Very simple c++ question regarding cout

Hi DanRaposo,

You need to convert the ASCII to hex display.  fprintf(), sprintf(), Format(), or any of many available tools will do this for you.


Good Luck,
Kent

Here is an exmple using streams.  - I compiled this using Visual Studio 2003


#include "stdafx.h"
#include <iostream>
#include <tchar.h>
#include <iostream>
#include <iomanip>

using namespace std;


//Showing the base of integer values
int _tmain(int argc, _TCHAR* argv[])
{
   int a=148;
   cout.setf(ios::showbase); // show the base of all integral output:
                             // leading 0x means hexadecimal,
                             // leading 01 to 07 means octal,
                             // leading 1 to 9 means decimal
   cout.setf(ios::oct,ios::basefield);
                             // change format state to octal
   cout << a << '\n';
   cout.setf(ios::dec,ios::basefield);
                             // change format state to decimal
   cout << a << '\n';
   cout.setf(ios::hex,ios::basefield);
                             // change format state to hexadecimal
   cout << a << '\n';
   cout << oct << a << '\n'; // Parameterized manipulators clear the
   cout << dec << a << '\n'; // basefield, then set the appropriate
   cout << hex << a << '\n'; // flag within basefield.

   return 0;
}

wow... so pretty much just stick to the printf... it doesn't such a nice job compared to having to do anything in c++

>>>> C++ seems to get confused when you're outputting char data.
Don't think it is confusion.

There is an overloaded template function  operator<< (ostream& os, char c) which atually cannot print hex output. It simply ignores the 'hex' formatter stored in the istream. By casting it to a int, operator<< (ostream& os, int i) comes to play which does the required.

Note, printf has an advantage: when specifiying %x it expects an integer. Yu only pass a byte but that doesn't matter cause the arguments were aligned at a 32bit address padded with zeroes. So, the char passed with printf actually turns to a 32bit integer on the stack. Fortunately, cause otherwise the conversion could fail.

Regards, Alex

casting to int worked just fine

cout << hex << (int)readBuf[1] << endl;

Thank you

Hi Alex,

This is perhaps the ONLY sane argument for little-endian.  casting (int) to (char) is a NOOP.   :)


Kent

>>>> compared to having to do anything in c++
today I've seen 3 errors regarding printf and and scanf.

C++ streaming has the advantage that it is checked by the compiler while the printf and scanf are responsible for a big part of the security leaks we can encounter every enw day.

>>>> the ONLY sane argument for little-endian

Actually little-endian and big-endian do not differ regarding all kind of integers (char, short, int, long long). It doesn't matter whether you call the most 'left' or the most 'right' bit the 'low' bit. So, problems only arise if you need to convert the one to the other ...