std::complex<long double> division

the following code is not working:

std::complex<long doublex(1.,1.);
std::complex<long doubleresult(0.,0.);

result = 1./x;

std::cout << "x = " << x << std::endl;
std::cout << "r = " << result << std::endl;

with std::complex<doubleit works. anyone knows how to get rid of this?
i need long double for my calculations. thx.

Fredy Halter napsal:

Quote:

the following code is not working:
>
std::complex<long doublex(1.,1.);
std::complex<long doubleresult(0.,0.);
>
result = 1./x;
>
std::cout << "x = " << x << std::endl;
std::cout << "r = " << result << std::endl;
>
>
with std::complex<doubleit works. anyone knows how to get rid of this?
i need long double for my calculations. thx.

Write it this way:
result = 1.L / x;

1. is double, 1.L is long double

Fredy Halter wrote:

Quote:

the following code is not working:
>
std::complex<long doublex(1.,1.);
std::complex<long doubleresult(0.,0.);
>
result = 1./x;
>
std::cout << "x = " << x << std::endl;
std::cout << "r = " << result << std::endl;
>
>
with std::complex<doubleit works. anyone knows how to get rid of this?

Get rid of what? Define "not working".

On Wed, 24 Jan 2007 14:56:08 +0100, Fredy Halter wrote:

Quote:

the following code is not working:

(In future please define "not working" and supply complete compilable code)

Quote:

std::complex<long doublex(1.,1.);
std::complex<long doubleresult(0.,0.);
>
result = 1./x;

result = std::complex<long double>(1.)/x;

There is no operator / (at least not on my system) that takes a lhs of
type double and a rhs of type std::complex<long double>.

[...]

--
Lionel B