1,For how to print the variable, I used the following sentence with windbg script, however it fail to print it.
r #$t0 = 123;
.printf #$t0
2,How to use the variable in if sentence, however I still found it failed.
r #$t0 = 123;
.if(#$t0 ==123)
{
.printf #$t0
}
How to modify them?
I think your way to use pseudo-register isn't wrong, but:
1) .printf() takes a format string as it does in C
2) By default, numbers are in Hex rather than decimals, use 0n or 0x to mention their base explicitly
3) Historically you shouldn't have used '#' to set a pseudo register with 'r', although I believe now it works in either way
r $t0 = 0n123;
.if ( #$t0 == 0n123 )
{
.printf "%d", #$t0
}
$$ sum of triangular numbers in windbg script
r eax=0;r $t0=0;.while(#$t0<10){r eax=#eax+#$t0;.printf "%2d=%3d\n",#$t0,#eax;r $t0=#$t0+1}
result
0= 0
1= 1
2= 3
3= 6
4= 10
5= 15
6= 21
7= 28
8= 36
9= 45
10= 55
11= 66
12= 78
13= 91
14=105
15=120
lets confirm f(n) = n * (n+1) /2
? 0n15 * (0n16/2)
Evaluate expression: 120 = 00000078
Related
Lets say I have this range of numbers, I want to expand these intervals. What is going wrong with my code here? The answer I am getting isn't correct :(
intervals are only represented with -
each 'thing' is separated by ;
I would like the output to be:
-6 -3 -2 -1 3 4 5 7 8 9 10 11 14 15 17 18 19 20
range_expansion('-6;-3--1;3-5;7-11;14;15;17-20 ')
function L=range_expansion(S)
% Range expansion
if nargin < 1;
S='[]';
end
if all(isnumeric(S) | (S=='-') | (S==',') | isspace(S))
error 'invalid input';
end
ixr = find(isnumeric(S(1:end-1)) & S(2:end) == '-')+1;
S(ixr)=':';
S=['[',S,']'];
L=eval(S) ;
end
ans =
-6 -2 -2 -4 14 15 -3
You can use regexprep to replace ;by , and the - that define ranges by :. Those - are identified by them being preceded by a digit. The result is a string that can be transformed into the desired output using str2num. However, since this function evaluates the string, for safety it is first checked that the string only contains the allowed characters:
in = '-6;-3--1;3-5;7-11;14;15;17-20 '; % example
assert(all(ismember(in, '0123456789 ,;-')), 'Characters not allowed') % safety check
out = str2num(regexprep(in, {'(?<=\d)-' ';'}, {':' ','})); % replace and evaluate
This has to be done in Perl:
I have integers on the order of e.g. 30_146_890_129 and 17_181_116_691 and 21_478_705_663.
These are supposedly made up of 6 bytes, where:
bytes 0-1 : value a
bytes 2-3 : value b
bytes 4-5 : value c
I want to isolate what value a is. How can I do this in Perl?
I've tried using the >> operator:
perl -e '$a = 330971351478 >> 16; print "$a\n";'
5050222
perl -e '$a = 17181116691 >> 16; print "$a\n";'
262163
But these numbers are not on the order of what I am expecting, more like 0-1000.
Bonus if I can also get values b and c but I don't really need those.
Thanks!
number >> 16 returns number shifted by 16 bit and not the shifted bits as you seem to assume. To get the last 16 bit you might for example use number % 2**16 or number & 0xffff. To get to b and c you can just shift before getting the last 16 bits, i.e.
$a = $number & 0xffff;
$b = ($number >> 16) & 0xffff;
$c = ($number >> 32) & 0xffff;
If you have 6 bytes, you don't need to convert them to a number first. You can use one the following depending on the order of the bytes: (Uppercase represents the most significant byte.)
my ($num_c, $num_b, $num_a) = unpack('nnn', "\xCC\xcc\xBB\xbb\xAA\xaa");
my ($num_a, $num_b, $num_c) = unpack('nnn', "\xAA\xaa\xBB\xbb\xAA\xaa");
my ($num_c, $num_b, $num_a) = unpack('vvv', "\xcc\xCC\xbb\xBB\xaa\xAA");
my ($num_a, $num_b, $num_c) = unpack('vvv', "\xaa\xAA\xbb\xBB\xcc\xCC");
If you are indeed provided with a number 0xCCccBBbbAAaa), you can convert it to bytes then extract the numbers you want from it as follows:
my ($num_c, $num_b, $num_a) = unpack('xxnnn', pack('Q>', $num));
Alternatively, you could also use an arithmetic approach like you attempted.
