意外的结果将十进制数转换为二进制

Question

# include <iostream>
# include <cmath>
using namespace std;
int main()
{

    int p;
    int n;
    int q;
    cin>>n;
    int r;
    r=0;
    for (int i=0,n; n>1; i=i+1,n=n/2)
    {

    p=n%2;
    q= p*(pow(10,i));
    r=r + q;
}

cout<<r;
system("pause");
return 0;
}

I am not supposed to use arrays. It compiles fine but when executed and a number is entered, it doesn't produce the desired results. For instance, when 22 is entered, it gives -2147483648 whereas the desired output would be 10110.

Answer 1

your way is limited and not effient in converting to binary

you should use string it's more helpful and the range is big enough for any number

this is my code for decimal-to-binary

#include<iostream>
#include<string>
#include<stack>

using namespace std;

int main()
{
   long long n;
   string s,bin;
   stack<string> res;

   cin>>n;
   while(n>0)
   {
     if(n%2==0)
         s='0';
     else
         s='1';
     res.push(s);
     n=n/2;
   }
   while(!res.empty())
   {
     bin=bin+res.top();
     res.pop();
   }
   cout<<bin<<endl;

return 0;
}

I hope it will help you.

Answer 2

int i=0,n;

should be

int i=0;

I don't know what you thought you were doing there, but what you are actually doing is declaring another variable n . Because the second n variable doesn't have a value the rest of the code doesn't work.

That's not the only problem with your code, but I'm sure you can figure out the rest.