Integer to hex string in C++

Question

How do I convert an integer to a hex string in C++ ?

I can find some ways to do it, but they mostly seem targeted towards C. It doesn't seem there's a native way to do it in C++. It is a pretty simple problem though; I've got an int which I'd like to convert to a hex string for later printing.

Answer 1

Use <iomanip> 's std::hex . If you print, just send it to std::cout , if not, then use std::stringstream

std::stringstream stream;
stream << std::hex << your_int;
std::string result( stream.str() );

You can prepend the first << with << "0x" or whatever you like if you wish.

Other manips of interest are std::oct (octal) and std::dec (back to decimal).

One problem you may encounter is the fact that this produces the exact amount of digits needed to represent it. You may use setfill and setw this to circumvent the problem:

stream << std::setfill ('0') << std::setw(sizeof(your_type)*2) 
       << std::hex << your_int;

So finally, I'd suggest such a function:

template< typename T >
std::string int_to_hex( T i )
{
  std::stringstream stream;
  stream << "0x" 
         << std::setfill ('0') << std::setw(sizeof(T)*2) 
         << std::hex << i;
  return stream.str();
}

Answer 2

To make it lighter and faster I suggest to use direct filling of a string.

template <typename I> std::string n2hexstr(I w, size_t hex_len = sizeof(I)<<1) {
    static const char* digits = "0123456789ABCDEF";
    std::string rc(hex_len,'0');
    for (size_t i=0, j=(hex_len-1)*4 ; i<hex_len; ++i,j-=4)
        rc[i] = digits[(w>>j) & 0x0f];
    return rc;
}

Answer 3

You can do it with C++20 std::format :

std::string s = std::format("{:x}", 42); // s == 2a

Until std::format is widely available you can use the {fmt} library , std::format is based on ( godbolt ):

std::string s = fmt::format("{:x}", 42); // s == 2a

Disclaimer : I'm the author of {fmt} and C++20 std::format .

Answer 4

Use std::stringstream to convert integers into strings and its special manipulators to set the base. For example like that:

std::stringstream sstream;
sstream << std::hex << my_integer;
std::string result = sstream.str();

Answer 5

Just print it as an hexadecimal number:

int i = /* ... */;
std::cout << std::hex << i;

Answer 6

#include <boost/format.hpp>
...
cout << (boost::format("%x") % 1234).str();  // output is: 4d2

Answer 7

You can try the following. It's working...

#include <iostream>
#include <fstream>
#include <string>
#include <sstream>
using namespace std;

template <class T>
string to_string(T t, ios_base & (*f)(ios_base&))
{
  ostringstream oss;
  oss << f << t;
  return oss.str();
}

int main ()
{
  cout<<to_string<long>(123456, hex)<<endl;
  system("PAUSE");
  return 0;
}

Answer 8

Since C++20, with std::format , you might do:

std::format("{:#x}", your_int);    // 0x2a
std::format("{:#010x}", your_int); // 0x0000002a

Demo

Answer 9

Just have a look on my solution, ^[1] that I verbatim ^[2] copied from my project. My goal was to combine flexibility and safety within my actual needs: ^[3]

• no 0x prefix added: caller may decide
• automatic width deduction : less typing
• explicit width control: widening for formatting, (lossless) shrinking to save space
• capable for dealing with long long
• restricted to integral types : avoid surprises by silent conversions
• ease of understanding
• no hard-coded limit

#include <string>
#include <sstream>
#include <iomanip>

/// Convert integer value `val` to text in hexadecimal format.
/// The minimum width is padded with leading zeros; if not 
/// specified, this `width` is derived from the type of the 
/// argument. Function suitable from char to long long.
/// Pointers, floating point values, etc. are not supported; 
/// passing them will result in an (intentional!) compiler error.
/// Basics from: http://stackoverflow.com/a/5100745/2932052
template <typename T>
inline std::string int_to_hex(T val, size_t width=sizeof(T)*2)
{
    std::stringstream ss;
    ss << std::setfill('0') << std::setw(width) << std::hex << (val|0);
    return ss.str();
}

