std::cout of the hexadecimal 0x00

Question

I've to print on the standard output some std::uint16_t values as hexadecimal with the following text formatting: 0x##. I found this code online which works fine for every value except 0:

std::cout << std::internal 
          << std::setfill( '0' ) 
          << std::hex 
          << std::showbase 
          << std::setw( 4 ) 
          << value << std::endl;

For some reason I don't understand, 0 is printed as 0000. All the other values are correctly printed as expected.

As seen in the [Notes section here](http://en.cppreference.com/w/cpp/io/manip/showbase#Notes), this was a deliberate choice. How to get around it? Not sure, I'm afraid, never needed to. Sounds like you might need to get into creating your own [`num_put`](http://en.cppreference.com/w/cpp/locale/num_put), which is a pain in the proverbials. — BoBTFish, Mar 10 '16 at 15:45
I'm fairly sure a programmer designed this system ("Zero is the same in every base, what's the problem?"). I think `... << "0x" << setw(2) << ...` should work. — molbdnilo, Mar 10 '16 at 16:12

score 1 · Answer 1 · edited May 23 '17 at 11:50

1

For your additional question. You only need to set the width again. The rest of the manipulators are persistent.

std::cout << std::internal 
      << std::setfill( '0' ) 
      << std::hex 
      << std::showbase ;

for(std::uint16_t i =1;i<255;++i){
      std::cout<< std::setw( 4 )<<i<<"\n";
}

To overcome the setw issue here are some workarounds: “Permanent” std::setw

edited May 23 '17 at 11:50

Community

1
1

answered Mar 10 '16 at 15:58

MagunRa

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std::cout of the hexadecimal 0x00

1 Answers1