Comparison Of Pointers - c++

I want to compare the memory address and pointer value of p, p + 1, q , and q + 1.
I want to understand, what the following values actually mean. I can't quite wrap my head around whats going on.
When I run the code:
I get an answer of 00EFF680 for everytime I compare the adresss p with another pointer.
I get an answer of 00EFF670 for everytime I compare the address of q with another pointer.
I get an answer of 15726208 when I look at the pointer value of p.
And I get an answer of 15726212 When I look at the pointer value of p + 1.
I get an answer of 15726192 when I look at the pointer value of q
And I get an answer of 15726200 Wehn I look at the pointer value of q + 1.
Code
#include <iostream>
#include <string>
using namespace std;
int main()
{
int val = 20;
double valD = 20;
int *p = &val;
double *q;
q = &valD;
cout << "Memory Address" << endl;
cout << p == p + 1;
cout << endl;
cout << q == q + 1;
cout << endl;
cout << p == q;
cout << endl;
cout << q == p;
cout << endl;
cout << p == q + 1;
cout << endl;
cout << q == p + 1;
cout << endl;
cout << "Now Compare Pointer Value" << endl;
cout << (unsigned long)(p) << endl;
cout << (unsigned long) (p + 1) << endl;
cout << (unsigned long)(q) << endl;
cout << (unsigned long) (q + 1) << endl;
cout <<"--------" << endl;
return 0;
}

There are a few warnings and/or errors.
The first is that overloaded operator << has higher precedence than the comparison operator (on clang++ -Woverloaded-shift-op-parentheses is the flag).
The second is that there is a comparison of distinct pointer types ('int *' and 'double *').
For the former, parentheses must be placed around the comparison to allow for the comparison to take precedence. For the latter, the pointers should be cast to a type that allows for safe comparison (e.g., size_t).
For instance on line 20, the following would work nicely.
cout << ((size_t) p == (size_t) (q + 1));
As for lines 25-28, this is standard pointer arithmetic. See the explanation here.

As to your question:
I want to compare p, p +1 , q , and q + 1. And Understand what the results mean.
If p is at address 0x80000000 then p+1 is at address 0x80000000 + sizeof(*p). If *p is int then this is 0x80000000 + 0x8 = 0x80000008. And the same reasoning applies for q.
So if you do p == p + 1 then compiler will first do the additon: p+1 then comparison, so you will have 0x80000000 == 0x80000008 which results in false.
Now to your code:
cout << p == p + 1;
is actually equivalent to:
(cout << p) == p + 1;
and that is because << has higher precedence than ==. Actually you should get a compilation error for this.
Another thing is comparision of pointers of non related types like double* with int*, without cast it should not compile.

In C and C++ pointer arithmetic is very closely tied with array manipulation. The goal is that
int array[3] = { 1, 10, 100 };
int *ptr = { 1, 10, 100 };
std::cout << array[2] << '\n';
std::cout << *(ptr + 2) << '\n';
outputs two 100s. This allows the language to treat arrays and pointers as equivalent - that's not the same thing as "the same" or "equal", see the C FAQ for clarification.
This means that the language allows:
int array[3] = { 1, 10, 100 };
int *ptr = { 1, 10, 100 };
And then
std::cout << (void*)array << ", " << (void*)&array[0] << '\n';
outputs the address of the first element twice, the first array behaves like a pointer.
std::cout << (void*)(array + 1) << ", " << (void*)&array[1] << '\n';
prints the address of the second element of array, again array behaving like a pointer in the first case.
std::cout << ptr[2] << ", " << *(ptr + 2) << '\n';
prints element #3 of ptr (100) twice, here ptr is behaving like an array in the first use,
std::cout << (void*)ptr << ", " << (void*)&ptr[0] << '\n';
prints the value of ptr twice, again ptr behaving like an array in the second use,
But this can catch people unaware.
const char* h = "hello"; // h points to the character 'h'.
std::cout << (void*)h << ", " << (void*)(h+1);
This prints the value of h and then a value one higher. But this is purely because the type of h is a pointer to a one-byte-sized data type.
h + 1;
is
h + (sizeof(*h)*1);
If we write:
const char* hp = "hello";
short int* sip = { 1 };
int* ip = { 1 };
std::cout << (void*)hp << ", " << (void*)(hp + 1) << "\n";
std::cout << (void*)sip << ", " << (void*)(sip + 1) << "\n";
std::cout << (void*)ip << ", " << (void*)(ip + 1) << "\n";
The first line of output will show two values 1 byte (sizeof char) apart, the second two values will be 2 bytes (sizeof short int) apart and the last will be four bytes (sizeof int) apart.
The << operator invokes
template<typename T>
std::ostream& operator << (std::ostream& stream, const T& instance);
The operator itself has very high precedence, higher than == so what you are actually writing is:
(std::cout << p) == p + 1
what you need to write is
std::cout << (p == p + 1)
this is going to print 0 (the result of int(false)) if the values are different and 1 (the result of int(true)) if the values are the same.

