C# - Is it possible to make divide-by-zeros return a number, instead of throwing an exception?

I have an expr开发者_JAVA技巧ession that includes divisions, for which some of the denominators are sometimes zero. However, in those cases I would like that division to result in 1, instead of throwing an exception. Is there any straightforward way about doing this, or am I forced to do some if statements and changing the inputs to the expression to get this desired effect?

---

Although I must question the motives here, if you need to do it, make a function.

double SafeDiv(double num, double denom) {
   if(denom == 0) {
      return 1;
   }
   return num/denom;
}

---

You could write a function for dividing. Here's an extension sample idea:

public static float DivideBy(this float toBeDivided, float divideBy){
    try
    {
    return toBeDivided / divideBy;
    }
    catch(DivideByZeroException)
    {
    return 1.0;
    }
}

---

Since it is not possible to redefine the division operator for built in types, you need to implement your version of division in a function.

---

Something like this?

public T divide<T>(T dividend, T divisor) {
    return ((divisor == 0) ? (1) : (dividend / divisor));
}

or perhaps...

public T divide<T>(T dividend, T divisor) {
    try {
        return dividend / divisor;
    }
    catch (DivideByZeroException) {
        return 1;
    }
}

Personally, if you know that the divisor may be 0 in some cases, I wouldn't consider that case "exceptional" and thus use the first approach (which can also, quite conveniently, be inlined manually if you are so inclined).

That said, I agree with what Chris Marasti-Georg wrote, and question the motives for doing this.

---

No. If I understand your question, you would like to change the behavior of the divide operation and have it return 1 instead of throw. There is no way to do that.

---

Check for the denominator first if not equal to zero perform the operation

public double ReturnValueFunction(int denominator, int numerator)
{
    try
    {
       if(denominator!=0)
       {
             return  numerator/ denominator; 

         /* or your code */
        }

  else
   {
            /*your code*/
        return someDecimalnumber;
   }
}
    catch(Exception ex)
    {

    }
}

---

If either divider or divisor are double and divisor is 0 (or 0.0) result is positive or negative infinity. It will pass through all subsequent operations without throwing, and instead returning ±Infinity or NaN, depending on the other operand. Later you can check if final value is something meaningful.

If it is for some reason necessary to return 1 when dividing by zero, then you have several options:

• Custom generic function
• Custom type with overloaded operators
• Check if divisor is 0 before the division

This is not something that is usually done, as far as I know, so you might rethink why would you do that..

---

Maybe you could try this

    double res = a / (b == 0 ? a : b);