#lang plait
number: DIGIT number
eval(0) = 0
eval(1) = 1
eval(2) = 2
...
eval(9) = 9
eval(DIGIT) = DIGIT
eval(DIGIT number) = 10*eval(DIGIT) + eval(number)
Documentation from Racket states that eval can only be used in DrRacket's interaction windows. How can I write this in definition window?
Related
In OCaml, how can I compute the length of a string that may have unicode encodings ? To give an example, here is my problem:
utop # "\u{02227}";;
- : string = "∧"
utop # Caml.String.length "\u{02227}";;
- : int = 3
utop # Base.String.length "\u{02227}";;
- : int = 3
and I would like to obtain the obvious answer: 1.
If you want to count the number of extended grapheme clusters (aka a graphical character), you can use uuseg. For instance
let len = Uuseg_string.fold_utf_8 `Grapheme_cluster (fun x _ -> x + 1) 0
let n = len "∧";;
returns
val n : int = 1
Introduction: Hi, everyone! I'm new to VB6! How can I make a code that tells VB6 that tells VB6 that you're inputting something other than Whole numbers?
Details: I'm making an arithmetic progression calculator (I think the code is not needed? but I'll just provide just in case.) Here is my code:
Option Explicit
Private Sub btCalc_Click()
Dim A As Long
Dim N As Long
Dim D As Long
Dim R As Long
Dim F As Long
A = Val(txtInitterm.Text)
N = Val(txtTermint.Text)
D = Val(txtFinterm.Text)
R = Val(txtTermint.Text)
F = N / 2 * (2 * A + (N - 1) * D)
lblOutput.Caption = F
End Sub
and I wanna notify or tell VB6 that I'm putting in a fraction, not an integer and uses that fraction to do operations.
NOTE: String Fraction to Value in VBA this doesn't answer my question... :D
Thank you for helping me, everyone! it's much appreciated.
There is no Application.Evaluate(...) in Vb6 like in VBA, so you have to do it like the "question" in "String Fraction to Value in VBA". Extract the logic to a function for re-use, and replace the Val(...) calls with the function for use.
Something like below would likely work, although you may want to provide better error handling in the obvious bad-math cases. I simply return zero and mark them with a comment.
Option Explicit
Private Sub btCalc_Click()
Dim A As Long, N As Long, D As Long, R As Long, F As Long
A = GetFrac(txtInitterm)
N = GetFrac(txtTermint)
D = GetFrac(txtFinterm)
R = GetFrac(txtTermint)
F = N / 2 * (2 * A + (N - 1) * D)
lblOutput.Caption = F
End Sub
Public Function GetFrac(ByVal S As String) As Double
GetFrac = 0 ' default return on error
If InStr(S, "/") = 0 Then GetFrac = Val(S): Exit Function
Dim P() As String, N As Double, D As Double
P = Split(S, "/")
If UBound(P) <> 1 Then Exit Function ' bad input -- multiple /'s
N = Val(P(0))
D = Val(P(1))
If D = 0 Then Exit Function ' div by 0
GetFrac = N / D
End Function
How can I remove multiline sections with Perl?
I have such wiki test code:
{|
|-
| colspan="2"|
: <math>
[\underbrace{\color{Red}4,2}_{4 > 2},5,1,7] \rightarrow
[2,\underbrace{\color{OliveGreen}4,5}_{4 < 5},1,7] \rightarrow
[2,4,\underbrace{\color{Red}5,1}_{5 > 1},7] \rightarrow
[2,4,1,\underbrace{\color{OliveGreen}5,7}_{5 < 7}]
</math>
|-
|
: <math>
[\underbrace{\color{OliveGreen}2,4}_{2 < 4},1,5,{\color{Blue}7}] \rightarrow
[2,\underbrace{\color{Red}4,1}_{4 > 1},5,{\color{Blue}7}] \rightarrow
[2,1,\underbrace{\color{OliveGreen}4,5}_{4 < 5},{\color{Blue}7}]
</math>
: <math>
[\underbrace{\color{Red}2,1}_{2 > 1},4,{\color{Blue}5},{\color{Blue}7}] \rightarrow
[1,\underbrace{\color{OliveGreen}2,4}_{2 < 4},{\color{Blue}5},{\color{Blue}7}]
</math>
: <math>
[\underbrace{\color{OliveGreen}1,2}_{1 < 2},{\color{Blue}4},{\color{Blue}5},{\color{Blue}7}]
</math>
|}
And I want to remove from this code all how to do it? I have done such code:
cat math-text.txt | perl -e 'while(<>) { s/<math>.+?<\/math>//gs; print $_; }'
It is not works but should since documentation explains that . will much new lines. How to do it?
