try catch unexpected result?

[adam2016]

hi guys so when I enter a valid number that is not 0 the try and catch block still gets called and it prints cannot divide by zero even when x and y are not equal to zero,what is happening? and how do i fix it

 #include <iostream>

using namespace std;


int main()
{
   try{
   int x;
   int y;
   int result;

   cin >> x;
   cin >> y;

    result = x / y;
   if(x || y == 0){

    throw 99;

   }
   }catch(int x){

       cout << "cannot divide by 0 :" << x;

}

}

[Thomas1965]

if(x || y == 0)
{
    throw 99;
}

This is true if x != 0. In C++ false == 0 , != 0 means true.

[adam2016]

not too sure what you mean Tom,But I thought x !=0 and y !=0 therefore the catch block should not be called :s

[Moschops]

if(x || y == 0)
Perhaps you meant
if (x ==0 || y == 0)

Shouldn't you be checking for divide-by-zero before you try to actually do the division? Not much point checking afterwards.

[E0p9LyTq]

All you need to check for is the divisor is not zero. Zero divided by any non-zero number is zero.

#include <iostream>

int main()
{
   int x;
   int y;

   std::cout << "x: ";
   std::cin >> x;
   std::cout << "y: ";
   std::cin >> y;

   try
   {
      if (y == 0)
      {
         throw "error";
      }
      int result = x / y;
      std::cout << "result: " << result << "\n";
   }

   catch(...)
   {
      std::cout << "\ncannot divide by 0:\n";
   }
}

[adam2016]

thanks guys just a quick question on the same note how come I cannot put result = x/y; after the try catch block? it tells me that it's not declared in this scope

#include <iostream>

using namespace std;


int main()
{
   try{
   int x;
   int y;
   int result;

   cin >> x;
   cin >> y;

   if(x == 0 || y == 0){

    throw 99;

   }

   
   }catch(int x){

       cout << "cannot divide by 0 :" << x;

}

result = x /y; // error
}

[E0p9LyTq]

result is being declared inside your try block, you are accessing it outside the try block.

[Moschops]

I see that in this code, even if you detect that y is zero and that you are thus going to try to divide by zero, you still actually try to do a divide by zero.

If you detect that there is going to be a divide-by-zero, you should then not do the divide-by-zero.

[adam2016]

Thanks guys that makes sense much appreciated