Why couldn't I swap the values of two variables with *temp?

[hefese]

My teacher gave us a homework about exchanging the values of two variables with three different methods. These are callByValue, callByPointer and callByReference. I did all of them as you see the following code. However, I can't understand a point in the program. When I tried to use pointer temp in the function replace_CallByPointer, the compiler gave a warning. The warning was that:

erlerini takaslama\main.cpp||In function 'void replace_CallByPointer(int*, int*)':|
erlerini takaslama\main.cpp|55|warning: 'temp' is used uninitialized in this function|
||=== Build finished: 0 errors, 1 warnings ===|

Thereupon, I put a value 0 to the pointer temp. So the compiler compiled without error and warning. But the program didn't run properly again.

I noticed that when I used variable temp instead of pointer temp, I can run the program properly. But why isn't the program working properly when I'm using pointer temp? I can't understand this point. Actually a variable temp is sufficient to accomplish the process, but yet it should run properly with pointer temp. Can anybody explain this case to me?

#include <iostream>


using namespace std;


void replace_CallByValue(int,int);
void replace_CallByPointer(int*,int*);
void replace_CallByReference(int&,int&);


int main()
{
    int x,y;


    cout << "Enter a value for x : ";
    cin >> x;


    cout << "Enter a value for y : ";
    cin >> y;


    cout << endl << endl << endl;


    cout << "Values of x and y before the functions is no called." << endl;
    cout << "x : " << x << endl << "y : " << y << endl << endl;


    replace_CallByValue(x,y);


    cout << "Values of x and y after the function 'callByValue'." << endl;
    cout << "x : " << x << endl << "y : " << y << endl << endl;


    replace_CallByPointer(&x,&y);


    cout << "Values of x and y after the function 'callByPointer'." << endl;
    cout << "x : " << x << endl << "y : " << y << endl << endl;


    replace_CallByReference(x,y);


    cout << "Values of x and y after the function 'callByReference'." << endl;
    cout << "x : " << x << " y : " << y << endl;
}


void replace_CallByValue(int a,int b)
{
    int temp;


    temp = a;
    a = b;
    b = temp;
}

void replace_CallByPointer(int *c,int *d)
{
    int *temp;


    *temp = *c;
    *c = *d;
    *d = *temp;
}

//void replace_CallByPointer(int *c,int *d)
//{
//    int temp;
//
//
//    temp = *c;
//    *c = *d;
//    *d = temp;
//}


void replace_CallByReference(int &e,int &f)
{
    int temp;


    temp = e;
    e = f;
    f = temp;
}

[laserlight]

Because, as the message says, you did not initialise temp. Actually temp should not be a pointer.

[std10093]

As laserlight said,temp should not be a pointer

I would suggest to initialize variable temp when declaring it It is equivalent to say int x=5;

[hefese]

Because, as the message says, you did not initialise temp.

Why do I have to initialise temp? If I don't remember as wrong, I use pointer variable in any program without initialising a lot of times before. Is there a situation about difference of C and C++ or else?

Actually temp should not be a pointer.

Yes, but at the same time does the program need not be able to run properly with pointer?

As laserlight said,temp should not be a pointer

I would suggest to initialize variable temp when declaring it It is equivalent to say int x=5;

So you mean ;

void replace_CallByPointer(int *c,int *d)
{
    int temp=*c;
 
    *c = *d;
    *d = temp;
}

Note: I editted my message for I hurry.

[std10093]

So you mean ;

yes

Also remember that if i have *variable as an argument and i want to take the reference to it i will apply the operator &,so i would have &*variable,and operators & * kill each other,so finally i would have variable

[hefese]

When I try to ask you, I learn why I shouldn't use pointer variable.

As laserlight says, if I initialise pointer temp;

void replace_CallByPointer(int *c,int *d)
{
    int *temp=c; // temp = &x;


    *c = *d;     // x = y;
    *d = *temp;  // y = x;
}

[laserlight]

Why do I have to initialise temp?

Because you use temp, in particular, by dereferencing it.

If I don't remember as wrong, I use pointer variable in any program without initialising a lot of times before.

Then you have made mistakes a lot of times before (unless you mean that you did not initialise it at the point of declaration, but later you assigned it a value before use).

Is there a situation about difference of C and C++ or else?

No difference here.

Yes, but at the same time does the program need not be able to run properly with pointer?

In this case, you need another integer variable to perform the swap. So, even if you create a pointer, that won't help as the pointer has to point to something.

[hefese]

yes

Also remember that if i have *variable as an argument and i want to take the reference to it i will apply the operator &,so i would have &*variable,and operators & * kill each other,so finally i would have variable

Yes, I saw the reason why I should use variable temp, when I was going to your line you say. (std10093:I would suggest to initialize variable temp when declaring it)

Thank you std10093 for your hint, and thank you laserlight for your interest,answers.

[Elysia]

A pointer can be used in two ways.
You can assign it a memory address (ie change where it points to). This is fine. You don't need to initialize it to do this.
You can also reference it, which tells the compiler to get or change what it points to, and if you haven't initialized the pointer, then the compiler will try to get or set the value it points to, which is... well, I think you get the point.

[std10093]

I thought that we were not supposed to use pointers in C++ as in C though...

[Elysia]

They still have their uses--especially where references cannot be used!

[std10093]

They still have their uses--especially where references cannot be used!

I know you are right because i have run into this situation a year ago ,but could you please give me a simple example,so that i can clearly remind the purpose?

-don't know if should open new thread

[Elysia]

Linked lists, dynamic memory, binary trees, situations where you need to be able to point to a different variable/address, etc.

[hefese]

Then you have made mistakes a lot of times before (unless you mean that you did not initialise it at the point of declaration, but later you assigned it a value before use).

Yes, that's what I mean.

A pointer can be used in two ways.
You can assign it a memory address (ie change where it points to). This is fine. You don't need to initialize it to do this.
You can also reference it, which tells the compiler to get or change what it points to, and if you haven't initialized the pointer, then the compiler will try to get or set the value it points to, which is... well, I think you get the point.

OK, thank you so much.

void replace_CallByPointer(int *c,int *d)
{
    int *temp;


    *temp = *c;
    *c = *d;
    *d = *temp;
}

I understand that I cannot dereference the pointer temp. Because, temp has no include any address value. So because it has no address value, it isn't point any variable. Then because it isn't point any variable, any value cannot be assigned to temp points to variable which is unavailable.

Thank you so much everybody again.

[hefese]

When I try to ask you, I learn why I shouldn't use pointer variable.

As laserlight says, if I initialise pointer temp;

void replace_CallByPointer(int *c,int *d)
{
    int *temp=c; // temp = &x;


    *c = *d;     // x = y;
    *d = *temp;  // y = x;
}

Meanwhile, I didn't write the diagram exactly correct. It should have been that;