my $num_a = $num & 0xFFFF;
my $num_b = ( $num >> 16 ) & 0xFFFF;
my $num_c = $num >> 32;
While the previous two solutions required a Perl built to use 64-bit integers, the following will work with any build of Perl:
my $num_a = $num % 2**16;
my $num_b = ( $num / 2**16 ) % 2**16;
my $num_c = int( $num / 2**32 );
Let's look at ( $num >> 16 ) & 0xFFFF in detail.
Original number: 0x0000CCccBBbbAAaa
After shifting: 0x00000000CCccBBbb
After masking: 0x000000000000BBbb
I am struggling with a text file that I have to read in. In this file, there are two types of line:
133 0102764447 44 11 54 0.4 0 0.89 0 0 8 0 0 7 Attribute_Name='xyz' Type='string' 02452387764447 884
134 0102256447 44 1 57 0.4 0 0.81 0 0 8 0 0 1 864
What I want to do here is to textscan all the lines and then try to determine the number of 'xyz' (and the total number of lines).
I tried to use:
fileID = fopen('test.txt','r') ;
data=textscan(fileID, %d %d %d %d %d %d %d %d %d %d %d %d %d %s %s %d %d','\n) ;
And then I will try to access data{i,16} to count how many are equal to Attribute_Name='xyz', it doesnt seem to be an efficient though.
what will be a proper way to read the data(what interests me is to count how many Attribute_Name='xyz' do I have)? Thanks
You could simply use count which is referenced here.
In your case you could use it in this way:
filetext = fileread("test.txt");
A = count(filetext , "xyz")
fileread will read the whole text file into a single string. Afterwards you can process that string using count which will return the occurrences from the given pattern.
An alternative when using older versions of MATLAB is this one. It will work with R2006a and above.
filetext = fileread("test.txt");
A = length(strfind(filetext, "xyz");
strfind will return an array which length represents the amount of occurrences of the specified string. The length of that array can be accessed by length.
There is the option of strsplit. You may do something like the following:
count = 0;
fid = fopen('test.txt','r');
while ~feof(fid)
line = fgetl(fid);
words = strsplit( line )
ind = find( strcmpi(words{:},'Attribute_Name=''xyz'''), 1); % Assume only one instance per line, remove 1 for more and correct the rest of the code
if ( ind > 0 ) then
count = count + 1;
end if
end
So at the end count will give you the number.
I am trying to convert AD maxpwdAge (a 64-bit integer) into a number of days.
According to Microsoft:
Uses the IADs interface's Get method to retrieve the value of the domain's maxPwdAge attribute (line 5).
Notice we use the Set keyword in VBScript to initialize the variable named objMaxPwdAge—the variable used to store the value returned by Get. Why is that?
When you fetch a 64-bit large integer, ADSI does not return one giant scalar value. Instead, ADSI automatically returns an IADsLargeInteger object. You use the IADsLargeInteger interface's HighPart and LowPart properties to calculate the large integer's value. As you may have guessed, HighPart gets the high order 32 bits, and LowPart gets the low order 32 bits. You use the following formula to convert HighPart and LowPart to the large integer's value.
The existing code in VBScript from the same page:
Const ONE_HUNDRED_NANOSECOND = .000000100 ' .000000100 is equal to 10^-7
Const SECONDS_IN_DAY = 86400
Set objDomain = GetObject("LDAP://DC=fabrikam,DC=com") ' LINE 4
Set objMaxPwdAge = objDomain.Get("maxPwdAge") ' LINE 5
If objMaxPwdAge.LowPart = 0 Then
WScript.Echo "The Maximum Password Age is set to 0 in the " & _
"domain. Therefore, the password does not expire."
WScript.Quit
Else
dblMaxPwdNano = Abs(objMaxPwdAge.HighPart * 2^32 + objMaxPwdAge.LowPart)
dblMaxPwdSecs = dblMaxPwdNano * ONE_HUNDRED_NANOSECOND ' LINE 13
dblMaxPwdDays = Int(dblMaxPwdSecs / SECONDS_IN_DAY) ' LINE 14
WScript.Echo "Maximum password age: " & dblMaxPwdDays & " days"
End If
How can I do this in Perl?
