About the End of Multi-level Pointer

Just as the code below:

int a = 1;
int *p0 = &a;
int **p1 = &p0;
......

Then, where is the end of the pointer.
a is not a pointer, but a "normal" variable.

p0 is a pointer that is pointing to a, i.e. it "contains" the memory address of a.

p1 is a pointer that is pointing to p0, i.e. it "contains" the memory address of p0. It's a pointer to a pointer.


Hence all of these will print the same, i.e. the value of a:
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printf("%d\n", a);
printf("%d\n", *p0);
printf("%d\n", **p1);


Also, all of these will print the memory address of a:
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printf("%p\n", &a);
printf("%p\n", p0);
printf("%p\n", *p1);

Last edited on
pointers don't have ends. you are responsible for knowing how big it is.
pointers can be null, if you chain them (eg, in a linked list), and you can use THAT to stop but it takes effort to set that up.

here,
p0 simply points to a. It is the address, or 'array index into the giant array we call computer ram', of the variable a.
p1 is the address of p0. Nothing more, nothing special. The fact that p0 is also a pointer is irrelevant **

the 'end', so to speak, -- these both have 1 item only.
so p1[0] is fine, p1[1] is off the end (that being 2, and you only have 1)
po[0] is fine, p0[1] or p0[100] are bad as above.

p0++ will make p0 invalid.

** it makes the syntax a little odd looking until you get used to seeing * as part of a type. But syntax aside, its not special.
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> p0++ will make p0 invalid.

After p0++, becomes a pointer past the end of an object.
ie p0 points to a fictitious object (it can't be dereferenced), but it is still valid (it is a well defined pointer).

For purposes of pointer arithmetic and comparison, a pointer past the end of the last element of an array x of n elements is considered to be equivalent to a pointer to a hypothetical array element n of x and an object of type T that is not an array element is considered to belong to an array with one element of type T.
https://eel.is/c++draft/basic.compound#3.4


This is fine:
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#include <iostream>

int main()
{
    int a = 5 ;
    int* p0 = &a ;
    p0++ ;
    std::cout << ( &a - p0 ) << '\n' // -1
              << std::boolalpha << ( p0 > &a ) << '\n' // true
              << ( (p0-1) == &a ) << '\n' ; // true
}

http://coliru.stacked-crooked.com/a/7bd8705d101e797e
Last edited on
Thanks all the answers.
I got it.
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