Re: [Haskell-cafe] Begginer question about alignment in Storable
Hello Maurício,
Saturday, November 15, 2008, 11:29:23 PM, you wrote:
struct {
double w[2];
}
Can I assume that its function 'alignment' should
return the same as alignment (1::CDouble)? Or
should it be 2 * alignment (1::CDouble)?
secind - definitely not. first - probably yes
--
Best regards,
Bulatmailto:[EMAIL PROTECTED]
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Re: [Haskell-cafe] Begginer question about alignment in Storable
If you are using hsc2hs (if you are using Cabal this is easy; just
rename the file to *.hsc and Cabal will take care of the rest), then
there is a macro for making this easier and so you don't have to think
about it.
First, place the following at the top of your source file:
#let alignment t = %lu, (unsigned long)offsetof(struct {char x__; t
(y__); }, y__)
Then #{alignment c_type} will always be expanded by hsc2hs to the
correct alignment for the C type named c_type.
On Sat, Nov 15, 2008 at 3:29 PM, Maurício [EMAIL PROTECTED] wrote:
Hi,
I need to make the following structure into
an instance of Storable:
struct {
double w[2];
}
Can I assume that its function 'alignment' should
return the same as alignment (1::CDouble)? Or
should it be 2 * alignment (1::CDouble)?
(I've read the wikipedia article about memory alignment,
and what I understand is that there's a better alignment
for hardware efficiency but compilers and programmers can
do almost anything about it. What can I do when writing
a library wrapper if I don't know what kind of compiler
pragmas were used in all platforms that library was
compiled?)
Thanks,
Maurício
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Re: [Haskell-cafe] Begginer question
On 6 Jan 2005, at 01:37, Maurício wrote:
import Complex;
complex_root :: (Float, Float, Float) - (Complex Float, Complex Float)
complex_root (a,b,c) = (x1,x2) where {
delta = b * b - 4 * a * c :: Float;
sqr_delta = if delta = 0 then (sqrt delta) :+ 0 else 0 :+ (sqrt
delta) :: (Complex Float);
x1 = (b + sqr_delta)/(2 * a);
x2 = (b - sqr_delta)/(2 * a);
}
Couldn't match `Float' against `Complex Float'
Expected type: Float
Inferred type: Complex Float
In the second argument of `(+)', namely `sqr_delta'
In the definition of `x1': x1 = (b + sqr_delta)
Can you help me finding what is wrong? Shouldn't b be converted to
Complex Float and be summed to sqr_delta?
Haskell will not automatically convert b from Float to Complex Float.
The arguments of (+) should have the same type.
One alternative is to use b :+ 0 instead of b. (and similarly for a).
Another approach is to define a 'cast' function like:
toComplex x = (fromRational.toRational) x :: Complex Float
and then you can use toComplex b instead of b :+ 0. that's more
characters to type, though...
Note that sqr_delta isn't going to be defined as you expect, either.
Since delta has type Float, sqrt delta has type Float. (And sqrt -1 ::
Float is Not A Number). If you want to do it by hand this way, then you
want:
sqr_delta = if delta = 0 then (sqrt delta) :+ 0 else 0 :+ (sqrt
(-delta)) :: (Complex Float);
If delta was itself already Complex, then sqrt would do the right thing
automatically.
Jules
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Re: [Haskell-cafe] Begginer question
Maurício [EMAIL PROTECTED] writes:
complex_root :: (Float, Float, Float) - (Complex Float, Complex Float)
complex_root (a,b,c) = (x1,x2) where {
delta = b * b - 4 * a * c :: Float;
sqr_delta = if delta = 0 then (sqrt delta) :+ 0 else 0 :+
(sqrt delta) :: (Complex Float);
x1 = (b + sqr_delta)/(2 * a);
x2 = (b - sqr_delta)/(2 * a);
}
Couldn't match `Float' against `Complex Float'
Expected type: Float
Inferred type: Complex Float
In the second argument of `(+)', namely `sqr_delta'
In the definition of `x1': x1 = (b + sqr_delta)
The error message says it all, really.
If you examine the definition of x1, you see that (+) is applied to
two variables. If you check these, you will discover that they have
different types, which is the cause of the error.
Note that Haskell doesn't automatically convert arguments for you --
this is a feature.
-kzm
--
If I haven't seen further, it is by standing in the footprints of giants
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