Endianness may come into this, but you may be able to say
#!/usr/bin/perl
use strict;
use warnings;
my $num = -37_108_517_437_440;
my $binary = sprintf "%064b", $num;
my ($high, $low) = $binary =~ /(.{32})(.{32})/;
$high = oct "0b$high";
$low = oct "0b$low";
my $together = unpack "q", pack "LL", $low, $high;
print "num $num, low $low, high $high, together $together\n";
Am I missing something? As far as I can tell from your question, your problem has nothing at all to do with 2’s complement. As far as I can tell, all you need/want to do is
use Math::BigInt;
use constant MAXPWDAGE_UNIT_PER_SEC => (
1000 # milliseconds
* 1000 # microseconds
* 10 # 100 nanoseconds
);
use constant SECS_PER_DAY => (
24 # hours
* 60 # minutes
* 60 # seconds
);
my $maxpwdage_full = ( Math::BigInt->new( $maxpwdage_highpart ) << 32 ) + $maxpwdage_lowpart;
my $days = $maxpwdage_full / MAXPWDAGE_UNIT_PER_SEC / SECS_PER_DAY;
Note that I deliberately use 2 separate constants, and I divide by them in sequence, because that keeps the divisors smaller than the range of a 32-bit integer. If you want to write this another way and you want it to work correctly on 32-bit perls, you’ll have to keep all the precision issues in mind.
Does anyone know why, using SQLServer 2005
SELECT CONVERT(DECIMAL(30,15),146804871.212533)/CONVERT(DECIMAL (38,9),12499999.9999)
gives me 11.74438969709659,
but when I increase the decimal places on the denominator to 15, I get a less accurate answer:
SELECT CONVERT(DECIMAL(30,15),146804871.212533)/CONVERT(DECIMAL (38,15),12499999.9999)
give me 11.74438969
For multiplication we simply add the number of decimal places in each argument together (using pen and paper) to work out output dec places.
But division just blows your head apart. I'm off to lie down now.
In SQL terms though, it's exactly as expected.
--Precision = p1 - s1 + s2 + max(6, s1 + p2 + 1)
--Scale = max(6, s1 + p2 + 1)
--Scale = 15 + 38 + 1 = 54
--Precision = 30 - 15 + 9 + 54 = 72
--Max P = 38, P & S are linked, so (72,54) -> (38,20)
--So, we have 38,20 output (but we don use 20 d.p. for this sum) = 11.74438969709659
SELECT CONVERT(DECIMAL(30,15),146804871.212533)/CONVERT(DECIMAL (38,9),12499999.9999)
--Scale = 15 + 38 + 1 = 54
--Precision = 30 - 15 + 15 + 54 = 84
--Max P = 38, P & S are linked, so (84,54) -> (38,8)
--So, we have 38,8 output = 11.74438969
SELECT CONVERT(DECIMAL(30,15),146804871.212533)/CONVERT(DECIMAL (38,15),12499999.9999)
You can do the same math if follow this rule too, if you treat each number pair as
146804871.212533000000000 and 12499999.999900000
146804871.212533000000000 and 12499999.999900000000000
To put it shortly, use DECIMAL(25,13) and you'll be fine with all calculations - you'll get precision right as declared: 12 digits before decimal dot, and 13 decimal digits after.
Rule is: p+s must equal 38 and you will be on safe side!
Why is this?
Because of very bad implementation of arithmetic in SQL Server!
Until they fix it, follow that rule.
I've noticed that if you cast the dividing value to float, it gives you the correct answer, i.e.:
select 49/30 (result = 1)
would become:
select 49/cast(30 as float) (result = 1.63333333333333)
We were puzzling over the magic transition,
P & S are linked, so:
(72,54) -> (38,29)
(84,54) -> (38,8)
Assuming (38,29) is a typo and should be (38,20), the following is the math:
i. 72 - 38 = 34,
ii. 54 - 34 = 20
i. 84 - 38 = 46,
ii. 54 - 46 = 8
And this is the reasoning:
i. Output precision less max precision is the digits we're going to throw away.
ii. Then output scale less what we're going to throw away gives us... remaining digits in the output scale.
Hope this helps anyone else trying to make sense of this.
Convert the expression not the arguments.
select CONVERT(DECIMAL(38,36),146804871.212533 / 12499999.9999)
Using the following may help:
SELECT COL1 * 1.0 / COL2