Question

What’s the fastest way to convert a string represented by (const char*, size_t) to an int?

The string is not null-terminated.
Both these ways involve a string copy (and more) which I’d like to avoid.

And yes, this function is called a few million times a second. :p

int to_int0(const char* c, size_t sz)
{
    return atoi(std::string(c, sz).c_str());
}

int to_int1(const char* c, size_t sz)
{
    return boost::lexical_cast<int>(std::string(c, sz));
}

Answer 1

Fastest:

int to_int(char const *s, size_t count)
{
     int result = 0;
     size_t i = 0 ;
     if ( s[0] == '+' || s[0] == '-' ) 
          ++i;
     while(i < count)
     {
          if ( s[i] >= '0' && s[i] <= '9' )
          {
              //see Jerry's comments for explanation why I do this
              int value = (s[0] == '-') ? ('0' - s[i] ) : (s[i]-'0');
              result = result * 10 + value;
          }
          else
              throw std::invalid_argument("invalid input string");
          i++;
     }
     return result;
} 

Since in the above code, the comparison (s[0] == '-') is done in every iteration, we can avoid this by calculating result as negative number in the loop, and then return result if s[0] is indeed '-', otherwise return -result (which makes it a positive number, as it should be):

int to_int(char const *s, size_t count)
{
     size_t i = 0 ;
     if ( s[0] == '+' || s[0] == '-' ) 
          ++i;
     int result = 0;
     while(i < count)
     {
          if ( s[i] >= '0' && s[i] <= '9' )
          {
              result = result * 10  - (s[i] - '0');  //assume negative number
          }
          else
              throw std::invalid_argument("invalid input string");
          i++;
     }
     return s[0] == '-' ? result : -result; //-result is positive!
} 

That is an improvement!

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In C++11, you could however use any function from std::stoi family. There is also std::to_string family.