无法在c ++中显示反向数字

Question

我是编程的新手，正在寻找一些指导。我无法打印反向号码。该算法正在按预期工作，以反转用户的输入但无法正确显示：

#include <iostream>
#include <stdio.h>
using namespace std;

int main()
{

 int rem, OriValue, InValue2 = 0, InValue = 0;//rem = remainder,InValue = User input
 bool neg = false; // Boolean Variable to remember input value

 //Request customer to enter a int value
 cout << "Enter any decimal integer value >= 0:";
 cin >> InValue;
 OriValue = InValue;
 if(InValue < 0)
  {
   neg = true;
   InValue = -InValue;
   cout << "Input value is negative :" << InValue <<"\n";
   }
   else if (InValue > 0 )

   cout << "Input value is positive:"<< InValue <<"\n";
   do
   {
   rem = (InValue % 10);
   cout << "Remainder value:"<< rem << "\n";
   InValue = InValue / 10;

   }
   while (InValue != 0);
   cout << OriValue << " in reverse is " << InValue << "\n";
   // Here is an example of the output:
   // -123 in reverse 321
   return 0;
}

Answer 1

这应该做的工作。检查代码中的注释

#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{

 int rem, OriValue, InValue2 = 0, InValue = 0;//rem = remainder,InValue = User input
 int reverseVal = 0; // Add this variable to store the reverse number
 bool neg = false; // Boolean Variable to remember input value

 //Request customer to enter a int value
 cout << "Enter any decimal integer value >= 0:";
 cin >> InValue;
 OriValue = InValue;
 if(InValue < 0)
  {
   neg = true;
   InValue = -InValue;
   cout << "Input value is negative :" << InValue <<"\n";
   }
   else if (InValue > 0 )

   cout << "Input value is positive:"<< InValue <<"\n";
   do
   {
           rem = (InValue % 10);
           cout << "Remainder value:"<< rem << "\n";
           InValue = InValue / 10;
           // Add this to store
           reverseVal = reverseVal*10 + rem;

   }
   while (InValue != 0);
   cout << OriValue << " in reverse is " << reverseVal << "\n";
   // Here is an example of the output:
   // -123 in reverse 321
   return 0;
}

Answer 2

你没有将反转变量存储在任何地方，而n == 0就是你的循环终止时。当您在循环结束后打印n时，它必须为0，如您所见。解决方案是在分割n时存储余数。以下是实现这一目标的几种无限方法。

使用stringstream：

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

int main() {
    int original_value, n;
    stringstream ss;

    cout << "Enter an integer >= 0:";
    cin >> original_value;
    n = original_value;

    if (n < 0) {
        n = -n;
        ss << '-';
    }

    while (n > 0) {
        ss << n % 10;
        n /= 10;
    }

    cout << original_value << " in reverse is " << ss.str() << "\n";
    return 0;
}

使用反向：

#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    int n = -1234;
    string rev = to_string(n);
    reverse(rev.begin() + (rev[0] == '-' ? 1 : 0), rev.end());

    cout << rev << "\n";
    return 0;
}

样品运行：

Enter an integer >= 0: -1234
-1234 in reverse is -4321

如果你需要一个整数，你可以使用atoi，istringstream或stoi从字符串中始终使用parse it。