hex char to decimal and store output as char - c++

Question

I am writing a hex to dec conversion function. The input is a single character which is converted to hex and returned back as a char.
Here is the function
char hex2dec(char inp)
{
char out;
cout << "inp:" << inp;
if(inp >= '0' && inp <='9')
{
out = (inp - '0');
cout << " out " << out;
}
else
{
out = (toupper(inp) - 'A' + 10);
cout << " out " << out;
}
return out;
}
When i pass '0' and 'A' to the function, the print i get is
inp:0 out
inp:A out
i.e nothing is printed in out.
I am not able to find the issue..Can anyone help?

Answer 1

What you are actually trying to print is ascii characters with codes [0-15] which are not printable characters ie you want to print 15 but you print "\x0f" etc
use:
cout << (int)out;
and you'll force cout to invoke method printing ints not chars - this will solve your issue.
..or more 'c++++ ish' ;-)
cout << static_cast(out);
..or this which for most looks weird:
cout << int(out);

Answer 2

Use int out instead of char out. cout prints char as a character not an integer. The ASCII values 0-15 are unprintable control characters.

Answer 3

the reason why u got an 'A' from print is that out is char ,and after 'A' - 'A' + 10, out is a character whose ascii value is 10 instead of integer 10. So u will get a character whose ascii value is 10 instead of getting 'A'.After checking the ascii table, that character is null,which can explain why u get nothing in the output.

Answer 4

Simply use std::stringstream for this purpose.
#include <iostream>
#include <sstream>
int main (int argc, char ** argv)
{
// use hex formatting
std::stringstream stream;
stream << std::hex << 'A';
// retrieve value of the hex input character
int value;
stream >> value;
std::cout << value << std::endl; // prints 10
// reset the stream
stream.str ();
stream.clear();
// also works for strings
stream << "0xABC" << std::endl;
stream >> value;
std::cout << value << std::endl; // prints 2748
return 0;
}