VBA - Check whether a parameter is set using the bit key - excel

i have a dialog with a few parameters (arround 20).
Every parameter gets a value like 1,2,4,8,16,32 and so on.
By closing dialog an integer will be set with the sum of the parameter like 2+8+16+64.
Now i have a few options to run my program.
For example:
The first option needs to run the parameter 2,8 and 16 so i need to check if 2,8 and 16 will be checked in my integer.
I know that there is a way but not really how.
Maybe u can help.
Thanks
Jassin
The first option need

Apply the And operator on a mask (m, interesting bits set) and your data (d, e). If the result is equal to m, then all the bits set in m are set in the data.
>> d = 4 + 2 + 1
>> e = 4 + 1
>> m = 4 + 2 + 1
>> WScript.Echo 0, CStr(m = (d And m))
>> WScript.Echo 1, CStr(m = (e And m))
>>
0 True
1 False

Related

Error in reading data into 3x3 matrix in Fortran [duplicate]

I would like to read and store scientific formatted numbers from a txt file, which is formatted and the numbers are separated by tabulator.
This is what I have so far:
IMPLICIT NONE
REAL,ALLOCATABLE,DIMENSION(2) :: data(:,:)
INTEGER :: row,column
INTEGER :: j,i
CHARACTER(len=30) :: filename
CHARACTER(len=30) :: format
filename='data.txt'
open(86,file=filename,err=10)
write(*,*)'open data file'
read(86, *) row
read(86, *) column
allocate(data(row,column))
format='(ES14.7)'
do i=1,row
read(86,format) data(i,:)
enddo
close(86)
This is how the txt file looks like:
200
35
2.9900E-35 2.8000E-35 2.6300E-35 2.4600E-35 2.3100E-35 2.1600E-35 ...
The problem is that it doesn't read and store the correct values from the txt to the data variable. Is it the format causing the problem?
I would also like to know how to count the number of columns in this case. (I can count the rows by using read(86,*) in a for loop.)
Yes, your format is not good for the data you show. Better one should be like that read(99,'(6(E11.4,X))') myData(i,:).
However, I am not sure if you really need to use format at your reading at all.
Following example pretty close to what you are trying to do, and it is working bot with and without format.
program readdata
implicit none
real, allocatable :: myData(:,:)
real :: myLine
integer :: i, j, myRow, myColumn
character(len=30) :: myFileName
character(len=30) :: myFormat
myFileName='data.dat'
open(99, file=myFileName)
write(*,*)'open data file'
read(99, *) myRow
read(99, *) myColumn
allocate(myData(myRow,myColumn))
do i=1,myRow
read(99,*) myData(i,:)
!read(99,'(6(E11.4,X))') myData(i,:)
print*, myData(i,:)
enddo
close(99)
end program readdata
To test, I assumed that you have rows and columns always in the file, as you give, so my test data was following.
2
6
2.9900E-35 2.8000E-35 2.6300E-35 2.4600E-35 2.3100E-35 2.1600E-35
2.9900E-35 2.8000E-35 2.6300E-35 2.4600E-35 2.3100E-35 2.1600E-35
If you are really interested to read your files with a format and if the number of columns are not constant you may need a format depending on a variable, please see related discussions here.
Though there are no direct command to count the number of items in a line, we can count the number of periods or (E|e|D|d) by using the scan command. For example,
program main
implicit none
character(100) str
integer n
read( *, "(a)" ) str
call countreal( str, n )
print *, "number of items = ", n
contains
subroutine countreal( str, num )
implicit none
character(*), intent(in) :: str
integer, intent(out) :: num
integer pos, offset
num = 0
pos = 0
do
offset = scan( str( pos + 1 : ), "." ) !! (1) search for periods
!! offset = scan( str( pos + 1 : ), "EeDd" ) !! (2) search for (E|e|D|d)
if ( offset > 0 ) then
pos = pos + offset
num = num + 1
print *, "pos=", pos, "num=", num !! just for check
else
return
endif
enddo
endsubroutine
end
Please note that pattern (1) works only when all items have periods, while pattern (2) works only when all items have exponents:
# When compiled with (1)
$ echo "2.9900 2.8000E-35 2.6300D-35 2.46 2.31" | ./a.out
pos= 2 num= 1
pos= 10 num= 2
pos= 22 num= 3
pos= 34 num= 4
pos= 40 num= 5
number of items = 5
# When compiled with (2)
$ echo "2.9900E-35 2.8000D-35 2.6300e-35 2.4600d-35" | ./a.out
pos= 7 num= 1
pos= 19 num= 2
pos= 31 num= 3
pos= 43 num= 4
number of items = 4
For more general purposes, it may be more convenient to write a custom "split()" function that separate items with white spaces (or use an external library that supports a split function).

