c ++从字符串中解析int

c++ parse int from string(c ++从字符串中解析int)

问题描述

可能重复:
如何在 C++ 中将字符串解析为 int?

我做了一些研究，有些人说要使用 atio，有些人说它不好，反正我无法让它工作.

I have done some research and some people say to use atio and others say it's bad, and I can't get it to work anyways.

所以我只想问清楚，将字符串转换为 int 的正确方法是什么.

So I just want to ask flat out, whats the right way to convert a string to a int.

string s = "10";
int i = s....?

谢谢！

推荐答案

• 在 C++11 中，使用 std::stoi 为:

   std::string s = "10";
   int i = std::stoi(s);
  

  请注意，如果无法执行转换，std::stoi 将抛出 std::invalid_argument 类型的异常，或者 std::out_of_range> 如果转换导致溢出(即字符串值对于 int 类型来说太大).您可以使用 std::stol 或std:stoll 虽然以防万一 int 对于输入字符串来说似乎太小了.

  Note that std::stoi will throw exception of type std::invalid_argument if the conversion cannot be performed, or std::out_of_range if the conversion results in overflow(i.e when the string value is too big for int type). You can use std::stol or std:stoll though in case int seems too small for the input string.

  在 C++03/98 中，可以使用以下任何一种:

  In C++03/98, any of the following can be used:

   std::string s = "10";
   int i;
  
   //approach one
   std::istringstream(s) >> i; //i is 10 after this
  
   //approach two
   sscanf(s.c_str(), "%d", &i); //i is 10 after this
  

请注意，对于输入 s = 10jh"，上述两种方法都会失败.他们将返回 10 而不是通知错误.因此，安全可靠的方法是编写自己的函数来解析输入字符串，并验证每个字符以检查它是否为数字，然后相应地工作.这是一个强大的实现(虽然未经测试):

Note that the above two approaches would fail for input s = "10jh". They will return 10 instead of notifying error. So the safe and robust approach is to write your own function that parses the input string, and verify each character to check if it is digit or not, and then work accordingly. Here is one robust implemtation (untested though):

int to_int(char const *s)
{
     if ( s == NULL || *s == '' )
        throw std::invalid_argument("null or empty string argument");

     bool negate = (s[0] == '-');
     if ( *s == '+' || *s == '-' ) 
         ++s;

     if ( *s == '')
        throw std::invalid_argument("sign character only.");

     int result = 0;
     while(*s)
     {
          if ( *s < '0' || *s > '9' )
            throw std::invalid_argument("invalid input string");
          result = result * 10  - (*s - '0');  //assume negative number
          ++s;
     }
     return negate ? result : -result; //-result is positive!
} 

此解决方案是我的另一个解决方案的略微修改版本.

This solution is slightly modified version of my another solution.

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