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

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int var   20  cout  lt  lt                            amp var  lt  lt  endl  cout  lt  lt                       int  amp var  lt  lt  endl  cout  lt  lt  std  hex  lt  lt   0x   lt  lt   int  amp var  lt  lt  endl  lt  lt  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  amp 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

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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 here GIST

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Just print it as an hexadecimal number   int i              std  cout  lt  lt  std  hex  lt  lt  i

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To make it lighter and faster I suggest to use direct filling of a string   template  lt typename I gt  std  string n2hexstr I w  size t hex len   sizeof I  lt  lt 1        static const char  digits    0123456789ABCDEF       std  string rc hex len  0        for  size t i 0  j  hex len-1  4   i lt hex len    i j- 4          rc i    digits  w gt  gt j   amp  0x0f       return rc

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You can try the following  It s working      include  lt iostream gt   include  lt fstream gt   include  lt string gt   include  lt sstream gt  using namespace std   template  lt class T gt  string to string T t  ios base  amp    f  ios base amp        ostringstream oss    oss  lt  lt  f  lt  lt  t    return oss str       int main        cout lt  lt to string lt long gt  123456  hex  lt  lt endl    system  PAUSE      return 0

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Use  lt iomanip gt  s  std  hex  If you print  just send it to std  cout  if not  then use std  stringstream  std  stringstream stream  stream  lt  lt  std  hex  lt  lt  your int  std  string result  stream str        You can prepend the first  lt  lt  with  lt  lt   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  lt  lt  std  setfill   0    lt  lt  std  setw sizeof your type  2           lt  lt  std  hex  lt  lt  your int    So finally  I d suggest such a function   template lt  typename T  gt  std  string int to hex  T i       std  stringstream stream    stream  lt  lt   0x             lt  lt  std  setfill   0    lt  lt  std  setw sizeof T  2             lt  lt  std  hex  lt  lt  i    return stream str

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This question is old  but I m surprised why no one mentioned boost  format   cout  lt  lt   boost  format   x     1234  str        output is  4d2

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Use std  stringstream to convert integers into strings and its special manipulators to set the base  For example like that   std  stringstream sstream  sstream  lt  lt  std  hex  lt  lt  my integer  std  string result   sstream str

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int num   30  std  cout  lt  lt  std  hex  lt  lt  num  lt  lt  endl     This should give you hexa- decimal of 30

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My solution  Only integral types are allowed    Update  You can set optional prefix 0x in second parameter   definition h      include   lt iomanip gt   include  lt sstream gt   template  lt class T  class T2   typename std  enable if lt std  is integral lt T gt   value gt   type gt  static std  string ToHex const T  amp  data  bool addPrefix   true      template lt class T  class gt  inline std  string Convert  ToHex const T  amp  data  bool addPrefix        std  stringstream sstream      sstream  lt  lt  std  hex      std  string ret      if  typeid T     typeid char     typeid T     typeid unsigned char     sizeof T   1                sstream  lt  lt  static cast lt int gt  data           ret   sstream str            if  ret length    gt  2                        ret   ret substr ret length   - 2  2                       else               sstream  lt  lt  data          ret   sstream str              return  addPrefix   u8 0x    u8      ret      main cpp     include  lt definition h gt  int main         std  cout  lt  lt  ToHex lt unsigned char gt  254   lt  lt  std  endl      std  cout  lt  lt  ToHex lt char gt  -2   lt  lt  std  endl      std  cout  lt  lt  ToHex lt int gt  -2   lt  lt  std  endl      std  cout  lt  lt  ToHex lt long long gt  -2   lt  lt  std  endl       std  cout lt  lt  std  endl      std  cout  lt  lt  ToHex lt unsigned char gt  254  false   lt  lt  std  endl      std  cout  lt  lt  ToHex lt char gt  -2  false   lt  lt  std  endl      std  cout  lt  lt  ToHex lt int gt  -2  false   lt  lt  std  endl      std  cout  lt  lt  ToHex lt long long gt  -2  false   lt  lt  std  endl      return 0      Results  0xfe 0xfe 0xfffffffe 0xfffffffffffffffe    fe fe  fffffffe fffffffffffffffe

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Code for your reference    include  lt iomanip gt   include  lt sstream gt      string intToHexString int intValue         string hexStr           integer value to hex-string     std  stringstream sstream      sstream  lt  lt   0x               lt  lt  std  setfill   0    lt  lt  std  setw 2       lt  lt  std  hex  lt  lt   int intValue       hexStr  sstream str        sstream clear         clears out the stream-string      return hexStr

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for fixed number of digits  for instance 2      static const char  digits    quot 0123456789ABCDEF quot    dec 2 hex digits positional map     char value hex 3    2 digits   terminator     value hex 0    digits  int value  gt  gt  4   amp  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  amp  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