Excel VBA: Larger than behaves as larger or equal

I use the follow code in Excel VBA:
If ((B - A) > 180) Then
result = 1
ElseIf ((B - A) < -180) Then
result = 2
Else
result = 0
End If
In case (B - A) = 180, VBA returns result = 1.
The debugger shows the first statement of the If as True even if (B - A) = 180
Is it VBA acting funny or more probably me missing something?
Any help will be much appreciated.

How do I convert a 4 digit number into individual digits?

I need to write logic to break down a 4 digit number into individual digits.
On a reply here at SO to a question regarding 3 digits, someone gave the math below:
int first = 321/100;
int second = (321/10)-first*10;
int third = (321/1)-first*100-second*10;
Can someone help me?
Thank you in advance!
Well, using the sample you found, we can quite easily infer a code for you.
The first line says int first = 321/100;, which returns 3 (integer division is the euclidian one). 3 is indeed the first integer in 321 so that's a good thing. However, we have a 4 digit number, let's try replacing 100 with 1000:
int first = 4321/1000;
This does return 4 !
Let's try adapting the rest of your code (plus I put your four digit number in the variable entry).
int entry = 4321;
int first = entry/1000;
int second = entry/100 - first*10;
int third = entry/10 - first*100 - second*10;
int fourth = entry - first*1000 - second*100 - third*10;
second will be entry/100 (43) minus first*10 (40), so we're okay.
third is then 432 - 400 - 30 which turns to 2. This also works till fourth.
For more-than-four digits, you may want to use a for-loop and maybe some modulos though.
This snip of code counts the number of digits input from the user
then breaks down the digits one by one:
PRINT "Enter value";
INPUT V#
X# = V#
DO
IF V# < 1 THEN
EXIT DO
END IF
D = D + 1
V# = INT(V#) / 10
LOOP
PRINT "Digits:"; D
FOR L = D - 1 TO 0 STEP -1
M = INT(X# / 10 ^ L)
PRINT M;
X# = X# - M * 10 ^ L
NEXT
END

Fibonacci Sequence logic in Python [duplicate]

This question already has answers here:
How to write the Fibonacci Sequence?
(67 answers)
Closed 9 years ago.
A tutorial I am going through had the following program
# This program calculates the Fibonacci sequence
a = 0
b = 1
count = 0
max_count = 20
while count < max_count:
count = count + 1
old_a = a # we need to keep track of a since we change it
print(old_a,end=" ") # Notice the magic end=" " in the print function arguments that
# keeps it from creating a new line
a = b
b = old_a + b
print() # gets a new (empty) line
The code is perfect. However, I am not able to figure out how the sequence is calculated.
How are the values changed to create the sequence?
It'll make more sense if you remove all of that extraneous code:
while count < max_count:
old_a = a
a = b
b = old_a + b
The old_a is probably confusing you. It's the long way of writing this:
a, b = b, a + b
Which swaps a with b and (at the same time), b with a + b. Note that it isn't the same as writing:
a = b
b = a + b
Because by the time you re-define b, a already holds its new value, which is equal to b.
I'd also run through the code manually by writing it out on paper.
This code works fine:
a, b = 0, 1
for _ in range(20):
print a
a, b = b, a+b

Equivalent of adding strings to a loop, for strings (matlab)?

How would I be able to do the equivalent of this with strings:
a = [1 2 3; 4 5 6];
c = [];
for i=1:5
b = a(1,:)+i;
c = [c;b];
end
c =
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
Basically looking to combine several strings into a Matrix.
You're growing a variable in a loop, which is a kind of sin in Matlab :) So I'm going to show you some better ways of doing array concatenation.
There's cell strings:
>> C = {
'In a cell string, it'
'doesn''t matter'
'if the strings'
'are not of equal lenght'};
>> C{2}
ans =
doesn't matter
Which you could use in a loop like so:
% NOTE: always pre-allocate everything before a loop
C = cell(5,1);
for ii = 1:5
% assign some random characters
C{ii} = char( '0'+round(rand(1+round(rand*10),1)*('z'-'0')) );
end
There's ordinary arrays, which have as a drawback that you have to know the size of all your strings beforehand:
a = [...
'testy' % works
'droop'
];
b = [...
'testing' % ERROR: CAT arguments dimensions
'if this works too' % are not consistent.
];
for these cases, use char:
>> b = char(...
'testing',...
'if this works too'...
);
b =
'testing '
'if this works too'
Note how char pads the first string with spaces to fit the length of the second string. Now again: don't use this in a loop, unless you've pre-allocated the array, or if there really is no other way to go.
Type help strfun on the Matlab command prompt to get an overview of all string-related functions available in Matlab.
You mean storing a string on each matrix position? You can't do that, since matrices are defined over basic types. You can have a CHAR on each position:
>> a = 'bla';
>> b = [a; a]
b <2x3 char> =
bla
bla
>> b(2,3) = 'e'
b =
bla
ble
If you want to store matrices, use a cell array (MATLAB reference, Blog of Loren Shure), which are kind of similar but using "{}" instead of "()":
>> c = {a; a}
c =
'bla'
'bla'
>> c{2}
ans =
bla

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