[email protected]> wrote:
>
>> PROBLEM 2: First all the expressions are evauated from right to left in
>> printf() and then results printed with tha final value of i.
>>
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>
>
>
> --
> Saurabh Singh
> B.Tech (Computer Science)
> MNNIT ALLAHABAD
>
>
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