@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
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
I have taken form book...i am writing exact code
#includestdio.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
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 xbut will it reflect address of a and x ???...NO
so if you print the address p and q ...before and after the
The purpose of typedef is to assign alternative names to existing type.
In your case statement typedef int (*pFunc) (int); just assign a new
name pFunc to pointer to function declaration int (*pFunc) (int);. This
statement is not define any variable. Please see below working code.
#include
