Re: How to use std::cout to output a char as a number?

From:
"Alf P. Steinbach" <alfps@start.no>
Newsgroups:
comp.lang.c++
Date:
Tue, 02 Oct 2007 01:47:03 +0200
Message-ID:
<13g31npj3kadv61@corp.supernews.com>
* dan.furlani@earthling.net:

On Apr 30, 7:52 am, Dancefire <Dancef...@gmail.com> wrote:

Hi, everyone

It might be a simple question, but I really don't know the answer.

charc = '1';cout<< c;

The above code will only output a '1' rather than 0x31;


A quick and dirty solution:

  class MyOs {
  public:
    MyOs(ostream &os) : os_(os) { }

    template <typename T>
    ostream & operator<<(T thing) { os_ << thing; return os_; }

    ostream & operator<<(char ch) // treat as unsigned.
    { os_ << static_cast<unsigned short>(static_cast<unsigned
char>(ch)); return os_; }
    ostream & operator<<(signed char ch)
    { os_ << static_cast<signed short>(ch); return os_; }
    ostream & operator<<(unsigned char ch)
    { os_ << static_cast<unsigned short>(ch); return os_; }

  private:
    ostream &os_;
  };

  enum Fix_Char { fix_char };
  MyOs operator<<(ostream &os, Fix_Char f) { return MyOs(os); }

Now you can do this:

  template <typename T>
  void foo(T arg) {
    cout << "test " << fix_char << arg << endl;
  }

I think that does what Dancefire wants but it sure is ugly. Can
someone please suggest the correct way to implement this?


Uh,

   char c = '1';
   cout << '0x' << hex << c+0;

should do the trick.

"+0" causes a conversion to int.

Cheers, & hth.,

- Alf

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