complex_double_get function

[ShikharJ]

Relevant: Issue #1088 Feature 3.
For the course of development refer: PR #1122.

[ShikharJ]

@isuruf A review?

[isuruf]

Can you fix the commit message? Otherwise +1 to merge

[ShikharJ]

Done.

[certik]

Sorry for coming to this PR so late. I think we should rather just use the native C complex type. Here is an example how to use it:

#include <stdio.h>
#include <complex.h>
#include <tgmath.h>
 
int main(void)
{
    double complex z1 = I * I;     // imaginary unit squared
    printf("I * I = %.1f%+.1fi\n", creal(z1), cimag(z1));
 
    double complex z2 = pow(I, 2); // imaginary unit squared
    printf("pow(I, 2) = %.1f%+.1fi\n", creal(z2), cimag(z2));
 
    double PI = acos(-1);
    double complex z3 = exp(I * PI); // Euler's formula
    printf("exp(I*PI) = %.1f%+.1fi\n", creal(z3), cimag(z3));
 
    double complex z4 = 1+2*I, z5 = 1-2*I; // conjugates
    printf("(1+2i)*(1-2i) = %.1f%+.1fi\n", creal(z4*z5), cimag(z4*z5));
}

[certik]

Instead of defining our own complex type, why cannot we use the C complex type?

http://en.cppreference.com/w/c/numeric/complex

[ShikharJ]

@certik This is exactly what was planned on implementing initially. Using double complex. The problem that arised was that using I from C99 gave an error like this
error: unable to find numeric literal operator ‘operator""iF’
d = i.imag()*_Complex_I + i.real();
( _Complex_I is a macro for I)
This is a well known error, and can still be resolved by using -fext-numeric-literals flag. But again, this flag only worked locally (as my system uses gcc 5.4.0, and Travis uses previous versions). The builds failed, and as per isuruf's suggestion, it would be easier to return a structure instead of using double complex. Hence this implementation.

[certik]

@ShikharJ would you mind creating a new issue for this? Let's discuss it over there, there must be some solution to that.

[certik]

I created an issue for it here: #1131