about "int const *p" and "const int *p " - c++

#include <iostream>
using namespace std;
int main(int argc, char* argv[])
{
int i1 = 0;
int i2 = 10;
const int *p = &i1;
int const *p2 = &i1;
const int const *p3 = &i1;
p = &i2;
p2 = &i2;
p3 = &i2;
cout << *p << endl
<< *p2 <<endl
<< *p3 <<endl;
return 0;
}
The code can be compiled with both VC6.0 and VC2010.
But I have questions as blow:
const int *p = &i1;
It means what the "p" points can not be modified,but p can not be modified,am I right?
so
p = &i2;
this line can be complied,yes?
This line:
int const *p2 = &i1;
In my mind,this means p2 can not be modified while what p2 points can be changed, am i right?
Why the
p2 = &i2;
can be compiled?
About this line:
const int const *p3 = &i1;
p3 = &i2;
Oh,god... I am crazy. I have no idea why this line can be compiled with no error...
Can any body help me?
Another code which confused me is here:
class Coo2
{
public:
Coo2() : p(new int(0)) {}
~Coo2() {delete p;}
int const * getP() const
{
*p = 1;
return this->p;
}
private:
int* p;
};
why this code can be compiled?
In
int const * getP() const
I have change the value or *p !

Here we consider 4 types of pointers declarations:
int * w;
It means that w is a pointer to an integer type value. We can modify both the pointer and its content. If we initialize w while declaration as below:
int * w = &a;
Then, both of below operations are viable:
w = &b;(true)
*w = 1;(true)
int * const x;
It means x is a constant pointer that points to an integer type value. If we initialize x while declaration as below:
int * const x = &a;
Then, we cannot do: x = &b;(wrong) because x is a constant pointer and cannot be modified.
However, it is possible to do: *x = 1;(true), because the content of x is not constant.
int const * y; //both mean the same
const int * y;
It means that y is a pointer that points to a constant integer value. If we initialize y while declaration as below:
int const * y = &a;
Then, it is possible to do: y=&b;(true) because y is a non-constant pointer that can point to anywhere.
However, we cannot do: *y=1;(wrong) because the variable that y points to is a constant variable and cannot be modified.
int const * const z; //both mean the same
const int * const z;
It means that z is a constant pointer that points to a constant integer value. If we initialize z while declaration as below:
int const * const z = &a;
Therefore, non of below operations are viable:
z = &b;(wrong)
*z = 1;(wrong)

With the help of pointer, you can actually do two things.
You can change the data it is pointing to but cannot point to a different memory location.
You can point it to a different memory location but cannot change the data it is pointing to.
Now, when you say, int const* ptr or int const* ptr, it falls under first category. It's same as -
const int num = 5; // Both mean the same.
int const num = 5;
To, actually not able to change to a different location, i.e., pointer to a constant location but be able to modify the data, the semantics should be int* const. Since the content of the pointer is a constant, it should be initialized while declaration.
int num = 5;
int* const ptr; // Wrong
ptr = &num; // Wrong
int* const ptr = &num;
*ptr = 100;
However, there is a third kind. Constant pointer to a constant location, which can neither point to a different memory location nor change the data it is pointing to. ( i.e., const int* const )
And now answering the questions, the first two can be compiled because they are not pointing to constant locations. So, they can be modified at later stages too.
const int const *p3 = &i1;
p3 = &i2; // Wrong
In the above snippet, p3 is a constant pointer to a constant location. So, it cannot be modified.
const at the end of a member function says it is not going to change the state of the object. When you say *p = 1;, you are not changing the state of the object. p still points to the same memory location. This is not allowed to do -
int const * Coo2::getP() const
{
*p = 1; // State of `p` is still not modified.
p = new int ; // Error: Changing the memory location to which p points.
// This is what changing the state of object mean and
// is not allowed because of `const` keyword at the end of function
return this->p;
}
Hope, now you understand why the program compiles :)

int const * p; and const int * p are the same. It is when the const comes after the * that
the semantics of the expression change.
I know, it's crazy.

const int *p = &i1;
int const *p2 = &i1;
These both declare non-const pointers to const data.
That is, using p, you cannot change the data it points to. However, you can change the pointer itself, for example, by assigning as p = &i2 which is legal. But *p = 87987 is illegal, as the data p points to is const!
--
int * const p = &i1;
This declares const pointer to non-const data. That is, p=&i2 is illegal, but *p = 98789 is legal.
--
const int * const p = &i1;
This declares const pointer to const data. That is, now both p=&i2 and *p=87897 are illegal.

