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

[Dancefire]

Hi, everyone

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

char c = '1';
cout << c;

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

If I use int cast, it can show the number:

cout << (int) c;

however, if the c is 0x80, which might be a part of MBCS(multibyte
character set) string, such as :

const char* cc = "\xba\xba";
cout << hex << "0x" << (int)cc[0] << ", 0x" << (int)cc[1] << endl;

it will extend negative bit, and will output as:

0xFFFFFFBA, 0xFFFFFFBA

I currently use a stupid way to avoid extending the negative bit and
show the number of a simple char by double casting:

const char* cc = "\xba\xba";
cout << hex << "0x" << (unsigned short)((unsigne d char)cc[0]) << ",
0x" << (unsigned short)((unsigne d char)cc[1]) << endl;

This time the output looks ok. But it doesn't make any sense, since I
only want to output the char as a number, should not so complicated by
double casting.

Did I miss anything? What is the most directly way to use std::cout
output a char as a number?

Thanks.

[Victor Bazarov]

Dancefire wrote:

Hi, everyone

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

char c = '1';
cout << c;

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

If I use int cast, it can show the number:

cout << (int) c;

however, if the c is 0x80,[..]

See, this is where the logic fails you. An 8-bit char cannot
have the value 0x7f, unless it's implemented as unsigned. If
it's a signed char (as you seem to show), 0x7f is its highest
possible value.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Dancefire]

If I use int cast, it can show the number:

>

cout << (int) c;

however, if the c is 0x80,[..]

See, this is where the logic fails you. An 8-bit char cannot
have the value 0x7f, unless it's implemented as unsigned. If
it's a signed char (as you seem to show), 0x7f is its highest
possible value.

Yes, define the 'unsigned char' works for above case. Thank you.

But I still have the problem. The original problem is raised when I
try to print a std::string's content as number for each character. The
standard define the string is basic_string<ch ar, ...>, is not
bacic_string<un signed char, ...>. Currently, I have to use the
following loop to show each character as a number:

string test("\xba\xba" );
for(std::string ::iterator iter = test.begin(); iter != test.end(); +
+iter)
{
cout << "0x" << hex << (unsigned short)((unsigne d char) *iter) <<
" ";
}

The *iter is char, not unsigned char. So, I have to use double casting
to do the job. If I use c-style function, I can easily specify "%x" in
the format string of printf() to print the char without any
unnecessary casting, it looks much straightforward . In C++'s
std::cout, there should be a way that easily specify whether I want to
print a character, or a numeric value, rather than determine by cout
itself, since the 'char' is really special, sometimes it's character,
sometime, it means a 8-bit integer number.

[Gennaro Prota]

On 30 Apr 2007 04:52:22 -0700, Dancefire wrote:

[...]

>const char* cc = "\xba\xba";
cout << hex << "0x" << (unsigned short)((unsigne d char)cc[0]) << ",
0x" << (unsigned short)((unsigne d char)cc[1]) << endl;

This time the output looks ok. But it doesn't make any sense, since I
only want to output the char as a number, should not so complicated by
double casting.

Did I miss anything? What is the most directly way to use std::cout
output a char as a number?

What do you mean by most direct way? You could write something like
the following:

unsigned
as_unsigned( char c )
{
return static_cast< unsigned char >( c );
}

(note: if c is negative, the cast to unsigned char will yield the
mathematical value UCHAR_MAX + c, regardless of whether char uses a
two's complement representation or not)

--
Gennaro Prota
https://sourceforge.net/projects/breeze/

[Gennaro Prota]

On Mon, 30 Apr 2007 16:25:12 +0200, Gennaro Prota wrote:

>(note: if c is negative, the cast to unsigned char will yield the
mathematical value UCHAR_MAX + c, regardless of whether char uses a
two's complement representation or not)

Of course I meant to write UCHAR_MAX + 1 + c, sorry.

--
Gennaro Prota
https://sourceforge.net/projects/breeze/

[Dancefire]

>

Did I miss anything? What is the most directly way to use std::cout
output a char as a number?