Perhaps a picture will help (For a 64bit machine)
p is a 64bit pointer to a 32bit (4byte) int. The green pointer p takes up 8 bytes. The data pointed to by p, the yellow int val takes up 4 bytes. Adding 1 to p goes to the address just after the 4th byte of val.
Similar for pointer q, which points to a 64bit (8byte) double. Adding 1 to q goes to the address just after the 8th byte of valD.

If you want to print the value of a pointer, you can cast it to void *, for example:
cout << static_cast<void*>(p) << endl;
A void* is a pointer of indefinite type. C code uses it often to point to arbitrary data whose type isn’t known at compile time; C++ normally uses a class hierarchy for that. Here, though, it means: treat this pointer as nothing but a memory location.
Adding an integer to a pointer gets you another pointer, so you want to use the same technique there:
cout << static_cast<void*>(p+1) << endl;
However, the difference between two pointers is a signed whole number (the precise type, if you ever need it, is defined as ptrdiff_t in <cstddef>, but fortunately you don’t need to worry about that with cout), so you just want to use that directly:
cout << (p+1) - p << endl;
cout << reinterpret_cast<char*>(p+1) - reinterpret_cast<char*>(p) << endl;
cout << (q - p) << endl;
That second line casts to char* because the size of a char is always 1. That’s a big hint what’s going on.
As for what’s going on under the hood: compare the numbers you get to sizeof(*p) and sizeof(*q), which are the sizes of the objects p and q point to.

The pointer values that are printed are likely to change on every execution (see why the addresses of local variables can be different every time and Address Space Layout Randomization)
I get an answer of 00EFF680 for everytime I compare the adresss p with another pointer.
int val = 20;
double valD = 20;
int *p = &val;
cout << p == p + 1;
It is translated into (cout << p) == p + 1; due to the higher precedence of operator << on operator ==.
It print the hexadecimal value of &val, first address on the stack frame of the main function.
Note that in the stack, address are decreasing (see why does the stack address grow towards decreasing memory addresses).
I get an answer of 00EFF670 for everytime I compare the address of q with another pointer.
double *q = &valD;
cout << q == q + 1;
It is translated into (cout << q) == q + 1; due to the precedence of operator << on operator ==.
It prints the hexadecimal value of &valD, second address on the stack frame of the main function.
Note that &valD <= &val - sizeof(decltype(valD) = double) == &val - 8 since val is just after valD on the stack. It is a compiler choice that respects some alignment constraints.
I get an answer of 15726208 when I look at the pointer value of p.
cout << (unsigned long)(p) << endl;
It just prints the decimal value of &val
And I get an answer of 15726212 When I look at the pointer value of p + 1.
int *p = &val;
cout << (unsigned long) (p + 1) << endl;
It prints the decimal value of &val + sizeof(*decltype(p)) = &val + sizeof(int) = &val + 4 since on your machine int = 32 bits
Note that if p is a pointer to type t, p+1 is p + sizeof(t) to avoid memory overlapping in array indexing.
Note that if p is a pointer to void, p+1 should be undefined (see void pointer arithmetic)
I get an answer of 15726192 when I look at the pointer value of q
cout << (unsigned long)(q) << endl;
It prints the decimal value of &valD
And I get an answer of 15726200 Wehn I look at the pointer value of q + 1.
cout << (unsigned long) (q + 1) << endl;
It prints the decimal value of &val + sizeof(*decltype(p)) = &valD + sizeof(double) = &valD + 8

