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Need help with an array problem.

P: n/a
Basically I think the problem is in converting from a 32 bit integer to
a 16 bit integer.

I have two arrays:
import array

a = array.array('L', [65537])
b = array.array('H', [0])

b[0] = a[0]

Which gives an overflow message....
So can't I truncate the long by discaring the upper bits ..
Like b[0] = 0x0000FFFF & a[0]

How does one normally cast an object from long to short?

Oct 2 '06 #1
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P: n/a
SpreadTooThin wrote:
Basically I think the problem is in converting from a 32 bit integer to
a 16 bit integer.

I have two arrays:
import array

a = array.array('L', [65537])
b = array.array('H', [0])

b[0] = a[0]

Which gives an overflow message....
So can't I truncate the long by discaring the upper bits ..
Like b[0] = 0x0000FFFF & a[0]

How does one normally cast an object from long to short?
Take the modulo 65536?

pyarray.array('H', (array.array('L', [65537%65536])))
array('H', [1])
James
--
James Stroud
UCLA-DOE Institute for Genomics and Proteomics
Box 951570
Los Angeles, CA 90095

http://www.jamesstroud.com/
Oct 2 '06 #2

P: n/a

SpreadTooThin wrote:
Basically I think the problem is in converting from a 32 bit integer to
a 16 bit integer.

I have two arrays:
import array

a = array.array('L', [65537])
b = array.array('H', [0])

b[0] = a[0]

Which gives an overflow message....
As it should.
So can't I truncate the long by discaring the upper bits ..
Like b[0] = 0x0000FFFF & a[0]
Any reason why you don't you just try it?

| >>import array
| >>a = array.array('L', [65537])
| >>b = array.array('H', [0])
| >>a[0]
| 65537L
| >>a[0] & 0xffff
| 1L
| >>b[0] = a[0] & 0xffff
| >>b[0]
| 1
| >>>

Oct 2 '06 #3

P: n/a
To your question on casting long to short. This is how:
a=1234L # long
b=int(a) # int (short)

John Machin skrev:
SpreadTooThin wrote:
Basically I think the problem is in converting from a 32 bit integer to
a 16 bit integer.

I have two arrays:
import array

a = array.array('L', [65537])
b = array.array('H', [0])

b[0] = a[0]

Which gives an overflow message....

As it should.
So can't I truncate the long by discaring the upper bits ..
Like b[0] = 0x0000FFFF & a[0]

Any reason why you don't you just try it?

| >>import array
| >>a = array.array('L', [65537])
| >>b = array.array('H', [0])
| >>a[0]
| 65537L
| >>a[0] & 0xffff
| 1L
| >>b[0] = a[0] & 0xffff
| >>b[0]
| 1
| >>>
Oct 2 '06 #4

P: n/a
ja*****@gmail.com wrote:
To your question on casting long to short. This is how:
a=1234L # long
b=int(a) # int (short)
No, a Python int is a C long. A Python long is an arbitrary-precision number and
does not correspond to any C type.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

Oct 2 '06 #5

P: n/a

Robert Kern wrote:
ja*****@gmail.com wrote:
To your question on casting long to short. This is how:
a=1234L # long
b=int(a) # int (short)

No, a Python int is a C long. A Python long is an arbitrary-precision number and
does not correspond to any C type.

--
So there is no short(number) casting?

Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco
Oct 3 '06 #6

P: n/a
SpreadTooThin wrote:
Robert Kern wrote:
>ja*****@gmail.com wrote:
>>To your question on casting long to short. This is how:
a=1234L # long
b=int(a) # int (short)
No, a Python int is a C long. A Python long is an arbitrary-precision number and
does not correspond to any C type.

So there is no short(number) casting?
Not in core Python, no, since C short ints have no Python type directly
corresponding to them. As John Machin pointed out, if you had tried what you
proposed, it would have worked just fine.

If you find that you keep needing to deal with the various C integer and
floating point types (and arrays of such), you might want to consider using numpy.

http://numpy.scipy.org
In [15]: import numpy

In [16]: a = numpy.array([65537], dtype=numpy.uint32)

In [17]: a
Out[17]: array([65537], dtype='uint32')

In [18]: b = a.astype(numpy.uint16)

In [19]: b
Out[19]: array([1], dtype='uint16')

In [20]: numpy.uint16(a[0])
Out[20]: 1
--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

Oct 3 '06 #7

P: n/a

SpreadTooThin wrote:
Robert Kern wrote:
ja*****@gmail.com wrote:
To your question on casting long to short. This is how:
a=1234L # long
b=int(a) # int (short)
No, a Python int is a C long. A Python long is an arbitrary-precision number and
does not correspond to any C type.

--

So there is no short(number) casting?
It depends on what you think you mean by "short(number) casting". I
thought I'd answered your original question -- if
short = long & 0xffff
doesn't do what you want to do, please give examples of what "long"
means to you, and what you expect the result of "short(number) casting"
each example to be.

Oct 3 '06 #8

P: n/a
Dennis Lee Bieber <wl*****@ix.netcom.comwrote:
> Of course, there is also the confusion between "type cast" and "type
conversion" -- at least, for me...
cast taking the bit-pattern of a value in one "type" and
interpreting the same bit-pattern as if it were a different "type"
conversion taking the value of a bit-pattern in one "type" and
converting it to the bit pattern of the equivalent value in another
"type"
From where have you learned those definitions? If it's from C,
then you have read the wrong books. A cast in C is a type
conversion, with the same semantics as you write under "conversion"
above. The C standard (ISO/IEC 9899:1999) says:

6.5.4 Cast operators
[...]
Semantics
Preceding an expression by a parenthesized type name converts the
value of the expression to the named type. This construction is
called a cast. A cast that specifies no conversion has no effect
on the type or value of an expression.

("A cast that specifies no conversion" refers to when you cast
from one type to the same type, e.g. '(int)x' if x is of the type
'int'.)

You may also try this program:

#include <stdio.h>
#include <string.h>

int main(void)
{
/* Assumption: sizeof(float)==sizeof(int). This is the most
* common case on modern computers. */
float f = -17.0;
int i = -23;
float fjunk;
int ijunk;

printf("Cast: %d %10.6f\n", (int)f, (float)i);
memcpy(&fjunk, &i, sizeof i);
memcpy(&ijunk, &f, sizeof f);
printf("Bitcopy: %d %10.6f\n", ijunk, fjunk);
return 0;
}

If a cast had been what you believed, both printf() statements
above would give the same output. Unless your C compiler uses
some really strange floating point representation, they will
print rather different things. The first one must print

Cast: -17 -23.000000

showing very clearly that a cast is a type conversion. The
second printf() will print some seemingly random numbers, showing
that those bit-patterns will be interpreted very differently when
interpreted as another type.

What you might have been confused by, is dereferencing a casted
pointer. Add the following statement:

printf("Pointer cast: %d %10.6f\n", *(int*)&f, *(float*)&i);

to the program. It should output the same numbers as the
"Bitcopy" printf(). But what is cast here is the *address* of
the variables, not the actual contents of them. It is the
*dereferencing* of those casted pointers that interpret the bit
patterns in the variables as if they were another type, not the
casting itself.
--
Thomas Bellman, Lysator Computer Club, Linköping University, Sweden
"I don't think [that word] means what you ! bellman @ lysator.liu.se
think it means." -- The Princess Bride ! Make Love -- Nicht Wahr!
Oct 3 '06 #9

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