string中的to_string()和stoi/atoi

std::to_string

string to_string (int val);
string to_string (long val);
string to_string (long long val);
string to_string (unsigned val);
string to_string (unsigned long val);
string to_string (unsigned long long val);
string to_string (float val);
string to_string (double val);
string to_string (long double val);
将数值转换为字符串
返回以val表示的字符串。 所使用的格式与printf为相应类型打印的格式相同：

// to_string example
#include <iostream>   // std::cout
#include <string>     // std::string, std::to_string

int main ()
{
  std::string pi = "pi is " + std::to_string(3.1415926);
  std::string perfect = std::to_string(1+2+4+7+14) + " is a perfect number";
  std::cout << pi << '\n';
  std::cout << perfect << '\n';
  return 0;
}

//输出
pi is 3.141593
28 is a perfect number

std::stoi

int stoi (const string& str, size_t* idx = 0, int base = 10);
int stoi (const wstring& str, size_t* idx = 0, int base = 10);

参数说明

str
String object with the representation of an integral number.
idx
Pointer to an object of type size_t, whose value is set by the function to position of the next character in str after the numerical value.
This parameter can also be a null pointer, in which case it is not used.
base
Numerical base (radix) that determines the valid characters and their interpretation.
If this is 0, the base used is determined by the format in the sequence (see strtol for details). Notice that by default this argument is 10, not 0.
用法：
stoi（字符串，起始位置，n进制），将 n 进制的字符串转化为十进制

stoi(str, 0, 2); //将字符串 str 从 0 位置开始到末尾的 2 进制转换为十进制

// stoi example
#include <iostream>   // std::cout
#include <string>     // std::string, std::stoi

int main ()
{
  std::string str_dec = "2001, A Space Odyssey";
  std::string str_hex = "40c3";
  std::string str_bin = "-10010110001";
  std::string str_auto = "0x7f";

  std::string::size_type sz;   // alias of size_t

  int i_dec = std::stoi (str_dec,&sz);
  int i_hex = std::stoi (str_hex,nullptr,16);
  int i_bin = std::stoi (str_bin,nullptr,2);
  int i_auto = std::stoi (str_auto,nullptr,0);

  std::cout << str_dec << ": " << i_dec << " and [" << str_dec.substr(sz) << "]\n";
  std::cout << str_hex << ": " << i_hex << '\n';
  std::cout << str_bin << ": " << i_bin << '\n';
  std::cout << str_auto << ": " << i_auto << '\n';

  return 0;
}

2001, A Space Odyssey: 2001 and [, A Space Odyssey]
40c3:  16579
-10010110001: -1201
0x7f: 127

atoi()的参数是 const char* ,因此对于一个字符串str我们必须调用 c_str()的方法把这个string转换成 const char类型的,而stoi()的参数是const string,不需要转化为 const char*；

stoi()会做范围检查，默认范围是在int的范围内的，如果超出范围的话则会runtime error！而atoi()不会做范围检查，如果超出范围的话，超出上界，则输出上界，超出下界，则输出下界；
代码：

//myfirst.cpp--displays a message   
 
#include "stdafx.h"  
#include <iostream>  
#include <set>  
#include <string>  
using namespace std;
int main()
{
	string s1 = "2147482", s2 = "-214748";
	string s3 = "214748666666663", s4 = "-21474836488";
	cout << stoi(s1) << endl;
	cout << stoi(s2) << endl;
	cout << atoi(s3.c_str()) << endl;
	cout << atoi(s4.c_str()) << endl;
	return 0;
}

本文地址：https://blog.csdn.net/CS654/article/details/107512981