Floating point constants in templated code

Question

I have a template function operating on a floating point argument. The function is templated so that a caller can use either float, double or any other floating point data type.

At one point in my code, I compare a value with zero (or any other floating-point constant). Should I use 0.0 or 0.0f for the comparison?

template<T> void f(T a){
  //  should I use 0.0 or 0.0f in the following line?
  if(a == 0.0){
  }
}

While this is not causing any problems at the moment, I'd like to know what the usual practice is.

Alexandre C. · Accepted Answer · 2011-05-28 14:45:16Z

13

I'd suggest

if (a == T(0)) ...

answered May 28, 2011 at 14:45

Alexandre C.

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1 Comment

Lightness Races in Orbit Over a year ago

+1 for the solution itself, but I wish the answer had the explanation of the others!

Ben Voigt · Accepted Answer · 2011-05-28 14:47:55Z

7

I would suggest simply 0. According to the promotion rules for numeric types, 0 will get promoted to the type of the floating-point operand a. Promotion of a constant is a compile-time transformation, it won't slow your program down at all.

On the other hand, using 0.0 will force a run-time conversion of the other operand to double, which probably is a non-issue, as the operand is most likely passed in an FPU register anyway. 0.0f will not cause conversion of floating-point operands, but if the template was ever used with an integral type, you'd get run-time conversion to float.

edited May 28, 2011 at 14:47

answered May 28, 2011 at 14:42

Ben Voigt

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7 Comments

Agnel Kurian Over a year ago

Thanks. What about other constant values?

Ben Voigt Over a year ago

@Agnel: Non-integral constant values? Don't use floating-point equality tests then.

Agnel Kurian Over a year ago

You are right. I would use an epsilon. But even then the question remains: do I use a float literal or a double literal?

Ben Voigt Over a year ago

@Agnel: A double, unless the literal can be exactly expressed as a float. You can still force compile-time conversion to T using static_cast (or the constructor notation shown by @Alexandre).

embert Over a year ago

According to the promotion rules for numeric types, 0 will get promoted to the type of the floating-point operand a. How does the compiler handle cases involving a cast like in (float)a * 100. As casts have precedence, what type does it promote for 100 here?

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Simone · Accepted Answer · 2011-05-28 14:54:44Z

1

You should not compare for equality on floating point number with a simple

if (value == 0.0) // or 0.0f, doesn't matter

because most of the time it won't yield the result you are expecting. You should check if value is enough close to the number you are expecting. That is:

if (abs(value - 0.0) < epsilon)

where epsilon is something little enough for you application domain.

answered May 28, 2011 at 14:54

Simone

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