CODEWITHSUNDEEP
c++variablespointersmemory-address
Jeffrey Lebowski
Jeffrey Lebowski

Reputation: 291

c++: pointer value different than address of the pointed variable

I just made a tiny stupid program about passing a variable the value contained in another using a pointer, just as an introduction to pointers themselves. I printed, before and after the assignation, the value and position of all three variables involved. However, I get, as the value contained inside the pointer, an address different from the variable it is pointing to's one, and I just can't understand why.

This is my main program:

#include <iostream>
#include "01.Point.h"

using namespace std;

int main() 

{

    int a,b;

    cout << "Insert variable's value: ";

    cin >> a;

    int * point;

    cout << "Before assignment:" << endl;

    printeverything (a,b,point);

    point = &a;

    b = * point;

    cout << "After assignment:" << endl;

    printeverything (a,b,point);

    cout << endl;

}

And this is my function's implementation:

#include <iostream>
#include "01.Point.h"

using namespace std;

void printeverything (int a, int b, int * c)    {

    cout << "First variable's value: " << a << "; its address: " << &a << endl;

    cout << "Second variable's value: " << b << "; its address: " << &b << endl;

    cout << "Pointer's value: " << c << "; its address: " << &c << endl;

}

b successfully gets a's value, so everything works right, but this is the complete output:

Before assignment:
First variable's value: 5; its address: 0x7fffee49b77c
Second variable's value: 0; its address: 0x7fffee49b778
Pointer's value: 0x7fffee49b8a0; its address: 0x7fffee49b770
After assignment:
First variable's value: 5; its address: 0x7fffee49b77c
Second variable's value: 5; its address: 0x7fffee49b778
Pointer's value: 0x7fffee49b7ac; its address: 0x7fffee49b770

I mean, if variable x is at position 3285, and i do p = &x, pointer p should contain value 3285, right? So why is this happening?

Upvotes: 1

Views: 748

Answers (1)

psmears
psmears

Reputation: 28000

In your printeverything function, the parameter/local variable a is an entirely different variable from the one you passed to it. It happens to have the same name, but that's (as far as the compiler is concerned) entirely a coincidence - it's a different variable with a different address; you can see this by assigning a value to a within the function, and then printing it afterwards - you'll see that the "outer" a will remain unchanged.

Upvotes: 6

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