Division by complex in clang++ versus g++

Question

When I compile the following code with g++ (4.8.1 or 4.9.0) or clang++ (3.4) I get different outputs.

#include 
#include 

int mai         


        

Answer 1

Ok so I have this wild guess, that in clang the complex number division is implemented like described on wiki: http://en.wikipedia.org/wiki/Complex_number#Multiplication_and_division.

One can see that the denominator is in the form c^2 + d^2. So 1.e-162 squared actually falls out of the IEE754 representable range for double which is std::numeric_limits::min() - 2.22507e-308, and we have an underflow.

gcc somehow works this out, but if clang does simple square, as per @40two's standard quotation it enters into UB, and treats it as 0 after performing 1.e-162^2 + 0.0^2.

I tested clang and gcc for a number that should not result with underflow when squared.

#include 
#include 

int main() {
  std::complex c = {1.e-104,0};
  std::cout << 1.0/c << std::endl;
  return 0;
}

Results are fine:

luk32:~/projects/tests$ g++ --std=c++11 ./complex_div.cpp 
luk32:~/projects/tests$ ./a.out 
(1e+104,0)
luk32:~/projects/tests$ clang++ --std=c++11 ./complex_div.cpp 
luk32:~/projects/tests$ ./a.out 
(1e+104,0)

Not sure if this is a bug. But I think this is what is going on.

Addendum:

(inf,-nan) is also consistent if one evaluates those expressions by hand

We get:

real = (ac+bd) / (o)  - real part 
imag = (bc-ad) / (o)  - imaginary part

{a,b} = {1.0, 0.0}
{c,d} = {1.e-104, 0.0}

o is (c^2 + d^2) and we assume it is 0.0.

real / o = (1.e-104 * 1.0 + 0.0 * 0.0) / o = 1/o = inf
imag / o = (0.0 * 1.e-104 - 1.0 * 0.0) / o = -0.0 / o = -nan

I am just not absolutetly sure about the sign of -0.0 and -nan, I don't know IEE754 enough to evaluate (0.0 * 1.e-104 - 1.0 * 0.0). But everything seems consistent.