---

^[1] based on the answer by Kornel Kisielewicz
^[2] Only the German API doc was translated to English.
^[3] Translated into the language of CppTest , this is how it reads:

TEST_ASSERT(int_to_hex(char(0x12)) == "12");
TEST_ASSERT(int_to_hex(short(0x1234)) == "1234");
TEST_ASSERT(int_to_hex(long(0x12345678)) == "12345678");
TEST_ASSERT(int_to_hex((long long)(0x123456789abcdef0)) == "123456789abcdef0");
TEST_ASSERT(int_to_hex(0x123, 1) == "123");
TEST_ASSERT(int_to_hex(0x123, 8) == "00000123");
// width deduction test as suggested by Lightness Races in Orbit:
TEST_ASSERT(int_to_hex(short(0x12)) == "0012");

Answer 10

Thanks to Lincoln's comment below, I've changed this answer.

The following answer properly handles 8-bit ints at compile time. It doees, however, require C++17. If you don't have C++17, you'll have to do something else (eg provide overloads of this function, one for uint8_t and one for int8_t, or use something besides "if constexpr", maybe enable_if).

template< typename T >
std::string int_to_hex( T i )
{
    // Ensure this function is called with a template parameter that makes sense. Note: static_assert is only available in C++11 and higher.
    static_assert(std::is_integral<T>::value, "Template argument 'T' must be a fundamental integer type (e.g. int, short, etc..).");

    std::stringstream stream;
    stream << "0x" << std::setfill ('0') << std::setw(sizeof(T)*2) << std::hex;

    // If T is an 8-bit integer type (e.g. uint8_t or int8_t) it will be 
    // treated as an ASCII code, giving the wrong result. So we use C++17's
    // "if constexpr" to have the compiler decides at compile-time if it's 
    // converting an 8-bit int or not.
    if constexpr (std::is_same_v<std::uint8_t, T>)
    {        
        // Unsigned 8-bit unsigned int type. Cast to int (thanks Lincoln) to 
        // avoid ASCII code interpretation of the int. The number of hex digits 
        // in the  returned string will still be two, which is correct for 8 bits, 
        // because of the 'sizeof(T)' above.
        stream << static_cast<int>(i);
    }        
    else if (std::is_same_v<std::int8_t, T>)
    {
        // For 8-bit signed int, same as above, except we must first cast to unsigned 
        // int, because values above 127d (0x7f) in the int will cause further issues.
        // if we cast directly to int.
        stream << static_cast<int>(static_cast<uint8_t>(i));
    }
    else
    {
        // No cast needed for ints wider than 8 bits.
        stream << i;
    }

    return stream.str();
}

---

Original answer that doesn't handle 8-bit ints correctly as I thought it did:

Kornel Kisielewicz's answer is great. But a slight addition helps catch cases where you're calling this function with template arguments that don't make sense (eg float) or that would result in messy compiler errors (eg user-defined type).

template< typename T >
std::string int_to_hex( T i )
{
  // Ensure this function is called with a template parameter that makes sense. Note: static_assert is only available in C++11 and higher.
  static_assert(std::is_integral<T>::value, "Template argument 'T' must be a fundamental integer type (e.g. int, short, etc..).");

  std::stringstream stream;
  stream << "0x" 
         << std::setfill ('0') << std::setw(sizeof(T)*2) 
         << std::hex << i;

         // Optional: replace above line with this to handle 8-bit integers.
         // << std::hex << std::to_string(i);

  return stream.str();
}

I've edited this to add a call to std::to_string because 8-bit integer types (eg std::uint8_t values passed) to std::stringstream are treated as char, which doesn't give you the result you want. Passing such integers to std::to_string handles them correctly and doesn't hurt things when using other, larger integer types. Of course you may possibly suffer a slight performance hit in these cases since the std::to_string call is unnecessary.

