Question

#include <iostream>
#include <iomanip>

using namespace std;

int main()
{
    char array[10];

    for(int i = 0; i<10;i++)
    {
        array[i] = 'a' + i;
    }

    char* test = array;

    printf("%x\n", test);
    cout << hex << test << endl;

}

The output for this is:

bffff94e
abcdefghijN???

Why is it not printing the same thing?

Answer 1

cout << hex << test << endl; 

It prints the string, not the address. It is because there is an overload of operator<< which takes char const* as argument and this overload treats the argument as string.

If you want to print the address, cast the argument to void* so that other overload of operator<< will be invoked which will print the address.

cout << hex << static_cast<void*>(test) << endl;

will print the address, in hexadecimal format.

Note that hex stream-manipulator is not needed here, as the address will be printed in hexadecimal format anway. So

cout << static_cast<void*>(test) << endl;

is enough.