Actual number when cout pointer

Question

#include<bits/stdc++.h>

using namespace std;

int main(){
int x;
int *y;

x=2;
y= &x;
cout << y << endl;
cout << y[0] << endl; #this cout shows the value of x ( 2 at this case)
cout << y[1] << endl;
cout << y[2] << endl;
cout << y[3] << endl;
return 0;
}

When I Try to cout only the pointer it comes in ) 0x(hex value), and I was trying to get only the hex value of it, why is that happening?

C++ have no bounds checking. Going out of bounds (which you do when indexing `y` with anything other than `0`) leads to [*undefined behavior*](https://en.wikipedia.org/wiki/Undefined_behavior). — Some programmer dude, Oct 16 '18 at 13:48
`y[0]` = `*(y + 0)`. Not sure what else were you expecting, though. — Algirdas Preidžius, Oct 16 '18 at 13:48
Welcome to Stack Overflow! Sounds like you could use a [good C++ book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) — NathanOliver, Oct 16 '18 at 13:48
And unrelated to your problem, please read [Why should I not #include ?](https://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h) — Some programmer dude, Oct 16 '18 at 13:48
What output do you expect for `cout << y[1]`, `cout << y[2]` and `cout << y[3]`? — Jabberwocky, Oct 16 '18 at 14:23

score 2 · Accepted Answer · answered Oct 16 '18 at 14:08

it comes in ) 0x(hex value), and I was trying to get only the hex value of it, why is that happening?

Because that is how the IO streams have been specified to behave.

If you want to get rid of the base prefix, you can achieve that by converting the pointer to an integer and using the std::noshowbase manipulator. Integers are shown in decimal base by default though, so if you want the hex value, then use the std::hex manipulator.

std::cout << std::noshowbase << std::hex << (std::uintptr_t)y << '\n';

Accessing y[1] and other indices greater than 0 cause your program to have undefined behaviour.

Actual number when cout pointer

1 Answers1