Note: I would have just added this in a comment to the original answer, but I don't have the rep to comment.

Answer 11

A new C++17 way: std::to_chars from <charconv> ( https://en.cppreference.com/w/cpp/utility/to_chars ):

char addressStr[20] = { 0 };
std::to_chars(std::begin(addressStr), std::end(addressStr), address, 16);
return std::string{addressStr};

This is a bit verbose since std::to_chars works with a pre-allocated buffer to avoid dynamic allocations, but this also lets you optimize the code since allocations get very expensive if this is in a hot spot.

For extra optimization, you can omit pre-initializing the buffer and check the return value of to_chars to check for errors and get the length of the data written. Note: to_chars does NOT write a null-terminator!

Answer 12

int num = 30;
std::cout << std::hex << num << endl; // This should give you hexa- decimal of 30

Answer 13

I do:

int hex = 10;      
std::string hexstring = stringFormat("%X", hex);  

Take a look at SO answer from iFreilicht and the required template header-file from hereGIST !

Answer 14

My solution. Only integral types are allowed.

You can test/run on https://replit.com/@JomaCorpFX/ToHex

Update. You can set optional prefix 0x in second parameter.

definition.h

#include  <iomanip>
#include <sstream>

template <class T, class T2 = typename std::enable_if<std::is_integral<T>::value>::type>
static std::string ToHex(const T & data, bool addPrefix = true);



template<class T, class>
inline std::string ToHex(const T & data, bool addPrefix)
{
    std::stringstream sstream;
    sstream << std::hex;
    std::string ret;
    if (typeid(T) == typeid(char) || typeid(T) == typeid(unsigned char) || sizeof(T)==1)
    {
        sstream << static_cast<int>(data);
        ret = sstream.str();
        if (ret.length() > 2)
        {
            ret = ret.substr(ret.length() - 2, 2);
        }
    }
    else
    {
        sstream << data;
        ret = sstream.str();
    }
    return (addPrefix ? u8"0x" : u8"") + ret;
}

main.cpp

#include <iostream>
#include "definition.h"

int main()
{
    std::cout << ToHex<unsigned char>(254) << std::endl;
    std::cout << ToHex<char>(-2) << std::endl;
    std::cout << ToHex<int>(-2) << std::endl;
    std::cout << ToHex<long long>(-2) << std::endl;
    
    std::cout<< std::endl;
    std::cout << ToHex<unsigned char>(254, false) << std::endl;
    std::cout << ToHex<char>(-2, false) << std::endl;
    std::cout << ToHex<int>(-2, false) << std::endl;
    std::cout << ToHex<long long>(-2, false) << std::endl;
    return 0;
}

Results:

0xfe
0xfe
0xfffffffe
0xfffffffffffffffe

fe
fe
fffffffe
fffffffffffffffe

Answer 15

I would like to add an answer to enjoy the beauty of C ++ language. Its adaptability to work at high and low levels. Happy programming.

public:template <class T,class U> U* Int2Hex(T lnumber, U* buffer)
{
    const char* ref = "0123456789ABCDEF";
    T hNibbles = (lnumber >> 4);

    unsigned char* b_lNibbles = (unsigned char*)&lnumber;
    unsigned char* b_hNibbles = (unsigned char*)&hNibbles;

    U* pointer = buffer + (sizeof(lnumber) << 1);

    *pointer = 0;
    do {
        *--pointer = ref[(*b_lNibbles++) & 0xF];
        *--pointer = ref[(*b_hNibbles++) & 0xF];
    } while (pointer > buffer);

    return buffer;
}

Examples:

char buffer[100] = { 0 };
Int2Hex(305419896ULL, buffer);//returns "0000000012345678"
Int2Hex(305419896UL, buffer);//returns "12345678"
Int2Hex((short)65533, buffer);//returns "FFFD"
Int2Hex((char)18, buffer);//returns "12"

wchar_t buffer[100] = { 0 };
Int2Hex(305419896ULL, buffer);//returns L"0000000012345678"
Int2Hex(305419896UL, buffer);//returns L"12345678"
Int2Hex((short)65533, buffer);//returns L"FFFD"
Int2Hex((char)18, buffer);//returns L"12"

