valarrays and regular arrays

[Sebastian Thiebes]

Hello all,

I am using valarrays of double in my code, but for using certain
functions I need to pass a normal array of double. What I did was
write something like this:

valarray<double> vad;
// somehow fill this valarray
double* da = new double[vad.size()];

for (long i = 0; i < vad.size(); ++i)
{
da = vad;
}

This works fine. However I feel this not to be an elegant solution to
the proplem, is there a way to maybe directly cast a valarray into a
normal array, or is there another way to do this conversion more
efficiently?

Thanks in advance,

Sebastian

 

[John Harrison]

Hello all,

I am using valarrays of double in my code, but for using certain
functions I need to pass a normal array of double. What I did was
write something like this:

valarray<double> vad;
// somehow fill this valarray
double* da = new double[vad.size()];

for (long i = 0; i < vad.size(); ++i)
{
da = vad;
}

This works fine. However I feel this not to be an elegant solution to
the proplem, is there a way to maybe directly cast a valarray into a
normal array, or is there another way to do this conversion more
efficiently?

Thanks in advance,

Sebastian

The simple &vad[0] does the trick.

some_function(&vad[0]);

john

 

[Sebastian Thiebes]

The simple &vad[0] does the trick.

some_function(&vad[0]);

Thanks a lot! This is what I was hoping for,
a simple and elegant solution.

Seb

 

[Marc Schellens]

Hello all,

I am using valarrays of double in my code, but for using certain
functions I need to pass a normal array of double. What I did was
write something like this:

valarray<double> vad;
// somehow fill this valarray
double* da = new double[vad.size()];

for (long i = 0; i < vad.size(); ++i)
{
da = vad;
}

This works fine. However I feel this not to be an elegant solution to
the proplem, is there a way to maybe directly cast a valarray into a
normal array, or is there another way to do this conversion more
efficiently?

Thanks in advance,

Sebastian

You could do:

double* da = &vad[0];

HDH,
marc