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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  lt class T class U gt  U  Int2Hex T lnumber  U  buffer        const char  ref    quot 0123456789ABCDEF quot       T hNibbles    lnumber  gt  gt  4        unsigned char  b lNibbles    unsigned char   amp lnumber      unsigned char  b hNibbles    unsigned char   amp hNibbles       U  pointer   buffer    sizeof lnumber   lt  lt  1         pointer   0      do            --pointer   ref   b lNibbles     amp  0xF            --pointer   ref   b hNibbles     amp  0xF         while  pointer  gt  buffer        return buffer     Examples  char buffer 100      0    Int2Hex 305419896ULL  buffer    returns  quot 0000000012345678 quot  Int2Hex 305419896UL  buffer    returns  quot 12345678 quot  Int2Hex  short 65533  buffer    returns  quot FFFD quot  Int2Hex  char 18  buffer    returns  quot 12 quot   wchar t buffer 100      0    Int2Hex 305419896ULL  buffer    returns L quot 0000000012345678 quot  Int2Hex 305419896UL  buffer    returns L quot 12345678 quot  Int2Hex  short 65533  buffer    returns L quot FFFD quot  Int2Hex  char 18  buffer    returns L quot 12 quot

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include  lt iostream gt    include  lt sstream gt     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  lt  lt   void  i            lt - FULL hex address using void pointer ss  gt  gt  str1                  giving address value of one given in decimals  ss clear                lt - Clear bits    HEX to INT ss  lt  lt  std  hex  lt  lt  str1        lt - Capitals doesn t matter so no need to do extra here ss  gt  gt  u1  ss clear     Adding 0x     INT to HEX with 0x ss  lt  lt   quot 0x quot   lt  lt   void  i        lt - Same as above but adding 0x to beginning ss  gt  gt  str2  ss clear       HEX to INT with 0x ss  lt  lt  std  hex  lt  lt  str2        lt - 0x is also understood so need to do extra here ss  gt  gt  u2  ss clear     Outputs  std  cout  lt  lt  str1  lt  lt  std  endl     004BCB7F std  cout  lt  lt  u1  lt  lt  std  endl       4967295 std  cout  lt  lt  std  endl  std  cout  lt  lt  str2  lt  lt  std  endl     0x004BCB7F std  cout  lt  lt  u2  lt  lt  std  endl       4967295   return 0

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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  lt iomanip gt   include  lt sstream gt    template lt  typename T  gt  std  string hexify T i        std  stringbuf buf      std  ostream os  amp buf         os  lt  lt   0x   lt  lt  std  setfill  0    lt  lt  std  setw sizeof T    2          lt  lt  std  hex  lt  lt  i       return buf str   c str        int someNumber   314159265  std  string hexified   hexify lt  int  gt  someNumber

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Just have a look on my solution  1  that I verbatim copied from my project  so there a German is API doc included  My goal was to combine flexibility and safety within my actual needs  2    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  lt string gt   include  lt sstream gt   include  lt iomanip gt       Vertextet einen Ganzzahlwert val im Hexadezimalformat      Auf die Minimal-Breite width wird mit f  hrenden Nullen aufgef  llt      falls nicht angegeben  wird diese Breite aus dem Typ des Arguments     abgeleitet  Funktion geeignet von char bis long long      Zeiger  Flie  kommazahlen u     werden nicht unterst  tzt  ihre       bergabe f  hrt zu einem  beabsichtigten   Compilerfehler      Grundlagen aus  http   stackoverflow com a 5100745 2932052 template  lt typename T gt  inline std  string int to hex T val  size t width sizeof T  2        std  stringstream ss      ss  lt  lt  std  setfill  0    lt  lt  std  setw width   lt  lt  std  hex  lt  lt   val 0       return ss str           1  based on the answer by Kornel Kisielewicz   2  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       with deduction test as suggested by Lightness Races in Orbit  TEST ASSERT int to hex short 0x12       0012

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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  e g  provide overloads of this function  one for uint8 t and one for int8 t  or use something besides  if constexpr   maybe enable if     template lt  typename T  gt  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 lt T gt   value   Template argument  T  must be a fundamental integer type  e g  int  short  etc             std  stringstream stream      stream  lt  lt   0x   lt  lt  std  setfill   0    lt  lt  std  setw sizeof T  2   lt  lt  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 lt std  uint8 t  T gt                            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  lt  lt  static cast lt int gt  i                     else if  std  is same v lt std  int8 t  T gt                    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  lt  lt  static cast lt int gt  static cast lt uint8 t gt  i              else                  No cast needed for ints wider than 8 bits          stream  lt  lt  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  e g  float  or that would result in messy compiler errors  e g  user-defined type    template lt  typename T  gt  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 lt T gt   value   Template argument  T  must be a fundamental integer type  e g  int  short  etc           std  stringstream stream    stream  lt  lt   0x             lt  lt  std  setfill   0    lt  lt  std  setw sizeof T  2             lt  lt  std  hex  lt  lt  i               Optional  replace above line with this to handle 8-bit integers               lt  lt  std  hex  lt  lt  std  to string i      return stream str        I ve edited this to add a call to std  to string because 8-bit integer types  e g  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

[c++] Integer to hex string in C++

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