Answer 16

I can see all the elaborate coding samples others have used as answers, but there is nothing wrong with simply having this in a C++ application:

    printf ("%04x", num);

for num = 128:

    007f

https://en.wikipedia.org/wiki/Printf_format_string

C++ is effectively the original C language which has been extended, so anything in C is also perfectly valid C++.

Answer 17

For those of you who figured out that many/most of the ios::fmtflags don't work with std::stringstream yet like the template idea that Kornel posted way back when, the following works and is relatively clean:

#include <iomanip>
#include <sstream>


template< typename T >
std::string hexify(T i)
{
    std::stringbuf buf;
    std::ostream os(&buf);


    os << "0x" << std::setfill('0') << std::setw(sizeof(T) * 2)
       << std::hex << i;

    return buf.str().c_str();
}


int someNumber = 314159265;
std::string hexified = hexify< int >(someNumber);

Answer 18

Code for your reference:

#include <iomanip>
#include <sstream>
...
string intToHexString(int intValue) {

    string hexStr;

    /// integer value to hex-string
    std::stringstream sstream;
    sstream << "0x"
            << std::setfill ('0') << std::setw(2)
    << std::hex << (int)intValue;

    hexStr= sstream.str();
    sstream.clear();    //clears out the stream-string

    return hexStr;
}

Answer 19

for fixed number of digits, for instance 2:

    static const char* digits = "0123456789ABCDEF";//dec 2 hex digits positional map
    char value_hex[3];//2 digits + terminator
    value_hex[0] = digits[(int_value >> 4) & 0x0F]; //move of 4 bit, that is an HEX digit, and take 4 lower. for higher digits use multiple of 4
    value_hex[1] = digits[int_value & 0x0F]; //no need to move the lower digit
    value_hex[2] = '\0'; //terminator

you can also write a for cycle variant to handle variable digits amount

benefits:

• speed: it is a minimal bit operation, without external function calls
• memory: it use local string, no allocation out of function stack frame, no free of memory needed. Anyway if needed you can use a field or a global to make the value_ex to persists out of the stack frame

Answer 20

_itoa_s

char buf[_MAX_U64TOSTR_BASE2_COUNT];
_itoa_s(10, buf, _countof(buf), 16);
printf("%s\n", buf);    // a

swprintf_s

uint8_t x = 10;
wchar_t buf[_MAX_ITOSTR_BASE16_COUNT];
swprintf_s(buf, L"%02X", x);

Answer 21

ANOTHER SIMPLE APPROACH

#include<iostream> 
#include<iomanip> // for setbase(), works for base 8,10 and 16 only
using namespace std;


int main(){
  int x = (16*16+16+1)*15;
  string ans;


  stringstream ss;
  ss << setbase(16) << x << endl;
  ans = ss.str();


  cout << ans << endl;//prints fff

Answer 22

With the variable:

char selA[12];

then:

snprintf(selA, 12, "SELA;0x%X;", 85);

will result in selA containing the string SELA;0x55;

Note that the things surrounding the 55 are just particulars related to the serial protocol used in my application.

