C++ stringstream >> int 返回零

Question

我正在编写一个函数来返回数字的反转，即将 int(1234) 转换为 int(4321)。这是我目前拥有的:

#include <iostream>
#include <cstdlib>
#include <string>
#include <sstream>

using namespace std;


int reverse(int num) {
  stringstream ss (stringstream::in | stringstream::out);
  string initial; 
  int reversed;

  // read the number in to a string stream
  ss << num;
  initial = ss.str();

  // flush the stringstream
  ss.str("");
  for(unsigned int i(0); i <= initial.size(); i++) {
    ss << initial[initial.size() - i];
  }
  ss >> reversed;

  return reversed;
}


int main(int argc, const char *argv[])
{
  int test = 9871;
  cout << "test = " << test << endl;
  cout << "reverse = " << reverse(test) << endl;

  return 0;
}

然而这只是输出:

test = 9871
reverse = 0

我很确定问题出在 ss >> reversed 行中 reversed 被设置为 0 > 而不是 ss 的值，但我无法弄清楚这段代码有什么问题，而且它看起来应该很简单，这令人气愤。谁能帮忙？

谢谢

Answer 1

i从0开始，然后是initial.size() - i超出字符串边界。

改为ss << initial[initial.size() - i - 1];并迭代 i从 1 到 initiali.size() - 1

for(unsigned i = 0; i != initial.size(); ++i) {
  ss << initial[initial.size() - i -1];
}

或者迭代i从 1 到 initial.size()

for(unsigned i = 1; i <= initial.size(); ++i) {
  ss << initial[initial.size() - i];
}