I have a query on selection of method to extract a byte from the word. My word is currently unsigned. It is simply collection of 32 bits.(1s and 0s). Both the follwing script I implement returns me same result. I wonder which is a better option to select.
Bytes0=(UINT8)((Word>>00 & 0x0000FF);
Bytes1=(UINT8)(Word>>08 & 0x0000FF);
Bytes2=(UINT8)(Word>>16 & 0x0000FF);
Bytes3=(UINT8)(Word>>24 & 0x0000FF);
or
Bytes0=(UINT8)((Word>>00);
Bytes1=(UINT8)(Word>>08 );
Bytes2=(UINT8)(Word>>16 );
Bytes3=(UINT8)(Word>>24 );
Am I missing something?
Thanks
DSP Guy
The cast to UINT8 is discarding all the bytes except for the lowest one. So in the first case, you convert the upper bytes to 0 and then discard them; in the second you simply discard them. The second option is clearly more efficient, assuming you are assigning to a UINT8.
Related
I'd like to understand how unpack is returning the sum in the given perl one-liner.
I've looked at pack man page and mostly understood that it is simply formatting the given array into a scalar of ten doubles.
However, I couldn't find proper documentation for unpack with %123. Looking for help here.
print unpack "%123d*" , pack( "d*", (1..10));
This line correctly outputs 55 which is 1+2+3+...+10.
From perldoc -f unpack:
In addition to fields allowed in pack(), you may prefix a field with a % to indicate that you want a <number>-bit checksum of the items instead of the items themselves.
Thus %123d* means to add all the input integers 1..10 and then take the first 123 bit of this result in order to construct the "<number>-bit checksum". Note that %8d* or just %d* (which is equivalent to %16d*) would suffice too given that the sum is small enough.
I have input with floats stored like 1000,50, ie. the decimal points are replaced by commas.
Is there an option in K to load these numbers directly into floats ?
When using
data:("SFF" ;";",";") 0:. filename
I get 0ns, of course, because the numbers are not recognized as floats.
I load them as strings now, and convert them using ssr like
c:.:' .q.ssr'[data;",";"."]
but that is extremely slow.
Is there an option somewhere to have K load these numbers in CCC,CC format as floats directly ? Normal format and ccc,cc format are not mixed, any file has just one of them.
If there is not, I imagine that it must by quite easy to replace a "." somewhere in the Q-binary where the load-function sits, with a ",", to get a version which loads these numbers. Has anybody tried that ? Or any other tip to load big files with these numbers in reasonable time ?
Cheers,
Co
If ssr' is slow for your task you may find this tiny function useful:
c2p:{c:-1_sums count each x;p:ss[r:raze x;","];r[p]:".";(0,c) _ r}
Update: an alternative version:
c2p:{p:ss[r:raze x;","];r[p]:".";(0,-1_sums count'[x])_r}
It concatenates all strings into a single long string, finds positions of commas, replaces commas with periods then splits that long string:
q)N:1000000
q)s:string[N?100000],'",",'string N?1000
q)\t r1:ssr'[s;",";"."]
4284
q)\t r2:c2p s
242
q)r1~r2
1b
I was thinking something like find (?) combined with indexing/applying
q)N:1000000
q)s:string[N?100000],'",",'string N?1000
q)\ts {s[x;y]:"."}./:flip(til count s;s?\:",")
967 52972144
q)s
"93912.794"
"57144.788"
"77809.659"
"7839.47"
"6363.523"
"44761.244"
"65699.712"
It's not perfect but that's the general idea. I'm sure there is an easier way...
I have data written into short data type. The data written is of 2's complement form.
Now when I try to print the data using %04x, the data with MSB=0 is printed fine for eg if data=740, print I get is 0740
But when the MSB=1, I am unable to get a proper print. For eg if data=842, print I get is fffff842
I want the data truncated to 4 bytes so expected output is f842
Either declare your data as a type which is 16 bits long, or make sure the printing function uses the right format for 16 bits value. Or use your current type, but do a bitwise AND with 0xffff. What you can do depends on the language you're doing it in really.
But whichever way you go, check your assumptions again. There seems to be a few issues in your question:
2s-complement applies to signed numbers only. There are no negative numbers in your question.
Assuming you mean C's short - it doesn't have to be 16 bits long.