The following is a python script which I use to extract all the mathematical formula from wikipedia dumps. Rather than using a multi-line regexp it scans for occurrences of <math> </math> and uses the position on the line to work out where the actual position on the line is and uses a finite state machine to find the actual equations, basically with two states determined by inEqn. It does a few other things like find the title and name space and attributes in the maths tags.
As dumps are in the order of 100MB using a line by line approach may well end up being more efficient than multi-line regexps.
import sys
import re
titleRE = re.compile('<title>(.*)</title>')
nsRE = re.compile('<ns>(.*)</ns>')
mathRE = re.compile('</?math(.*?)>')
pageEndRE = re.compile('</page>')
title =""
attr = ""
ns = -1
inEqn = 0
for line in sys.stdin:
m = titleRE.search(line)
if m :
title = m.group(1)
expression = ""
inEqn = 0
m = nsRE.search(line)
if m :
ns = m.group(1)
start = 0
pos = 0
m = mathRE.search(line,pos)
while m :
if m.group().startswith('<math'):
attr = m.group(1)
start = m.end()
pos = start
expression = ""
inEqn = 1
if m.group() == '</math>' :
end = m.start()
expression = ' '.join([expression,line[start:end]])
print title,'\t',attr,'\t',expression.lstrip().replace('<','<').replace('>','>').replace('&','&')
pos = m.end()
expression = ""
start = 0
inEqn = 0
m = mathRE.search(line,pos)
if start > 0 :
expression = line[start:].rstrip()
elif inEqn :
expression = ' '.join([expression,line.rstrip()])
Another option might be to consider an xml parser. A SAX or DOM based parser would be able to find the equations. This might be worth considering if you want to do more sophisticated analysis of the wiki-text.
In finding the values of x and y, if (x567) + (2yx5) = (71yx) ( all in base 8) I proceeded as under.
I assumed x=abc and y=def and followed.
(abc+010 def+101 110+abc 111+101)=(111 001 def abc) //adding ()+()=() and equating LHS=RHS.
abc=111-010=101 which is 5 in base 8 and then def=001-101 which is -4
so x=5 and y=-4
Now the Question is that the answer mentioned in my book is x=4 and y=3.
Is the above method correct.If so,then what's issue here ??
you can't compare the digits beginning with the most significant digit, because you don't know the carry from the digit below. Also a digit cannot have a negative value.
You can start with the least significant digit, because there is no carry:
7 + 5 = 14
so x = 4 with a carry of 1 at the next digit.
now you can rewrite your equation to:
(4567) + (2y45) = (71y4)
now you can look at the second least significant digit (the carry in mind):
6 + 4 + 1 (carry) = 13
so y = 3, also with a carry of 1.
the whole equation is:
(4567) + (2345) = (7134)
which is true for the octal system.
I noticed that Verilog rounds my real number results into integer results. For example when I look at simulator, it shows the result of 17/2 as 9. What should I do? Is there anyway to define something like a: output real reg [11:0] output_value ? Or is it something that has to be done by simulator settings?
Simulation only (no synthesis). Example:
x defined as a signed input and output_value defined as output reg.
output_value = ((x >>> 1) + x) + 5;
If x=+1 then output value has to be: 13/2=6.5.
However when I simulate I see output_value = 6.
Code would help, but I suspect your not dividing reals at all. 17 and 2 are integers, and so a simple statement like that will do integer division.
17 / 2 = 8 (not 9, always rounds towards 0)
17.0 / 2.0 = 8.5
In your second case
output_value = ((x >>> 1) + x) + 5
If x is 1, x >>> 1 is 0, not 0.5 because you've just gone off the bottom of the word.
output_value = ((1 >>> 1) + 1) + 5 = 0 + 1 + 5 = 6
There's nothing special about verilog here. This is true for the majority of languages.