Whats wrong with this pointer reasoning?

[Salad7]

Just when i try to understand pointers =/

int a = 5, b = 10; //A = 5, B = 10
int *p1, *p2;  //Create pointers that can hold addresses
p1 = &a;  //the p1 ptr will now hold the address of a so *p1 = 5;
p2 = &b;  //the p2 ptr will now hold the address of a so *p2 = 10;
*p1 = 10; //the value of *p1 now equals 10, so a = 10
p1 = p2;  //the address of p1 is now equal to p2 so a and b = 1
*p1 = 20; //the value of p1 is now 20 so a = 20 <--Doesnt this tell a it ='s 20?

//results are a = 10 and b = 20;
//I said a = 20 and b = 20 

[shadowCODE]

so what is your question?

The answers are a = 10 and b = 20. Do you doubt it?

[Salad7]

Where did i go wrong the comments on the side explain my thought process

[Lowest0ne]

Line 7 is where you go wrong. The value of p1 became the value of p2, which is the address of b. You didn't change anything about the values of a or b.

A pointer is like any other variable type, the trick is that its value is a memory address.

[novellof]

Error at line 8.

At line 7 you say:

p1 = p2;

so now the address for p1 is = to the address of p2.

If this is the case...p1 is does not contain the address of where the variable a is. p1 now contains the address of where variable b is.

therefore setting the value *p1 = 20 will change the value of "b" because p2 holds the address ove where the b variable is...not variable a.

[48T7M4Gy]

#include <iostream>

int main()
{
int a = 5, b = 10; //A = 5, B = 10
int *p1, *p2;  //Create pointers that can hold addresses

p1 = &a;  //the p1 ptr will now hold the address of a so *p1 = 5;
std::cout << "1. " << *p1 << std::endl;

p2 = &b;  //the p2 ptr will now hold the address of a so *p2 = 10;
std::cout << "2. " << *p2 << std::endl;

*p1 = 10; //the value of *p1 now equals 10, so a = 10
std::cout << "3. " << *p1 << " " << a << std::endl;

p1 = p2;  //the address of p1 is now equal to p2 so a and b = 10
std::cout << "4. " << a << " " << b << std::endl;

*p1 = 20; //the value of p1 is now 20 so a = 20 <--Doesnt this tell a it ='s 20?
std::cout << "5. " << *p1 << " " << a << std::endl; // *p1 is dereferenced p1 which is "*&a" which is a

//results are a = 10 and b = 20;
std::cout << "6. " << a << " " << b << std::endl;

//I said a = 20 and b = 20 
}

The best way to satisfy yourself is run it ;-)

[novellof]

Yes thats the best way, but he has to understand that p1 does not contain the address of "a" anymore but rather the address "b". because of the line p1 = p2;

[48T7M4Gy]

That's quite right. Both pointers point to the same address which happens to be the location of a. Also dereferecing the pointers doesn't change the value of the varaiable being pointed to which is what the printout shows. He should printout *p1 andf *p2 along with p1 and p2 and &a and &b to demonstrate the points you quite rightly raise.

[Salad7]

thanks