print 64bit c++ full memory address

Question

While I was writing code on a 64 bit machine for a c++ program,I noticed that printing the address of a variable (for example) returns just 12 hexadecimal characters, instead of 16. Here's an example code:

int a = 3 ;
cout sizeof(&a) << " bytes" << endl ;
cout << &a << endl ;

The output is:

8 bytes

0x7fff007bcce0

Obviously, the address of a variable is 8 byte (64 bit system). But when I print it, I get only 12 hexadecimal digits instead of 16.

• Why this? I think that is due to the fact that the 4 "lost" digits were leading zeroes, that were not printed. But this is only my thought, and I wish to have a definitive and technically correct answer.

• How could I print the entire address? Is there a built-in solution, or should I manually use "sizeof" in order to get the real lenght and then add to the address the right number of zeroes?

Forgive me, I googled for a day for an answer to my stupid question, but I wasn't able to find an answer. I'm just a newbie. (On stackoverflow I did not find any question/answer about what I needed to know, but maybe I'm wrong.)

Answer 1

I am currently writing a book on C++ and windows 32-bit programming for peeps such as you, but unfortunately I am not yet done with it :(

The following code demonstrates how you would display a 64-bit unsigned number using cout method:

// Define a 64-bit number. You may need to include <stdint.h> header file depending on your C++ compiler.

uint64_t UI64 = 281474976709632ULL; // Must include ULL suffix and this is C99 C++ compiler specific.

// unsigned __int64 UI64 = 281474976709632ULL; // Must include ULL suffix and this is Microsoft C++ compiler specific.

// Set decimal output.

cout << dec;

// Display message to user.

cout << "64-bit unsigned integer value in decimal is: " << UI64 << endl;

cout << "\n64-bit unsigned integer value in hexadecimal is: ";

// Set the uppercase flag to display hex value in capital letters.

cout << uppercase;

// Set hexadecimal output.

cout << hex;

// Set the width output to be 16 digits.

cout.width(16);

// Set the fill output to be zeros.

cout.fill('0');

// Set right justification for output.

right(cout);

// Display the 64-bit number.

cout << UI64 << endl;

You may need to (type) cast the address into a 64-bit unsigned value.
In this case, you can do the following:

// (type) cast pointer adddress into an unsigned 64-bit integer.

uint64_t UADD64 = (uint64_t)&UI64; // C99 C++ compiler specific.

Answer 2

Someone asks a pretty similar question here: c++ pointer on 64 bit machine

Hope this helps :)

To print the full 64bit address with leading zeros you can use:

std::cout
<< "0x"
<< std::hex
<< std::noshowbase
<< std::setw(16)
<< std::setfill('0')
<< n
<< std::endl ;

Got it from: How can I pad an int with leading zeros when using cout << operator?