stringstream string to int [英] stringstream string to int

问题描述

下面的C ++代码 int 到 string ， int 到 string 和 string 到 int 然后，它再次重复这些步骤。 stringstream 到 int 行 stream1> i3; 是打破代码。我在这里缺少什么？

The C++ code below does int to string, int to string and a string to int conversions. Then, it repeats these steps again. The stringstream to int line stream1 >> i3; is breaking the code. What am I missing here?

#include <iostream>
#include <string>
#include <sstream>
using namespace std;

int main()
{
 int i1 = 101;
 int i2 = 202;
 int i3 = 0;
 string s1 = "";
 string s2 = "";
 stringstream stream1;

 for (int i=0;i<2;i++)
 {

   //int to string
   stream1.str("");
   cout << "i1 " << i1 << endl;
   stream1 << i1;
   s1 = stream1.str();
   cout << "s1 " << s1 << endl;

   //int to string
   cout << "i2 " << i2 << endl;
   stream1.str("");
   stream1 << i2;
   s2 = stream1.str();
   cout << "s2 " << s2 << endl;

   //string to int
   stream1.str("");
   stream1.str(s2);
   stream1 >> i3;
   //line above causes s1 and s2 to get messed up during 2nd time in for loop
   cout << "i3-- " << i3 << endl;

  }
  return 0;
 }

推荐答案

再现问题。我解决了在.str（）之前插入一个.clear（）

I tested your code and could reproduce the problem. I solved it inserting a .clear() before .str("")

看看这里：如何重用ostringstream？

#include <iostream>
#include <string>
#include <sstream>
using namespace std;

int main()
{
 int i1 = 101;
 int i2 = 202;
 int i3 = 0;
 string s1 = "";
 string s2 = "";
 stringstream stream1;

 for (int i=0;i<2;i++)
 {

   //int to string
   stream1.clear();
   stream1.str("");
   cout << "i1 " << i1 << endl;
   stream1 << i1;
   s1 = stream1.str();
   cout << "s1 " << s1 << endl;

   //int to string
   cout << "i2 " << i2 << endl;
   stream1.clear();
   stream1.str("");
   stream1 << i2;
   s2 = stream1.str();
   cout << "s2 " << s2 << endl;

   //string to int
   stream1.clear();
   stream1.str("");
   stream1.str(s2);
   stream1 >> i3;
   //line above causes s1 and s2 to get messed up during 2nd time in for loop
   cout << "i3-- " << i3 << endl;

  }
  return 0;
 }

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