Answer 23

#include <iostream> 
#include <sstream>  

int main()
{
unsigned int i = 4967295; // random number
std::string str1, str2;
unsigned int u1, u2;

std::stringstream ss;

Using void pointer:

// INT to HEX
ss << (void*)i;       // <- FULL hex address using void pointer
ss >> str1;          //     giving address value of one given in decimals.
ss.clear();         // <- Clear bits
// HEX to INT
ss << std::hex << str1;   // <- Capitals doesn't matter so no need to do extra here
ss >> u1;
ss.clear();

Adding 0x:

// INT to HEX with 0x
ss << "0x" << (void*)i;   // <- Same as above but adding 0x to beginning
ss >> str2;
ss.clear();
// HEX to INT with 0x
ss << std::hex << str2;   // <- 0x is also understood so need to do extra here
ss >> u2;
ss.clear();

Outputs:

std::cout << str1 << std::endl; // 004BCB7F
std::cout << u1 << std::endl;   // 4967295
std::cout << std::endl;
std::cout << str2 << std::endl; // 0x004BCB7F
std::cout << u2 << std::endl;   // 4967295


return 0;
}

Answer 24

char_to_hex returns a string of two characters

const char HEX_MAP[] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};

char replace(unsigned char c)
{
    return HEX_MAP[c & 0x0f];
}

std::string char_to_hex(unsigned char c)
{
    std::string hex;

    // First four bytes
    char left = (c >> 4);
    // Second four bytes
    char right = (c & 0x0f);

    hex += replace(left);
    hex += replace(right);

    return hex;
} 

Answer 25

This question is quite old but the answers given are to my opinion not the best. If you are using C++20 then you have the option to use std::format which is a very good solution. However if you are using C++11/14/17 or below you will not have this option.

Most other answers either use the std::stringstream or implement their own conversion modifying the underlying string buffer directly by themselves. The first option is rather heavy weight. The second option is inherently insecure and bug prone.

Since I had to implement an integer to hex string lately I chose to do aa true C++ safe implementation using function overloads and template partial specialization to let the compiler handle the type checks. The code uses sprintf (which one of its flavors is generally used by the standard library for std::to_string ). And it relies on template partial specialization to correctly select the right sprintf format and leading 0 addition. It separately and correctly handles different pointer sizes and unsigned long sizes for different OSs and architectures. (4/4/4, 4/4/8, 4/8/8)

This answer targets C++11

H File:

#ifndef STRINGUTILS_H_
#define STRINGUTILS_H_

#include <string>

namespace string_utils
{
    /* ... Other string utils ... */
    
    std::string hex_string(unsigned char v);
    std::string hex_string(unsigned short v);
    std::string hex_string(unsigned int v);
    std::string hex_string(unsigned long v);
    std::string hex_string(unsigned long long v);
    std::string hex_string(std::ptrdiff_t v);

} // namespace string_utils

#endif

CPP File

#include "stringutils.h"

#include <cstdio>

namespace
{
    template <typename T, int Width> struct LModifier;

    template <> struct LModifier<unsigned char, sizeof(unsigned char)>
    {
        static constexpr char fmt[] = "%02hhX";
    };
    template <> struct LModifier<unsigned short, sizeof(unsigned short)>
    {
        static constexpr char fmt[] = "%04hX";
    };
    template <> struct LModifier<unsigned int, sizeof(unsigned int)>
    {
        static constexpr char fmt[] = "%08X";
    };
    template <> struct LModifier<unsigned long, 4>
    {
        static constexpr char fmt[] = "%08lX";
    };
    template <> struct LModifier<unsigned long, 8>
    {
        static constexpr char fmt[] = "%016lX";
    };
    template <> struct LModifier<unsigned long long, sizeof(unsigned long long)>
    {
        static constexpr char fmt[] = "%016llX";
    };
    template <> struct LModifier<std::ptrdiff_t, 4>
    {
        static constexpr char fmt[] = "%08tX";
    };
    template <> struct LModifier<std::ptrdiff_t, 8>
    {
        static constexpr char fmt[] = "%016tX";
    };

    constexpr char LModifier<unsigned char, sizeof(unsigned char)>::fmt[];
    constexpr char LModifier<unsigned short, sizeof(unsigned short)>::fmt[];
    constexpr char LModifier<unsigned int, sizeof(unsigned int)>::fmt[];
    constexpr char LModifier<unsigned long, sizeof(unsigned long)>::fmt[];
    constexpr char LModifier<unsigned long long, sizeof(unsigned long long)>::fmt[];
    constexpr char LModifier<std::ptrdiff_t, sizeof(std::ptrdiff_t)>::fmt[];

