if you think the your expanded version is incorrect.You are wrong ,
because int * will hold pointer but you are not allocating address of
x ..instead you are allocating x value as an address of x to *t.This
wont work.
so to make it work you need to save the address of x and y in temp pointers i.e
int *p.*q;
p=&x;
q=&y;
int t;
t=*p;
*p=*q;
*q=t;
now you can convert it into macro.
On 10/29/12, rahul sharma <rahul23111...@gmail.com> wrote:
> @atul...mistakenly i have put w at place of t in my last post...i wana say
>
>
>
> On Mon, Oct 29, 2012 at 10:07 AM, dCoder <bansal....@gmail.com> wrote:
>
>> Just replace your macro with its definition and you will understand.
>>
>> its not doing swapping of pointers...plz explain
>>
>
>
> @dCode i expanded..but its fine...please tell
>
>> #include<stdio.h>
>> #define swap(a,b,c) c t;t=a,a=b,b=t
>>
>> int main
>> int x=10,y=20;
>> int *p,*q;
>> swap(x,y,int*);
>>
> int * t;
> t=x;
> x=y;
> y=t;
>
>
> There is int* at the end..there is som problem with my
> keyborad...:(.........acc to me axpanded version is above..but not swapping
> two pointerss....plz comment
>
>
>
>
>> On Sun, Oct 28, 2012 at 9:16 PM, rahul sharma
>> <rahul23111...@gmail.com>wrote:
>>
>>> its now doing swapping of pointers...plz explain
>>>
>>>
>>> On Sun, Oct 28, 2012 at 8:08 PM, atul anand
>>> <atul.87fri...@gmail.com>wrote:
>>>
>>>> it should swap....
>>>>
>>>> On 10/28/12, rahul sharma <rahul23111...@gmail.com> wrote:
>>>> > Why the following code is not able to swap two macros???although it
>>>> > is
>>>> > easily swapping 2 variables
>>>> >
>>>> > #include<stdio.h>
>>>> > #define swap(a,b,c) c t;t=a,a=b,b=t
>>>> >
>>>> >
>>>> > int main
>>>> >
>>>> >
>>>> > int x=10,y=20;
>>>> > int *p,*q;
>>>> >
>>>> > swap(x,y,int);
>>>> >
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