[RESOLVED] Why address of an object is 6 bytes and not 8 bytes on 64 bit Linux?

Hello,

My code:

Code:

---

    int i12 = 1001;
    cout << i12 << " " << &i12 << endl;

---

gives the result: 0x7fff0d065098

It's 6 bytes, but I'd expect the address to be 8 bytes on my 64 bit machine. I am using Ubuntu 12.04, GNU compiler.

So, why the address is 6 bytes and not 8 bytes?

Thank you.

Quote:

---

Originally Posted by vincegata

Hello,

My code:

Code:

---

    int i12 = 1001;
    cout << i12 << " " << &i12 << endl;

---

gives the result: 0x7fff0d065098

It's 6 bytes, but I'd expect the address to be 8 bytes on my 64 bit machine. I am using Ubuntu 12.04, GNU compiler.

So, why the address is 6 bytes and not 8 bytes?

Thank you.

---

The code you posted does not give you the size of the pointer. To get the size of any type, then use sizeof().

Code:

---

#include <iostream>

int main()
{
    std::cout << " The sizeof(int*) is " << sizeof(int *) << std::endl;
}

---

So what gets outputted?

Regards,

Paul McKenzie

Right, it shows 8.

I am not looking for the size of the pointer or reference, but, let's say, I want to know the address where my object is stored.

Also, 0x7fff0d065098 is 140,733,411,905,688 decimal, which is 140 Terabyte, 733 MB, and so on. I have only 8GB of RAM on my laptop, how is that possible?

Quote:

---

Originally Posted by vincegata

I am not looking for the size of the pointer or reference,

---

Is this a different request? Your original post asked why the pointer is 6 bytes, but as the code shows, the pointer is clearly 8 bytes in size.

Quote:

---

but, let's say, I want to know the address where my object is stored.

---

You use the address-of operator (&).

Regards,

Paul McKenzie

I think it's just not showing two leading zeros in 0x7fff0d065098.

Quote:

---

Originally Posted by vincegata

Also, 0x7fff0d065098 is 140,733,411,905,688 decimal, which is 140 Terabyte, 733 MB, and so on. I have only 8GB of RAM on my laptop, how is that possible?

---

What does the pointer value have to do with the amount of RAM you have?

Regards,

Paul McKenzie

Quote:

---

Originally Posted by vincegata

I think it's just not showing two leading zeros in 0x7fff0d065098.

---

That's why you use the correct means of determining the size in bytes of any type, and that is to use sizeof().

Regards,

Paul McKenzie

> What does the pointer value have to do with the amount of RAM you have?
> Regards,
> Paul McKenzie

Because it looks like my object is located beyond the memory that I actually have.

Quote:

---

Originally Posted by vincegata

> What does the pointer value have to do with the amount of RAM you have?
> Regards,
> Paul McKenzie

Because it looks like my object is located beyond the memory that I actually have.

---

Look up the difference between a physical address and virtual/logical address. Most OS systems work this way.

http://en.wikipedia.org/wiki/Virtual_address_space

Regards,

Paul McKenzie

Thanks for your help.