CODEWITHSUNDEEP
c++pointers
Atraente Homem
Atraente Homem

Reputation: 15

Confusion on pointers in C++

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;

}

Upvotes: 0

Views: 97

Answers (3)

tillaert
tillaert

Reputation: 1845

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.

Upvotes: 1

Bilal Akil
Bilal Akil

Reputation: 4755

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

Upvotes: 2

Blindy
Blindy

Reputation: 67352

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.

Upvotes: 0

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