"I get is fffff842 I want the data truncated to 4 bytes" - fffff842 is 4 bytes long. f842 is 2 bytes long.
2-bytes long value 842 does not have the MSB set.
I'm assuming C (or possibly C++) as the language here.
Because of the default argument promotions involved when calling a variable argument function (such as printf), your use of a short will result in an integer promotion, which states that "If an int can represent all values of the original type (as restricted by the width, for a
bit-field), the value is converted to an int".
A short is converted to an int by means of sign-extension, and 0xf842 sign-extended to 32 bits is 0xfffff842.
You can use a bitwise AND to mask off the most significant word:
printf("%04x", data & 0xffff);
You could also add the h length specifier to state that you only want to print an (unsigned) short worth of bits from an int:
printf("%04hx", data);
I use sprintf for conversion to hex - example >>
$hex = sprintf("0x%x",$d)
But I was wondering, if there is some alternative way how to do it without sprintf.
My goal is convert a number to 4-byte hex code (e.g. 013f571f)
Additionally (and optionally), how can I do such conversion, if number is in 4 * %0xxxxxxx format, using just 7 bits per byte?
sprintf() is probably the most appropriate way. According to http://perldoc.perl.org/functions/hex.html:
To present something as hex, look into printf, sprintf, and unpack.
I'm not really sure about your second question, it sounds like unpack() would be useful there.
My goal is convert a number to 4-byte hex code (e.g. 013f571f)
Hex is a textual representation of a number. sprintf '%X' returns hex (the eight characters 013f571f). sprintf is specifically designed to format numbers into text, so it's a very elegant solution for that.
...But it's not what you want. You're not looking for hex, you're looking for the 4-byte internal storage of an integer. That has nothing to do with hex.
pack 'N', 0x013f571f; # "\x01\x3f\x57\x1f" Big-endian byte order
pack 'V', 0x013f571f; # "\x1f\x57\x3f\x01" Little-endian byte order
sprintf() is my usual way of performing this conversion. You can do it with unpack, but it will probably be more effort on your side.
For only working with 4 byte values, the following will work though (maybe not as elegant as expected!):
print unpack("H8", pack("N1", $d));
Be aware that this will result in 0xFFFFFFFF for numbers bigger than that as well.
For working pack/unpack with arbitrary bit length, check out http://www.perlmonks.org/?node_id=383881
The perlpacktut will be a handy read as well.
For 4 * %0xxxxxxx format, my non-sprintf solution is:
print unpack("H8", pack("N1",
(((($d>>21)&0x7f)<<24) + ((($d>>14)&0x7f)<<16) + ((($d>>7)&0x7f)<<8) + ($d&0x7f))));
Any comments and improvements are very welcome.
I have some code where I am converting some data elements in a flat file. I save the old:new values to a hash which is written to a file at the end of processing. On subsequence execution, I reload into a hash so I can reuse previously converted values on additional data files. I also save the last conversion value so if I encounter an unconverted value, I can assign it a new converted value and add it to the hash.
I had used this code before (back in Feb) on six files with no issues. I have a variable that is set to ZCKL0 (last character is a zero) which is retrieved from a file holding the last used value. I apply the increment operator
...
$data{$olddata} = ++$dataseed;
...
and the resultant value in $dataseed is 1 instead of ZCKL1. The original starting seed value was ZAAA0.
What am I missing here?
Do you use the $dataseed variable in a numeric context in your code?
From perlop:
If you increment a variable that is
numeric, or that has ever been used in
a numeric context, you get a normal
increment. If, however, the variable
has been used in only string contexts
since it was set, and has a value that
is not the empty string and matches
the pattern /^[a-zA-Z][0-9]\z/ , the
increment is done as a string,
preserving each character within its
range.
As prevously mentioned, ++ on strings is "magic" in that it operates differently based on the content of the string and the context in which the string is used.
To illustrate the problem and assuming:
my $s='ZCL0';
then
print ++$s;
will print:
ZCL1
while
$s+=0; print ++$s;
prints
1
NB: In other popular programming languages, the ++ is legal for numeric values only.
Using non-intuitive, "magic" features of Perl is discouraged as they lead to confusing and possibly unsupportable code.
You can write this almost as succinctly without relying on the magic ++ behavior:
s/(\d+)$/ $1 + 1 /e
The e flag makes it an expression substitution.