Related

Pointer Syntax Within Condition

After a pointer is initialized, do you have to use the * dereference operator to call the pointer in a condition?
Example:
int main()
{
int var = 10;
int *ptr = &var;
if(ptr) // does this need to be if(*ptr) ???
{.......}
}
And can I have a short explanation as to why?
Thank you.
if (ptr)
check if the pointer is not Null but
if (*ptr)
check if the value it points to is not zero (in this example is 10)
So for checking the value you shoud add *.
It depends on what you want to do.
if(ptr) checks if the pointer value is nullptr or not. Note that this is shorthand for if(ptr != nullptr).
if(*ptr) checks if what the pointer points to is nullptr (or 0) - and in that case, since you dereference (follow) the pointer to answer the question, the pointer itself had better not be nullptr in that case.
First of all, a pointer is only a variable. However, there are different contexts in which you can use it.
As any other variable you can access the pointers content (which is the adress of the underlying memory) as follows:
int i = 1;
int * p = &i;
std::cout << p << std::endl
this would output the adress of i since this is what is stored in p
If you however want to access the content of the underlying memory (the value of i), you need to dereference the pointer first by using the * operator:
std::cout << *p << std::endl;
This would print the value of iso 1.
of course you can also access the pointer's adress (since the adress of i is a numeric value as well and needs to be stored somewhere too):
std::cout << &p << std::endl;
That would output the adress of p so the adress where the adress of i is stored :)
As a little example try to run this code:
#include <iostream>
int main() {
int i = 1;
int * p = &i;
std::cout << "Value of i(i): " << i << std::endl
<< "Adress of i(&i): " << &i << std::endl
<< "Value of p(p): " << p << std::endl
<< "Dereferenced p(*p): " << *p << std::endl
<< "Adress of p(&p): " << &p << std::endl
<< "Dereferenced adress of p(*(&p)): " << *(&p) << std::endl;
}

What is the relationship between an array and its address?

The following code:
#include<iostream>
int main (void) {
int lista[5] = {0,1,2,3,4};
std::cout << lista << std::endl;
std::cout << &lista << std::endl;
std::cout << lista+1 << std::endl;
std::cout << &lista+1 << std::endl;
std::cout << lista+2 << std::endl;
std::cout << &lista+2 << std::endl;
std::cout << lista+3 << std::endl;
std::cout << &lista+3 << std::endl;
return (0);
}
Outputs:
0x22ff20
0x22ff20
0x22ff24
0x22ff34
0x22ff28
0x22ff48
0x22ff2c
0x22ff5c
I understood that an array is another form to express a pointer, but we cannot change its address to point anywhere else after declaration. I also understood that an array has its value as the first position in memory. Therefore, 0x22ff20 in this example is the location of the array's starting position and the first variable is stored there.
What I did not understand is: why the other variables are not stored in sequence with the array address? I mean, why lista+1 is different from &lista+1. Should not they be the same?
In pointer arithmetic, types matter.
It's true that the value is the same for both lista and &lista, their types are different: lista (in the expression used in cout call) has type int* whereas &lista has type int (*)[5].
So when you add 1 to lista, it points to the "next" int. But &lista + 1 points to the location after 5 int's (which may not be a valid).
Answering the question as asked:
std::cout << &lista+1 << std::endl;
In this code you take the address of array lista and add 1 to obtained answer. Given the sizeof of the array is sizeof(int) * 5, which means when you increment a pointer to it by 1 you add sizeof(int) * 5 to the pointer address, you end up with a number you see.

Pointer and char in struct

How to read the second letter in char with a pointer? I can read the whole message "carp" and first letter 'c' but I have no idea how to read second letter... Here is my example code:
#include <iostream>
#include <string>
using namespace std;
int main()
{
struct list {
char name[20];
int length;
};
list first ={
"carp",
6,
};
list *p = &first;
cout << p->name << endl; // "carp"
cout << *p->name << endl; // "c"
p = p + 1;
cout << *p->name << endl; // Not working...How to read a?
return 0;
}
Incrementing p with p++ or p = p+1 moves you to the next instance of struct list, which is not what you want (and it's not even there).
Instead, you want to move to the second letter of name, which can be done in several ways:
Use index on the name - cout << p->name[1] << endl;
Make a pointer to p->name and increment it, i.e. char *np = p->name; np++; cout << *np
Use pointer arithmetic instead of indexing, i.e. cout << *(p->name+1) << endl;
You can use index on name to access any character:
p->name[1] // gives 'a'
p->name[2] // gives 'r'
Note that arrays begin with an index of 0. So p->name[0] would give 'c'.
p + 1 actually increments p, which is a pointer to list. This essentially moves to the next instance of list, which is not even initialized in your code.
Use the array subscript operator with an index of 1:
p->name[1];
If you want to output the second character using pointers without the subscript operator then you can just write
cout << *( p->name + 1 ) << endl;
It is the same as
cout << p->name[1] << endl;
Or you can introduce an intermediate pointer. For example
for ( const char *q = p->name; *q != '\0'; ++q )
{
cout << *q;
}
cout << endl;

Why is this method for navigating through an array wrong?