The two are exactly the same. What matters is the position of the qualifier relative to the asterisk (*):
int const *p; // normal pointer to const int
const int *p; // ditto
int *const p; // const pointer to normal int (rarely useful)
int const * const p; // const pointer to const int

Here is a pointer to a constant:
const int* p;
The following statement is illegal, because it attempts to change the
value of a constant:
*p = 3;
But this one is legal, because the pointer itself is not a constant:
p = &x;
On the other hand, this declaration shows a constant pointer to a variable.
int* const p;
In that case, the following statement is legal, as the variable can be changed:
*p = 3;
but this one is not, because it attempts to change the value of a constant.
p = &x;
Reference link:http://faculty.cs.niu.edu/~freedman/241/const-ptrs.txt

No, the const keyword before the * means that the variable you are pointing to is a "const" variable and only it can not be modified.
If you want a pointer that can't be reassigned then you need to declare it as Foo* const p = &bar;
If you want a pointer that points to a "const" object that can't be reassigned declare it as const Foo* const p = &bar
It is perfectly fine to have a pointer of const int* foo be assigned to a pointer of const int* const bar just like it is fine to have an int's value assigned to a const int. Think of it in the same manner.

int const * is the same as const int *

Succinctly; each combination of read/write int & pointer;
int main() {
int a,b;
int* w; // read/write int, read/write pointer
w= &b; // good
*w= 1; // good
int* const x = &a; // read only pointer, read/write int
// x = &b; // compilation error
*x = 0; // good
int const * y; // read/write ptr, read only int
const int * y2; // " " "
y = &a; // good
// *y = 0; // compilation error
y2 = &a; // good
// *y2 = 0; // compilation error
int const * const z = &a; // read only ptr and read only int
const int * const z2 = &b; // " " " "
// *z = 0; // compilation error
// z = &a; // compilation error
// *z2 = 0; // compilation error
// z2 = &a; // compilation error
}

Related

What is a legal definition of a pointer to a pointer to a const object?

I know from this answer that a pointer const int** z is supposed to be read as
Variable z is [a pointer to [a pointer to a const int object]].
In my humble opinion, this would mean if z=&y then y should be a pointer to a const int object. However, the following code also compiles:
int x=0;
int const* y=&x;
const int** z=&y;
Why is an int const* object i.e. a const pointer to an int instead of a pointer to a const int acceptable to be the pointed object of z?
You are misunderstanding what the const refers to. A const always refers to the element to the left of it - unless it is the leftmost element itself, in which it refers to the element to the right.
This means that
int const * is a pointer to a const int, not a const pointer to int as you think. To get that, you would have to write int * const
int const * and const int * are two ways of writing exactly the same: a pointer to a const int.
You get it right if you read declarations from right to left. If const is the leftmost element, add a "that is" before it when reading. Ex:
const int *: pointer to int that is const.
int const *: pointer to const int.
int * const: const pointer to int.
const int * const: const pointer to int that is const.
int const * const: const pointer to const int.
Note that 1/2 are the same, and 4/5 are the same. 1 and 4 are called "west const" since const is on the west/left side, while 2 and 5 are called "east const".
Why is an int const* object i.e. a const pointer to an int
No.
int const * and const int * are the same type.
People sometimes prefer writing int const * because it reads right-to-left as "pointer to a const int", whereas const int * really reads as "pointer to an int (which is const)".
A const pointer to an int is int * const.
Try it:
int a = 42;
const int * y = &a;
int const * z = &a;
*y = 24; // compile error assigning to const
*z = 24; // compile error assigning to const
int b = 0;
y = &b;
z = &b; // re-pointing non-const pointers is fine
*z = 1; // still a compile error
int * const x = &a;
*x = 24; // fine, assigning via pointer to non-const
x = &b; // error reassigning a const pointer

int **const p does not behave like constant [duplicate]

This question already has answers here:
What is the difference between const int*, const int * const, and int const *?
(23 answers)
Closed 7 years ago.
We know that in int *const p, where p is a constant pointer
it means that address that p holds cannot changed but here in function foo we change the address.
How can it be possible?
int main(){
int i = 10;
int *p = &i;
foo(&p);
printf("%d ", *p);
printf("%d ", *p);
}
void foo(int **const p){
int j = 11;
*p = &j;
printf("%d ", **p);
}
int **const p means p is constant.
So following are not allowed
p++; // Bad
p += 10; // Bad
p = newp; // Bad
But following are fine:
if(p) *p = some_p;
if(p && *p) **p = some_int;
If you want *p should not be re-assigned, use the following
int * const *p;
If you want neither p nor *p should be changeable, use:
int * const * const p;
And following will make all p, *p and **p read-only
const int *const *const p;
// 1 2 3
1: **p is constant
2: *p is constant
3: p is constant
Use 1 or 2 or 3 or any combination as per your requirement.
cdecl page: How to read complex declarations like int ** const p and const int *const *const p
Related: c - what does this 2 const mean?
In your case, const is applied on p itself, not on * p or ** p. In other words, you can change *p and **p, but you cannot change p.
FWIW, try changing the value of p and see.
See it LIVE

