Can I have a pointer wihtout creating a new variable?

Development Programming & Scripting C++
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#1

So here’s what I’m thinking. Let’s say you have the following
So instead of, say,

int v = 5;
int* vptr = &v;
printf(*vptr)

I could do this:

int v = 5:
printf(*&v)

I believe the second code is a pointer to a reference, or vice versa, but is there a way to actually not have to initialize a new variable in order to dereference the memory address? This would help a lot with readability for me personally, plus saving a bit of memory. Thank you.

#2

What are you actually trying to accomplish?

Pointers are just values. They’re kind-of like integers, that happen to contain values that identify specific pieces of memory. (In fact, at the assembly level, they are exactly that.)

The *&v construct takes the address of the local variable v and then immediately dereferences it, creating a reference to that local variable. This is largely a long way of doing nothing (compared to just saying v) unless v has a more exciting type where you have overridden operator&().

You can create a pointer without a value:

int *vptr = nullptr;
if (vptr) {
    printf(*vptr);
}

You can create a value without a pointer:

int v = 5;
printf(v);

You can create a pointer, and initialize it to a value that will live longer than the current functions scope:

int *vptr = new int(5);
if (vptr) {
    printf(*vptr);
}
return vptr; // keeps living until you delete it

What, specifically, is it you’re trying to accomplish, though?

2 Likes
#3

Basically, i don’t want to have to declare a pointer variable when I want to point to the original variable. This is mainly for readability.

So in other words, I do a pointer to the memory address of v. If &v is rhe memory address of v, then I would like to do a pointer TO &v.

#4

Pointer-to-pointer is a thing.
However, anything you point to must have “storage.”
Rvalues don’t have “storage” so you can’t take the address of them.
&v returns a rvalue of type pointer, so you can’t directly take the address of that; you have to give it “storage,” which you do by storing it into the heap, or into a variable.

int v1 = 5;
int *pv = &v1;
int **ppv = &pv;
printf(**ppv); // prints 5;
int v2 = 6;
pv = &v2;
printf(**ppv); // prints 6;
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#5

Ok then, thanks.