Method that converts uint8_t to string

Question

I have this:

uint8_t key[] = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31};

How can I convert it to char or something so that I can read its contents? this is a key i used to encrypt my data using AES.

Help is appreciated. Thanks

There's more than one way you could convert a bunch of numbers to a string. What do you expect the result to look like? — user253751, Jul 14 '15 at 22:12

score 12 · Accepted Answer · edited Nov 22 '19 at 18:23

12

String converter(uint8_t *str){
    return String((char *)str);
}

edited Nov 22 '19 at 18:23

Rudi Strydom

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answered Aug 06 '16 at 11:56

TRiNE

5,020
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score 3 · Answer 2 · answered Jul 14 '15 at 22:30

If I have understood correctly you need something like the following

#include <iostream>
#include <string>
#include <numeric>
#include <iterator>
#include <cstdint>

int main()
{
    std::uint8_t key[] = 
    { 
        0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31
    };

    std::string s;
    s.reserve( 100 );

    for ( int value : key ) s += std::to_string( value ) + ' ';

    std::cout << s << std::endl;

}

The program output is

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

You can remove blanks if you not need them.

Having the string you can process it as you like.

score 1 · Answer 3 · answered Jul 14 '15 at 22:30

1

If the goal is to construct a string of 2-character hex values, you can use a string stream with IO manipulators like this:

std::string to_hex( uint8_t data[32] )
{
  std::ostringstream oss;
  oss << std::hex << std::setfill('0');
  for( uint8_t val : data )
  {
    oss << std::setw(2) << (unsigned int)val;
  }
  return oss.str();
}

This requires the headers:

<string>
<sstream>
<iomanip>

answered Jul 14 '15 at 22:30

paddy

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sstream already includes string and adding the setw flag to the stream every pass isn't needed too – Youka Jul 14 '15 at 22:37
2

@Youka setw isn't sticky, please see: http://stackoverflow.com/questions/1532640/which-iomanip-manipulators-are-sticky – Snooze Jul 26 '16 at 20:01

score 1 · Answer 4 · answered Jul 14 '15 at 22:33

#include <sstream>  // std::ostringstream
#include <algorithm>    // std::copy
#include <iterator> // std::ostream_iterator
#include <iostream> // std::cout

int main(){
    uint8_t key[] = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31};
    std::ostringstream ss;
    std::copy(key, key+sizeof(key), std::ostream_iterator<int>(ss, ","));
    std::cout << ss.str();
    return 0;
}

0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29, 30,31,

score 1 · Answer 5 · answered Jul 14 '15 at 22:37

You can use a stringstream:

 #include <sstream>

    void fnPrintArray (uint8_t key[], int length) {

         stringstream list;

         for (int i=0; i<length; ++i)
        {
            list << (int)key[i];
        }
        string key_string = list.str();
        cout << key_string << endl;
    }

Method that converts uint8_t to string

5 Answers5