CODEWITHSUNDEEP
c++stlstdmultiplicationinfinity
Alex Ni
Alex Ni

Reputation: 75

Regarding arithemetic operations with std::numeric_limits<T>::infinity()

I have a peculiar usecase where I have some edges w/ double weights set initially to std::numeric_limits<double>::infinity(). These weights will be set to something else later on in program execution.

Now that we have the context, here is the main issue. I need to compare these edge weights to the square of some weight that I compute, and to account for the squaring of my computed weight, I will also have to square the edge weights. Of course, this will result in an initial multiplying of infinity by infinity. I am wondering if multiplying a double set to std::numeric_limits<double>::infinity() by itself is defined behavior. Can I expect it to remain unchanged?

I could not find any documentation even on cppreference.

Upvotes: 4

Views: 311

Answers (1)

Bathsheba
Bathsheba

Reputation: 234795

Restricting this answer to IEEE754, using +/-Inf as some sort of starting value brings a fair bit of trouble.

Under IEEE754,

  1. Inf * 0.0 = NaN
  2. -Inf * 0.0 = Inf * -0.0 = -NaN
  3. Inf * Inf = -Inf * -Inf = Inf
  4. -Inf * Inf = -Inf

Inf multiplied by any positive floating point value (including a subnormal value) is Inf, and similarly for -Inf.

In other words, you need to treat +0.0 and -0.0 as special cases when multiplying, and is one of those rare cases where a signed negative zero produces a different result. Use std::isnan to test, if you cannot adopt some other scheme.

Upvotes: 5

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