Thread: double constructor & complex initialisation

Hi there,

In C++, if I do this:

Code:

double a;

Is 'a' guaranteed to be 0.0 according to the standard?

I ask because, I am interested in doing this:

Code:

std::complex<double> b = 5.0;

If the default constructor of double does not guarantee zero, then presumably the value of the imaginary component of 'b' could be anything?

>I don't believe C++ initializes variable.

Mmmm. That what I thought. However, if this is the case it makes the default constructors for the complex class dangerous.

It means I should be doing this to be safe:

Code:

complex<double> b(5.0, 0.0);

and this

Code:

if ( b == complex<double>(5.0, 0.0) ) ...

Thanks for the reply. I'm a little confused.

The complex constructor looks like so:

complex (const T& re_arg = T(), const T& im_arg = T());

If I were, to do this:

Code:

complex<double> b = 5.0;

presumably, the imaginary component is equal to whatever the default value for a double constructor assigns.

Are you suggesting:

Code:

double a;

does not initialise 'a' to zero, but:

Code:

double a();

would?

Follow up. I guess this explains things:

(from the C++ documentation)
complex<T>& operator=(const T& v);
Assigns v to the real part of itself, setting the imaginary part to 0.

Scrub my previous post & thanks for replying.

>Don't confuse
>> double a;
>with
>> double a = double();
>One is unititialized and the other is zero initialized per the standard.

That was something I didn't know & was very helpful.

Cheers to you both for the help.