I'm trying to understand C pointers, I set up the following example:
int main()
{
int temp[] = {45,67,99};
int* address;
address = &temp[0]; //equivalent to address = temp;
std::cout << "element 0: " << *address << std::endl;
address = (address + sizeof(int));
std::cout << "element 1: " << *address << std::endl;
std::getchar();
return 0;
}
Now, the first element is printed correctly while the second one is garbage. I know that I can just use:
address = (address + 1);
in order to move the pointer to the second element of the array, but I don't understand why the first method is wrong.
The operator sizeof returns the size of the object in bytes. So (depending on your compiler) you may have just done
address + sizeof(int) == address + 4
This will obviously access out of bounds when you dereference the pointer. The problem is that the way you are trying to use sizeof is already accounted for in the pointer arithmetic. When you add an integral (e.g. 1) to an int* it already knows to move 1 int over to the next address.
Incrementing a pointer by N will not increment it by N bytes, but by N * sizeof(pointed_type) bytes.
So when you do:
address = (address + sizeof(int));
You are incrementing the address by (probably) 4 times the size of int, which is past your original array.
What you intended to do is:
address = reinterpret_cast<int*>(reinterpret_cast<char*>(address) + sizeof(int));
Which is pretty horrendous.
An array is a group of elements of the Same Type mapped consecutively in memory one-next to the other.
The size of the array is:
sizeof(any element) * the number of elements
So the vice-vers to get the number of elements of an array you can:
Number of elements = sizeof(array) / sizeof(any element).
These elements are addressable and each one is addressed with the first byte's address.
int a[] = {1, 10, 100, 1000, 10000};
std::cout << sizeof(int) << std::endl; // 4
std::cout << sizeof(a) << std::endl; // 20: 5 * 4
The addresses:
cout << &a[0] << " : " << &a[1] << " : " <<
&a[2] << " : " << &a[3] << endl;
The outPut:
0018FF38 : 0018FF3C : 0018FF40 : 0018FF44
As you can see the address is incrementing by sizeof(int) // 4
To do it on your way:
for(int* tmp = array; tmp <= a + nElement; tmp++)
cout << tmp << " : ";
The result:
0018FF38 : 0018FF3C : 0018FF40 : 0018FF44
As you can see the result is identical.
To move from one element to another just increment / decrement the address as I did above:
tmp++; // incrementing the address by 1 means moving to the next element, element means an integer which means 4 bytes. 1 here is not a single byte but a unit or element which is 4 byte here;
To move to the n-element just:
tmp += n;
In your example:
address = (address + sizeof(int)); :
address = address + 4; // moving to the 0 + 4 element which means fifth element (here outbound).

Putting own address in variable

int main()
{
int x;
x = int(&x);
int* q;
q = &x;
int* p;
p = (int*)&x;
int* w;
w = (int*)x;
cout << *p << endl << x << endl << *q << endl << *w << endl << p << endl << q << endl << w << endl << &x;
return 0;
}
OUTPUT:
5762632
5762632
5762632
5762632
0057EE48
0057EE48
0057EE48
0057EE48
Is this step x = int(&x) valid ? Is there anything conceptually wrong about this?
How to visualize/interpret what happens when we write p = (int*)&x and w = (int*)x?
Also note that this program executes only when step 1 is written. If x does not contain its own address, then the program will show an error.
Is this step x = int(&x) valid ?
Is there anything conceptually wrong about this?
The statement may be valid and could provoke undefined behavior. All depends on the platform.
On many of the platforms I've worked with, the addresses were never signed. They were always unsigned because the sign bit was needed to represent the full range of addresses. In other words, I suggest:
unsigned int x = (unsigned int) &x;
Again it's about the sign bit and how addresses are represented.
How to visualize/interpret what happens when we write p = (int*)&x and w = (int*)x?
If you want to print an address, I suggest you use the pointer format:
cout << ((void *) &x) << endl;
or
printf("%p\n", &x);
I recommend using a debugger rather than printing out addresses. The debugger allows you to examine the values in a pointer. But there are times when printing out information is a better method (like when working with multiple tasks or multiple processors).