can't understand type aliases and const

i'm here because i need of help with type aliases and const.
i'm a real beginner in c++ and i'm studying from the book " c++ primer ".
my problem is that i can't understand why :
int ival = 10;
const int *p = &ival; // this is a pointer to const
int rval = 15;
int *const ppi = &rval; // this is a const pointer
typedef int *integer; // if i create an alias for the type int
int num = 50;
const integer pr = &num; // this is not a pointer to const but a pointer to const
// so my book is telling me that this : const int *pr = &num; is a wrong interpretation
can someone explain me why this ? i've studied pointer ( the basics ) and i can't undestand this
You should put the const to the right side, it would be easier to read. You do it from right to left.
const int *p = &ival;
This is equivalent to
int const *p = &ival.
You read it "pointer to const int". Pointer is not const, data it points to is const.
int * const ppi = &rval;
Here the const is already on the right side and you read "const pointer to int". The pointer is const, the data it points to is not const.
typedef int *integer;
int num = 50;
const integer pr = &num;
In this case you have
integer const pr = &num;
What is basically, if you replace integer with int*:
int* const pr = &num;
And that is read "const pointer to int", just like in the second case.

Assigning const int to a const pointer to int is illegal?

Why is the following illegal?
extern const int size = 1024;
int * const ptr = &size;
Surely a pointer to non-const data should be allowed to point to a const int (just not the other way around)?
This is from C++ Gotchas item #18
If you really meant one of
const int * const ptr = &size;
const int * ptr = &size;
that is legal. Yours is illegal. Because it it wasn't you could do
int * ptr const = &size;
*ptr = 42;
and bah, your const was just changed.
Let's see the other way around:
int i = 1234; // mutable
const int * ptr = &i; // allowed: forming more const-qualified pointer
*i = 42; // will not compile
We can't do harm on this path.
If a pointer to nonconst data was allowed to point to a const int, then you could use pointer to change the value of the const int, which would be bad. For example:
int const x = 0;
int * const p = &x;
*p = 42;
printf("%d", x); // would print 42!
Fortunately, the above isn't allowed.

const int = int const?

For example, is
int const x = 3;
valid code?
If so, does it mean the same as
const int x = 3;
?
They are both valid code and they are both equivalent. For a pointer type though they are both valid code but not equivalent.
Declares 2 ints which are constant:
int const x1 = 3;
const int x2 = 3;
Declares a pointer whose data cannot be changed through the pointer:
const int *p = &someInt;
Declares a pointer who cannot be changed to point to something else:
int * const p = &someInt;
Yes, they are the same. The rule in C++ is essentially that const applies to the type to its left. However, there's an exception that if you put it on the extreme left of the declaration, it applies to the first part of the type.
For example in int const * you have a pointer to a constant integer. In int * const you have a constant pointer to an integer. You can extrapolate this to pointer to pointers, and the English may get confusing but the principle is the same.
For another dicussion on the merits of doing one over the other, see my question on the subject. If you are curious why most folks use the exception, this FAQ entry of Stroustrup's may be helpful.
Yes, that is exactly the same. However, there is difference in pointers. I mean:
int a;
// these two are the same: pointed value mustn't be changed
// i.e. pointer to const value
const int * p1 = &a;
int const * p2 = &a;
// something else -- pointed value may be modified, but pointer cannot point
// anywhere else i.e. const pointer to value
int * const p3 = &a;
// ...and combination of the two above
// i.e. const pointer to const value
const int * const p4 = &a;
From "Effective C++" Item 21
char *p = "data"; //non-const pointer, non-const data
const char *p = "data"; //non-const pointer, const data
char * const p = "data"; //const pointer, non-const data
const char * const p = "data"; //const pointer, const data
const pointer: pointed value may be modified, but pointer cannot point
anywhere else
It is the same in meaning and validity.
As far as I know, const only get complex whenever it involves pointer.
int const * x;
int * const x;
are different.
int const * x;
const int * x;
are same.
I found that if always write const in right, it will becoming easier to read a complex pointer type, for fun.
For example:
// Just append const or pointer to
int x; // data
int const cx; // const data
int *px; // pointer to data
int const *pcx; //pointer to const data
int * const cpx; // const pointer to data
int const * const * * const cppcpcx; // const pointer to (pointer to (const pointer to const data))
I have an idea how to interpret (non)const pointers and data. Split it with *.
Data type
Star sign
Pointer type
Combine
Data
Pointer
int
*
p
int *p
non-const
non-const
const int
*
p
const int *p
const
non-const
int
*
const p
int *const p
non-const
const
const int
*
const p
const int *const p
const
const
// 1.
int x1 = 1;
int x2 = 2;
const int* p = &x1;
//*p = 3; // variable data is CAN NOT BE changed through pointer
p = &x2; // pointer CAN POINT to other variable
cout << *p << endl; // OUTPUT: 2
// 2.
int y1 = 1;
int y2 = 2;
int* const t = &y1;
*t = 3; // variable data is CAN BE changed through pointer
//t = &y2; // pointer CAN NOT POINT to other variable
cout << *t<<endl; // OUTPUT: 3