C Converting Hexadecimal to Decimal

Question

I m looking for a way to convert hex hexadecimal  to dec decimal  easily  I found an easy way to do this like    int k   0x265  cout  lt  lt  k  lt  lt  endl    But with that I can t input 265  Is there anyway for it to work like that   Input   265  Output   613  Is there anyway to do that    Note  I ve tried    int k   0x  b  cin  gt  gt  b  cout  lt  lt  k   b  lt  lt  endl    and it doesn t work

User · Answer

Here is a solution using strings and converting it to decimal with ASCII tables    include  lt iostream gt   include  lt string gt   include  math h  using namespace std  unsigned long hex2dec string hex        unsigned long result   0      for  int i 0  i lt hex length    i              if  hex i  gt  48  amp  amp  hex i  lt  57                        result     hex i -48  pow 16 hex length  -i-1             else if  hex i  gt  65  amp  amp  hex i  lt  70                result     hex i -55  pow 16 hex length   -i-1             else if  hex i  gt  97  amp  amp  hex i  lt  102                result     hex i -87  pow 16 hex length  -i-1                       return result     int main int argc  const char   argv          string hex str      cin  gt  gt  hex str      cout  lt  lt  hex2dec hex str   lt  lt  endl      return 0

User · Answer

Use std  hex manipulator    include  lt iostream gt   include  lt iomanip gt   int main         int x      std  cin  gt  gt  std  hex  gt  gt  x      std  cout  lt  lt  x  lt  lt  std  endl       return 0

User · Answer

This should work as well   include  lt ctype h gt   include  lt string h gt   template lt typename T   unsigned int gt  T Hex2Int const char  const Hexstr  bool  Overflow        if   Hexstr          return false      if  Overflow           Overflow   false       auto between      char val  char c1  char c2    return val  gt   c1  amp  amp  val  lt   c2         size t len   strlen Hexstr       T result   0       for  size t i   0  offset   sizeof T   lt  lt  3  i  lt  len  amp  amp   int offset  gt  0  i                  if  between Hexstr i    0    9                result   result  lt  lt  4   Hexstr i  -  0           else if  between tolower Hexstr i     a    f                result   result  lt  lt  4   tolower Hexstr i   -   a  - 10      Remove the decimal part          offset -  4            if    len     len   2     0    lt  lt  2   gt   sizeof T   lt  lt  3   amp  amp  Overflow           Overflow   true      return result     The  Overflow  parameter is optional  so you can leave it NULL  Example  auto result   Hex2Int  quot C0ffee quot   NULL   auto result2   Hex2Int lt long gt   quot DeadC0ffe quot   NULL

User · Answer

Well  the C way might be something like       include  lt stdlib h gt   include  lt stdio h gt   int main             int n          scanf   d    amp n           printf   X   n            exit 0

User · Answer

only use  cout  lt  lt  dec  lt  lt  0x

User · Answer

include  lt iostream gt   include  lt iomanip gt   include  lt sstream gt   int main         int x  y      std  stringstream stream       std  cin  gt  gt  x      stream  lt  lt  x      stream  gt  gt  std  hex  gt  gt  y      std  cout  lt  lt  y       return 0

User · Answer

std  cout  lt  lt   Enter decimal number          std  cin  gt  gt  input        std  cout  lt  lt   0x   lt  lt  std  hex  lt  lt  input  lt  lt    n      if your adding a input that can be a boolean or float or int it will be passed back in the int main function call     With function templates  based on argument types  C generates separate functions to handle each type of call appropriately   All function template definitions begin with the keyword template followed by arguments enclosed in angle brackets  lt  and     A single formal parameter T is used for the type of data to be tested     Consider the following program where the user is asked to enter an integer and then a float  each uses the square function to determine the square  With function templates  based on argument types  C generates separate functions to handle each type of call appropriately   All function template definitions begin with the keyword template followed by arguments enclosed in angle brackets  lt  and     A single formal parameter T is used for the type of data to be tested     Consider the following program where the user is asked to enter an integer and then a float  each uses the square function to determine the square    include  lt iostream gt   using namespace std  template  lt class T gt          function template T square T         returns a value of type T and accepts                  type T      int or float or whatever       void main     int x  y  float w  z  cout  lt  lt   Enter a integer      cin  gt  gt  x  y   square x   cout  lt  lt   The square of that number is      lt  lt  y  lt  lt  endl  cout  lt  lt   Enter a float      cin  gt  gt  w  z   square w   cout  lt  lt   The square of that number is      lt  lt  z  lt  lt  endl     template  lt class T gt          function template T square T u    accepts a parameter u of type T  int or float    return u   u     Here is the output   Enter a integer   5 The square of that number is   25 Enter a float   5 3 The square of that number is   28 09

User · Answer

I use this   template  lt typename T gt  bool fromHex const std  string amp  hexValue  T amp  result        std  stringstream ss      ss  lt  lt  std  hex  lt  lt  hexValue      ss  gt  gt  result       return  ss fail

[c++] C++: Converting Hexadecimal to Decimal

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