CODEWITHSUNDEEP
c++linuxpointers
Bankin
Bankin

Reputation: 797

C++ pointers value changing

Hello I have simple code like this

void func(int &x, int y, int* z){
    int a = 3;
    int b = 1;

    cout << *z << endl;
    x = *z + b;
    cout << *z << endl;
    y = *z + x;
    z = &a;

    cout << "func: x = " << x << endl;
    cout << "func: y = " << y << endl;
    cout << "func: *z = " << *z << endl;
    cout << "func: a = " << a << endl;
    cout << "func: b = " << b << endl;

    return;
}

int main(){
    int a = 2;
    int b = 4;

    func(b, a, &b);
    cout << "main: a = " << a << endl;
    cout << "main: b = " << b << endl;

    return 0;
}

And for output I get 

4
5
func: x = 5
func: y = 10
func: *z = 3
func: a = 3
func: b = 1
main: a = 2
main: b = 5

So my question is, for what reason the value of *z changes after x = *z + b because I can't figure it out by myself.

Thank you in advance

Upvotes: 0

Views: 2877

Answers (4)

Martin J.
Martin J.

Reputation: 5118

x = *z + b;

changes the value of z because:
- here, x is an int reference pointing to your caller-side variable b
- z is an int pointer containing the address of the same variable b
Therefore, at this point, x and z refer to the same thing.
Therefore, assigning a new value to x means assigning that value to b. Since z points to b, reading z afterwards gives you the value you assigned to x.

You can make sure of this, for instance by writing and comparing the address of x, and the value of z:

cout << "(1) z is: " << z << " - &x is: " &x
// should be the same
z = &a;
cout << "(2) z is: " << z << " - &x is: " &x
// should be the different

In your context, the two values will be identical until you reassign z.

Upvotes: 1

mekkanizer
mekkanizer

Reputation: 741

When you use & before an argument you give to a method, it becomes not copied but linked to it.

So that:

  1. You change x which is linked to b

  2. z changes corresponding to b's new value, because it's linked to it.

Upvotes: 1

marcadian
marcadian

Reputation: 2618

Because you changed it to point to local variable a

z = &a;

Hence when you dereference it, it gives you value of local variable a, which is 3

Upvotes: 0

Stack Exchange Broke The Law
Stack Exchange Broke The Law

Reputation: 58909

x is a reference to the b in main (not to be confused with the b in func, which is separate). z is a pointer to the b in main. References are basically pointers, but without the * or & syntax, and you can't make them point to something else (there's no equivalent to z = &a for references).

The line:

x = *z + b;

is basically doing:

b_in_main = b_in_main + b_in_func;

because both x and *z refer to the same variable.

Upvotes: 2

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