Yeah its working...actually it was written in book...and i did nt
compile....mistake of author....

On Tue, Oct 30, 2012 at 12:17 PM, Vikram Pradhan <[email protected]>wrote:

> @rahul : dude it's working fine ...check your printf() statement ....you
> are swapping the pointers and printing the original float values of a and x
> it should be printf("%f%f",*p,*q);
>
> or if you want to swap float values not the pointers then it should be
> swap(a,x,float);
> printf("%f%f",a,x);
>
>
> On Mon, Oct 29, 2012 at 11:30 PM, atul anand <[email protected]>wrote:
>
>> well they should not , if you see it closely .... pointer p and q
>> contain contain address of a and x.
>> and swap() macro will swap value these pointers are holding i.e adress
>> of a and x....but will it reflect address of a and x ???...NO
>> so if you print the address p and q ...before and after the swap then
>> you should see that after swap ....p will be holding address of x and
>> q will be holding address of a..that it
>>
>> On 10/29/12, rahul sharma <[email protected]> wrote:
>> > I have taken form book...i am writing exact code
>> >
>> > #include<stdio.h>
>> > #define swap(a,b,c) c t;t=a,a=b,b=t;
>> >
>> >
>> > int main()
>> > {
>> > float a,x;
>> > a=20.0;
>> > x=30.0;
>> > float *p,*q;
>> > p=&a,q=&x;
>> > swap(p,q,float*);
>> > printf("%f %f",a,x);
>> > getchar();
>> > }
>> >
>> > o/p=20.000 30.000
>> >
>> >
>> > why not swapped???
>> > On Mon, Oct 29, 2012 at 11:01 PM, atul anand
>> > <[email protected]>wrote:
>> >
>> >> 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 <[email protected]> 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 <[email protected]>
>> 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
>> >> >> <[email protected]>wrote:
>> >> >>
>> >> >>> its now doing swapping of pointers...plz explain
>> >> >>>
>> >> >>>
>> >> >>> On Sun, Oct 28, 2012 at 8:08 PM, atul anand
>> >> >>> <[email protected]>wrote:
>> >> >>>
>> >> >>>> it should swap....
>> >> >>>>
>> >> >>>> On 10/28/12, rahul sharma <[email protected]> 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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>
> --
> Vikram Pradhan | B.Tech| Computer Science & Engineering | NIT Jalandhar  |
>  9740186063 |
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