What do you mean by most direct way? You could write something like
the following:

unsigned
as_unsigned( char c )
{
return static_cast< unsigned char >( c );
}

(note: if c is negative, the cast to unsigned char will yield the
mathematical value UCHAR_MAX + c, regardless of whether char uses a
two's complement representation or not)

--
Gennaro Protahttps://sourceforge.net/projects/breeze/

Thanks for reply, do I have to use a function to do this? It involve
much more overhead. Not only the function call, but also I should put
the function to some where my every time cout << char can see the
code, which means I should put the function into my function lib.

And the above function still cannot output the number, it's will still
output the character, Unless I call another casting to cast the
unsigned char to unsigned short. It's not better than just double
casting the char at the place I output it. which is provide at
original post:

cout << " 0x" << hex << static_cast<uns igned
short>(static_c ast<unsigned char>(c));

But this form is not semantically correct, I only want to output the 8-
bit unsigned integer, rather than 16-bit unsigned integer. And I use 2
casting only for correct output the char as a number? Is there any
other way to do that? is there anything just like the printf("%x", c)
in cout, which means I set something, and then the output for char is
a number. Such as:

cout << " 0x" << hex << char_is_number << c;

or

cout << " 0x" << hex << static_cast<*a simple type here*>(c);

Thanks.

[Gennaro Prota]

On 30 Apr 2007 09:25:46 -0700, Dancefire wrote:

>Did I miss anything? What is the most directly way to use std::cout
output a char as a number?

What do you mean by most direct way? You could write something like
the following:

unsigned
as_unsigned( char c )
{
return static_cast< unsigned char >( c );
}

(note: if c is negative, the cast to unsigned char will yield the
mathematical value UCHAR_MAX + c, regardless of whether char uses a
two's complement representation or not)

[sig snipped...]

Thanks for reply, do I have to use a function to do this? It involve
much more overhead. Not only the function call,

You shouldn't be concerned with that. Try for instance googling for
"premature optimization".

>but also I should put
the function to some where my every time cout << char can see the
code, which means I should put the function into my function lib.

What's wrong with that?

>And the above function still cannot output the number, it's will still
output the character, Unless I call another casting to cast the
unsigned char to unsigned short.

No. First the function performs no output, just a (double) conversion.
Secondly, the return type is unsigned int, not unsigned char. Try

std::cout << as_unsigned( c );

>It's not better than just double casting the char at the place

It's better than double casting "in place" for at least two reasons:
a) it gives a name to the operation you perform b) it encapsulates the
way you do it: should you later discover that the way you did it is
incorrect, you just have to fix it in one place.

And, of course, you should anyway use new-style casts.

>I output it. which is provide at original post:

cout << " 0x" << hex << static_cast<uns igned
short>(static_ cast<unsigned char>(c));

But this form is not semantically correct,

You lost me here. Didn't you say in your original post that this gives
what you want?

>I only want to output the 8-
bit unsigned integer, rather than 16-bit unsigned integer.

What's the difference on output if your unsigned int has at most 8
bits on?

>And I use 2
casting only for correct output the char as a number? Is there any
other way to do that? is there anything just like the printf("%x", c)
in cout, which means I set something, and then the output for char is
a number. Such as:

cout << " 0x" << hex << char_is_number << c;

or

cout << " 0x" << hex << static_cast<*a simple type here*>(c);

You could obtain the latter, yes, though I don't see what advantage it
gives.

#include <ostream>

struct simple_type
{
typedef unsigned short number_type;

number_type m_n;

explicit simple_type( char c )
: m_n( static_cast< unsigned char >( c ) )
{
}
};

std::ostream &
operator<<( std::ostream & dest, const simple_type & s )
{
return dest << s.m_n;
}

--
Gennaro Prota
https://sourceforge.net/projects/breeze/

[Gennaro Prota]

On 30 Apr 2007 07:07:41 -0700, Dancefire wrote:

>Currently, I have to use the
following loop to show each character as a number:

string test("\xba\xba" );
for(std::strin g::iterator iter = test.begin(); iter != test.end(); +
+iter)
{
cout << "0x" << hex << (unsigned short)((unsigne d char) *iter) <<
" ";
}

The *iter is char, not unsigned char. So, I have to use double casting
to do the job. If I use c-style function, I can easily specify "%x" in
the format string of printf() to print the char without any
unnecessary casting, it looks much straightforward .

