Why is the division by zero not caught in a generic catch()

Question

I have the following code which should throw a division by zero exception while running. It seems that the exception is not be caught at all. I am using catch(...) to catch any exception from the try block. Is catch(...) a good approach to catch any exceptions from the try block??

try
{
    printf("Try Block");
    for (int i = 0; i < 10; i++)
    {
        printf("%d\n", 1 / i);
    }       
}

catch (...)
{
    printf("Division by zero");
}

No, it should not necessarily throw. If it does, that's due to the undefined behaviour, and catching it and continuing on will likely bring only pain. — chris, Mar 25 '15 at 14:02
This was just an example to catch any exceptions within the try block. — Ajit, Mar 25 '15 at 14:08

TartanLlama · Answer 1 · 2015-03-25T14:09:15.563

2

You can't rely on the behaviour of division by zero; it's undefined:

N4296 §5/4:

If during the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined.

edited Mar 25 '15 at 14:09

answered Mar 25 '15 at 14:04

TartanLlama

63,752
13
157
193

score 0 · Answer 2 · answered Mar 25 '15 at 14:07

If you need to make sure an exception is thrown when you try to divide by zero, you'll have to make your own wrapper that does it:

long double Divide(long double numerator, long double denominator)
{
  if (denominator == 0)
  {
    throw std::range_error("Division by zero!");
  }
  return numerator / denominator;
}

score 0 · Answer 3 · answered Mar 25 '15 at 14:07

If you want portable code that handles divide by zero by throwing an exception, then you could write a class that wraps a suitable numeric type and checks before divide and mod operations, and use it instead of the inbuilt numeric types.... The C++ library does not provide such a type. In reality, it's not particularly useful - you can just check in your code in the specific places that might do a division or mod operation with 0.

Why is the division by zero not caught in a generic catch()

3 Answers3