How To Turn Char of '2' into Integer 2 in C++

Question

I am Kind Of New To C++. But I Want To Turn The Character 2 into An integer 2. I tried using Casting But It Gives Me The ASCII Value Of The Char. How Can I Achieve This Iam Using Linux And Codeblocks

Answer 1

Just subtract the ASCII value of '0' from '2' to get the integer 2.

char c = '2';
int n = c - '0';

This is guaranteed to work even if the encoding is not ASCII, since the language requires that the encoding of the characters increases from '0' to '9'.

Answer 2

I Want To Turn The Character 2 into An integer 2 ...

C++ std::istream provides several stream operators. You probably should learn how to use "space delimited formatted extraction".

An example:

#include <iostream>
using std::cout, std::cerr, std::endl;
//using std::hex, std::dec, std::cin, std::flush; // c++17

#include <string>
using std::string;

#include <sstream>
using std::stringstream;

#ifndef                 DTB_PCKLRT_HH
#include "../../bag/src/dtb_pclkrt.hh"  // private library
using  DTB::PClk_t;  // posix clock access
#endif

#include <cstdint>
#include <cassert>


namespace // anonymous
{
   class F893_t // Functor ctor and dtor use compiler provided defaults
   {
      PClk_t  pclk; // posix clock

   public:
      int operator()(int argc, char* argv[]) { return exec(argc, argv);  }

   private:

   int exec(int , char** )
      {
         int retVal = 0;
         uint64_t start_ns = pclk.ns();

         // create a string -- white space delimits, but extra is ignored
         string s = "1 2 3   4  55555  666666  ";

         // use s to initialize a stringstream
         stringstream ss(s); 

         do
         {
           int n;    // this object will hold the extracted value
           // NOTE: it is wrong to test eof() here, it is unknown.

           ss >> n;  // n also identifies target translation , i.e. an "int"
           //           the default translation format of "to int" is from 'decimal text'
           //           non decimal text can trigger a failure

           // NOTE: ss status is not valid until AFTER a read attempt

           if (!ss.good()) // check ss error status
           {
             if (ss.eof()) { break; } // attempt to read past the end of file is an error
             // but typically, you don't want to fail your program because of it
             // instead, simply continue after the loop

             // getting here means that ss is not good() and not eof()
             // This can be caused by an invalid char, and other errors 
             // so at least let the user know
             // Test NOTE: to test "no eof()" and "not good"
             //  change string s to include a char that is not valid for decimal
             cerr << "\n\n  ss.bad() after 'n' formatted integer read attempt"
                  << "\n  ss.str(): " << ss.str() << endl;
             assert(0);    // it is possible to continue ...
             //               but not useful for this example
           }

           // at this point, n should have a value from the stringstream
           cout << "\n  n = " << n;

         } while (true);

         cout << "\n\n  --eof()-- detected, resumed after loop" << endl;

         auto  duration_ns = pclk.ns() - start_ns;
         cout << "\n  F893_t::exec() duration   "
              << duration_ns << " ns    (" <<  __cplusplus  
              << ")" << endl;
         return retVal;
      }

   }; // class F893_t

}  // anonymous namespace


int main(int argc, char* argv[]) { return F893_t()(argc, argv); }

// "Everything should be made as simple as possible, but no simpler"
//    attributed to Albert Einstein the New York times in 1950

Typical output:

  n = 1
  n = 2
  n = 3
  n = 4
  n = 55555
  n = 666666

  --eof()-- detected, resume after loop

  F893_t::exec() duration   149825 ns    (201709)