Pointing to a reference in C++

Question

So I am trying to understand pointers and references.

Lets say we have the following:

int val = 100;
int *ip = &val;

Why is that not the same as simply pointing to val like this:

int val = 100;
int *ip = val;

Answer 1

Easy way to understand why they are not the-same if you do this your self.

int val = 100;
cout<<val<<endl;

You shoud get 100 as output.

Now try:

int val = 100;
cout<<&val<<endl;

You should get a 0x7ffc201cab0c or something similar as the outout. If you access val with the "&", you are accessing the memory address of it.

If you access val without the "&", you are accessing the value of it.

Therefore if you do

int val = 100;
int *ip = val;

you will should get a warning or an error but assuming that you want to see it compile so you can figure out what would happen, this is what it you should do:

int val = 100;
int *ip;
*ip = val;

This is the-same as the previous code except that it will compile. You are creating a pointer and assigning a value to the pointer instead of assigning a memory address to it. The code is valid but has a run-time error. The pointer ip is NOT pointing anywhere before you assigned value to it.

To make it work properly, you can do:

int val = 100;
int *ip = new int;
*ip = val;

Now, you have a pointer with a memory address. The value of that pointer is val which is 100. That's the value NOT the memory address.

To conclude this, pointer must be initialized or it will have undefined behavior or even crash. You use the ref sign (&) or the new keyword to initialize pointers then you can assign assign values to it.

int val = 100;
int *ip = new int; //Create a new pointer and initilize it with a new memory address
*ip = val; //Assign the value from val(100) to the pointer

OR

int val = 100;
int *ip = &val; //Create a new pointer and initialize it by making it point to existing address of **val**


*ip = 200; //change the value in the **memory addresss to **200**

Now val changes to 200 too.

int otherValue = 500;
*ip = otherValue ; //change the value in the **memory address** to 500

Now val changes to 500 too.

CHANGE MEMORY ADDRESS WHERE ip IS POINTING

You can also change the memory address anytime. Lets make p point to otherValue.

*ip = &otherValue; //Point p to otherValue memory address not value
*ip = 5; //Change the value of p to 5(no memory address change) 

Now otherValue has 5 as the value. p points to otherValue memory address.

Answer 2

Neither one of your two examples contains a pointer to reference:

• The first example has a pointer to int, properly initialized using the address-of operator &
• The second example has a pointer to int, improperly initialized to the value of val itself

Generally, you do not assign values to pointers directly: normally, you want an address-of operator, an implicit pointer conversion, or some pointer arithmetic. Although assigning a well-known address on specific hardware is also possible, compiler toolchains often provide ways to keep this outside of your program. Hence, if you see an assignment int *ip = 100 it's almost certainly incorrect.

If you want a reference, declare it like this:

int val = 100;
int &ref = val;

Now you can make a pointer to that reference, like this:

int *ipRef = &ref;

Since ref represents an alias for val, ipRef would have the same value as ip from your example (demo).

Answer 3

Because pointers and integers are different data types.

int *ip = &val;

This initialises one int* with another int* (the value obtained by taking the address of val).

int *ip = val;

This is an attempt to initialise an int* with an int, which is not allowed.

Note that your code examples do not contain any references.

Answer 4

val is an integer, so you can not assign an int to a pointer that points to an int because is not the valid value, you will need the address (obtained with the & operator)