使用C中的*＆*＆p了解复杂的指针和地址运算符

Question

Can any one explain how the printf is printing hello in the following?

#include<stdio.h>

void main()
{
 char *p;
 p="hello";
 printf("%s",*&*&p);
}

I know that *&p...means value in p, ie the address of string "hello". What is happening in the initial *&

Answer 1

As you said, *&p means p , that means the consecutive * and & cancels out. Hence *&*&p becomes p too.

And as @Kerrek said (in the comment) that *&p produces an lvalue, so you take its address again .

---

Note that your code is not standard conformant. main() must have int as return type. And you cannot assign "hello" to a non-const char* . It must be const char* . A standard conformant code would be this:

#include<stdio.h>

int main()
{
   const char *p = "hello"; 
   printf("%s",*&*&p);
}

Answer 2

&p is the address of p .

*p is the thing pointed at by the address p .

*&p is *(&p) the thing pointed at by the address &p - which is p itself (ie, the thing pointed at by the address "address of p").

Thus it turns out that *&p is just p - the *& cancel each other out. You can repeat this: *&*&p will still be p . You can do this ad infinitum: *&*&*&*&*&*&*&*&*&p will also be p .

Answer 3

'*&' cancels each other. You are getting the address of p then dereferencing it again. So the end result will just be p.