BTW, where did you get that idea from? If you specify "%x" the
corresponding argument must have type unsigned int; otherwise you have
*undefined behavior*.

--
Gennaro Prota
https://sourceforge.net/projects/breeze/

[Gianni Mariani]

Dancefire wrote:
....

Did I miss anything? What is the most directly way to use std::cout
output a char as a number?

No, you didn't miss anything. There may be slightly more elegant
solutions but this is what you get when the compiler picks your output
format by overloading the << operator.

To make it a little more palatable, this is one alternative.

unsigned OutAsUnsigned( char val )
{
return 0xff & unsigned( val );
}
cout << hex << "0x" << OutAsUnsigned( cc[0] )
<< 0x" << OutAsUnsigned( cc[1] );

[Dancefire]

And the above function still cannot output the number, it's will still

output the character, Unless I call another casting to cast the
unsigned char to unsigned short.

No. First the function performs no output, just a (double) conversion.
Secondly, the return type is unsigned int, not unsigned char. Try

std::cout << as_unsigned( c );

oops, yes you're right, the function return "unsigned" which is
"unsigned int", that makes it work. But, actually, it's still a double
casting of the char, one explicit casting in as_unsigned() function,
one implicit casting during the return of the function. After
optimization by compiler (eliminate the function call), the result
will be the same as the in place double casting.

It's not better than just double casting the char at the place

It's better than double casting "in place" for at least two reasons:
a) it gives a name to the operation you perform b) it encapsulates the
way you do it: should you later discover that the way you did it is
incorrect, you just have to fix it in one place.

And, of course, you should anyway use new-style casts.

Yes, you are right, encapsulate the operation as a function can make
the debug much easier.

I output it. which is provide at original post:

cout << " 0x" << hex << static_cast<uns igned
short>(static_c ast<unsigned char>(c));

But this form is not semantically correct,

You lost me here. Didn't you say in your original post that this gives
what you want?

The original solution in my first post:

cout << " 0x" << hex << (unsigned short)((unsigne d char)c);

It works, (I modified them to static_cast<>() way now), I'm currently
using this way, but I think this form is not elegant and might be not
necessary. I thought there may be a flag or something like "hex" in
STL can be set on cout, so I can directly cout << c, without any
(logically) unnecessary casting in my code.

I only want to output the 8-
bit unsigned integer, rather than 16-bit unsigned integer.

What's the difference on output if your unsigned int has at most 8
bits on?

no, no different for the output, just might not elegant in the code.

And I use 2
casting only for correct output the char as a number? Is there any
other way to do that? is there anything just like the printf("%x", c)
in cout, which means I set something, and then the output for char is
a number. Such as:

cout << " 0x" << hex << char_is_number << c;

or

cout << " 0x" << hex << static_cast<*a simple type here*>(c);

You could obtain the latter, yes, though I don't see what advantage it
gives.

#include <ostream>

struct simple_type
{
typedef unsigned short number_type;

number_type m_n;

explicit simple_type( char c )
: m_n( static_cast< unsigned char >( c ) )
{
}
};

std::ostream &
operator<<( std::ostream & dest, const simple_type & s )
{
return dest << s.m_n;
}

--
Gennaro Protahttps://sourceforge.net/projects/breeze/

Yes, I can get the latter form by this code, but I don't see the
advantage of it either, and it actually do the same thing above,
double casting, and just became a struct form.

So, in conclusion, there is no way to avoid the double casting if I
want std::cout output a char as a number, right?

Thank you.