C IEEE-Floats inf equal inf - c

Question

In C, on a implementation with IEEE-754 floats, when I compare two floating point numbers which are NaN, it return 0 or "false". But why do two floating point numbers which both are inf count as equal?
This Program prints "equal: ..." (at least under Linux AMD64 with gcc) and in my opinion it should print "different: ...".
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
volatile double a = 1e200; //use volatile to suppress compiler warnings
volatile double b = 3e200;
volatile double c = 1e200;
double resA = a * c; //resA and resB should by inf
double resB = b * c;
if (resA == resB)
{
printf("equal: %e * %e = %e = %e = %e * %e\n",a,c,resA,resB,b,c);
}
else
{
printf("different: %e * %e = %e != %e = %e * %e\n", a, c, resA, resB, b, c);
}
return EXIT_SUCCESS;
}
A other example, why I think inf is not the same as inf, is: the numbers of natural numbers and rational numbers, both are infinite but not the same.
So why is inf == inf?

Answer 1

Infinities compare equal because that's what the standard says. From section 5.11 Details of comparison predicates:
Infinite operands of the same sign shall compare equal.

Answer 2

inf==inf for the same reason that almost all floating point numbers compare equal to themselves: Because they're equal. They contain the same sign, exponent, and mantissa.
You might be thinking of how NaN != NaN. But that's a relatively unimportant consequence of a much more important invariant: NaN != x for any x. As the name implies, NaN is not any number at all, and hence cannot compare equal to anything, because the comparison in question is a numeric one (hence why -0 == +0).
It would certainly make some amount of sense to have inf compare unequal to other infs, since in a mathematical context they're almost certainly unequal. But keep in mind that floating point equality is not the same thing as absolute mathematical equality; 0.1f * 10.0f != 1.0f, and 1e100f + 1.0f == 1e100f. Just as floating point numbers gradually underflow into denormals without compromising as-good-as-possible equality, so they overflow into infinity without compromising as-good-as-possible equality.
If you want inf != inf, you can emulate it: 1e400 == 3e400 evaluates to true, but 1e400 - 3e400 == 0 evaluates to false, because the result of +inf + -inf is NaN. (Arguably you could say it should evaluate to 0, but that would serve nobody's interest.)

Answer 3

Background
In C, according to the IEEE 754 binary floating point standard (so, if you use a float or a double) you're going to get an exact value that can be compared exactly with another variable of the same type. Well, this is true unless your computations result in a value that lies outside the range of integers that can be represented (i.e., overflow).
Why is Infinity == Infinity
resA and resB
The IEEE-754 standard tailored the values of infinity and negative infinity to be greater than or less than, respectively, all other values that may be represented according to the standard (<= INFINITY == 0 11111111111 0000000000000000000000000000000000000000000000000000 and >= -INFINITY == 1 11111111111 0000000000000000000000000000000000000000000000000000), except for NaN, which is neither less than, equal to, or greater than any floating point value (even itself). Take note that infinity and it's negative have explicit definitions in their sign, exponent, and mantissa bits.
So, resA and resB are infinity and since infinity is explicitly defined and reproducible, resA==resB. I'm fairly certain this is how isinf() is implemented.
Why is NaN != NaN
However, NaN is not explicitly defined. A NaN value has a sign bit of 0, exponent bits of all 1s (just like infinity and it's negative), and any set of non-zero fraction bits (Source). So, how would you tell one NaN from another, if their fraction bits are arbitrary anyways? Well, the standard doesn't assume that and simply returns false when two floating point values of this structure are compared to one another.
More Explanation
Because infinity is an explicitly defined value (Source, GNU C Manual):
Infinities propagate through calculations as one would expect
2 + ∞ = ∞
4 ÷ ∞ = 0
arctan (∞) = π/2.
However, NaN may or may not propagate through propagate through computations. When it does, it is a QNan (Quieting NaN, most significant fraction bit set) and all computations will result in NaN. When it doesn't, it is a SNan (Signalling NaN, most significant fraction bit not set) and all computations will result in an error.

Answer 4

There are many arithmetic systems. Some of them, including the ones normally covered in high school mathematics, such as the real numbers, do not have infinity as a number. Others have a single infinity, for example the projectively extended real line. Others, such as the IEEE floating point arithmetic under discussion, and the extended real line, have both positive and negative infinity.
IEEE754 arithmetic is different from real number arithmetic in many ways, but is a useful approximation for many purposes.
There is logic to the different treatment of NaNs and infinities. It is entirely reasonable to say that positive infinity is greater than negative infinity and any finite number. It would not be reasonable to say anything similar about the square root of -1.