String（const char *，size_t）to int？ [英] String (const char*, size_t) to int?

问题描述

将（const char *，size_t）表示的字符串转换为int的最快的方法是什么？

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

字符串不是 null终止。
这两种方式都涉及到我想避免的字符串副本（和更多）。

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

是的，这个函数被称为几百万次第二。 ：p

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));
}

推荐答案

最快：

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;
} 

因为在上面的代码中，比较 [0] ==' - '）是在每次迭代中完成的，我们可以通过在循环中计算 result 然后返回 result 如果 s [0] 确实' - '，否则返回 -result （这是一个正数，因为它应该是）：

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!
} 

这是一个改进！

在C ++ 11中，你可以使用 std :: stoi 家庭。还有 std :: to_string 家庭。

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

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