    template <typename T, std::size_t BUF_SIZE = sizeof(T) * 2U> std::string hex_string_(T v)
    {
        std::string ret(BUF_SIZE + 1, 0);
        std::sprintf((char *)ret.data(), LModifier<T, sizeof(T)>::fmt, v);
        return ret;
    }
} // anonymous namespace

std::string string_utils::hex_string(unsigned char v)
{
    return hex_string_(v);
}
std::string string_utils::hex_string(unsigned short v)
{
    return hex_string_(v);
}
std::string string_utils::hex_string(unsigned int v)
{
    return hex_string_(v);
}
std::string string_utils::hex_string(unsigned long v)
{
    return hex_string_(v);
}
std::string string_utils::hex_string(unsigned long long v)
{
    return hex_string_(v);
}
std::string string_utils::hex_string(std::ptrdiff_t v)
{
    return hex_string_(v);
}

Answer 26

All the answers I read are pretty slow, except one of them, but that one only works for little endian CPUs. Here's a fast implementation that works on big and little endian CPUs.

std::string Hex64(uint64_t number)
{
    static const char* maps = "0123456789abcdef";

    // if you want more speed, pass a buffer as a function parameter and return an std::string_view (or nothing)
    char  buffer[17]; // = "0000000000000000"; // uncomment if leading 0s are desired
    char* c          = buffer + 16;
    do
    {
        *--c = maps[number & 15];
        number >>= 4;
    }
    while (number > 0);

    // this strips the leading 0s, if you want to keep them, then return std::string(buffer, 16); and uncomment the "000..." above
    return std::string(c, 16 - (c - buffer));
}

Compared to std::format and fmt::format("{:x}", value) , I get between 2x (for values > (1ll << 60)) and 6x the speed (for smaller values).

Examples of input/output:

const std::vector<std::tuple<uint64_t, std::string>> vectors = {
    {18446744073709551615, "ffffffffffffffff"},
    { 4294967295u,         "ffffffff"},
    { 16777215,            "ffffff"},
    { 65535,               "ffff"},
    { 255,                 "ff"},
    { 16,                  "10"},
    { 15,                  "f"},
    { 0,                   "0"},
};

Answer 27

`

char sw(int num){
   char h[16] = {'0','1','2','3','4','5','6','7','8','9','a','b',' c','d','e','f'};
   return h[num];
}

String to_hexa(int num) {
  int i, j, exp = 0, temp = num;
  String hexa;
  String begin;


  while( temp >= 16 ){
    exp++;
    temp /= 16;
  }

  for ( i = 0; i <= exp; ++i){
    temp = num;
    for ( j = i; j <= exp; ++j){
      if ( j == exp ) hexa += String(sw(temp % 16));
      else temp /= 16;
    }
  }
  begin = "0x";
  for (int i=0; i < 3 - exp; i++){
    begin += "0";
  }  
  return begin + hexa;
}

`

Answer 28

int var = 20;
cout <<                          &var << endl;
cout <<                     (int)&var << endl;
cout << std::hex << "0x" << (int)&var << endl << std::dec; // output in hex, reset back to dec

0x69fec4 (address)
6946500 (address to dec)
0x69fec4 (address to dec, output in hex)

---

instinctively went with this...
int address = (int)&var;

saw this elsewhere...
unsigned long address = reinterpret_cast(&var);

comment told me this is correct...
int address = (int)&var;

speaking of well covered lightness, where are you at? they're getting too many likes!

Answer 29

You can define MACRO to use as one liner like this.

#include <sstream>
#define to_hex_str(hex_val) (static_cast<std::stringstream const&>(std::stringstream() << "0x" << std::hex << hex_val)).str()