Confusion on pointers in C++

Question

I was practising with pointers and came across this thing.
The following code gives the answer:

 1
 0
 3

when I change *a with *c the answer is same but on changing with *b=0, the answer is

1
2
0

Can you please what is going on behind the scenes in each of these processes?

#include<iostream>
using namespace std;
void func(int *a, int *b,int *c)
{
    a=b;
    b=c;
    c=a;
    *a=0;

}


int main()
{
    int a=1,b=2,c=3;
    func(&a,&b,&c);
    cout<<a<<endl<<b<<endl<<c;

}

Answer 1

a = 1
b = 2
c = 3

For easier reading, I'll rename the variables in func to pa , pb and pc . So what you're mixing up in func makes:

• pa point to what pb is pointing to ( b )
• pb point to what pc is pointing to ( c )
• pc point to what pa is pointing to (which is now b )

Therefore

*pa = 0

follows pa 's pointer (to b ) and sets that to 0, giving you:

a = 1
b = 0
c = 3

However

*pb = 0

instead follows pb 's pointer (to c ) and sets that to 0, giving you:

a = 1
b = 2
c = 0

Answer 2

You are first swapping pointers; the only relevant line is:

a=b;

So the variable a in func now points to the address of b in your main.

Then

*a=0;

sets the value of the variable a in func is pointing to (which is b in your main now) to 0 .

So, a , b , c are pointers to things, while a , b , c are "things". It become easier when you rename the variables in func to something else, for example: pa , pb and pc as "pointer-to-a". Then you can see you are swapping pointers and that *pa=0 is "setting the value pa is pointing to to 0 .

Keeping this in mind, it should become clear that modifying your code will swap pointers differently, and thus leads to different results.

Answer 3

when I change *a with *c the answer is same

Why would you expect any different? You wrote c=a; just above it.

changing with *b=0, the answer is 1 2 0

Indeed, you wrote b=c; before it, so writing 0 in *b will zero out the third parameter's contents.