How would I go about having multiple conditionals in an if statement?
Eg. The user is asked a set of questions by the program:
1.) Enter an altitude between 0 and 1000
(user types in data)
2.) Enter Velocity between 0 and 500
(user types in data)
3.) Enter Temperature between 0 and 200
(user types in data)
the program then prints back
altitude = user value
velocity = user value
temperature = user value //ignore those list numbers
I have set in my (.ads) file that each of these ranges has a criticle value.
I want to create an if statement that has multiple conditionals.
in pseudo : If velocity = critical velocity & temperature = critical temperature & altitude = critical altitude
then print ("some message")
else do nothing
The syntax of an if-statement is
if_statement ::=
if condition then
sequence_of_statements
{elsif condition then
sequence_of_statements}
[else
sequence_of_statements]
end if;
and the syntax of “condition” is
condition ::= boolean_expression
(that is, an expression which happens to be boolean); the syntax of “expression” is
expression ::=
relation {and relation} | relation {and then relation}
| relation {or relation} | relation {or else relation}
| relation {xor relation}
so your code would look like
if velocity = critical_velocity
and temperature = critical_temperature
and altitude = critical_altitude
then
print ("some message”);
else
null;
end if;
You can leave out the else clause, and you can say and then instead of plain and if for some reason you shouldn’t check the rest of the condition if the first part is already False. This is called short-circuit evaluation, and it’s not the default in Ada (it is in C).
if X /= 0 and Y / X > 2 then
evaluates Y / X even if X is 0.
In Ada you will use the and, or and not boolean operator:
if Velocity = Critical_Velocity
and Temperature = Critical_Temperature
and Altitude = Critical_Altitude
then
Ada.Text_IO.Put_Line ("Crash");
else
...
end if;
When evaluation order matters, you will use the and then or the or else syntax (otherwise the compiler can change order for optimization).
Expressions will be evaluated in the 'and then'/'or else' order.
if Velocity = Critical_Velocity
and then Temperature = Critical_Temperature
and then Altitude = Critical_Altitude
then
Ada.Text_IO.Put_Line ("Crash");
else
...
end if;
With an or else, you may write as follows:
if Velocity = Critical_Velocity
or else Temperature = Critical_Temperature
or else Altitude = Critical_Altitude
then
Ada.Text_IO.Put_Line ("Crash");
else
...
end if;
Beware that you cannot mix and and or together (because this leads to many confusion for developers).
You have to use parenthesis arround them if you do it.
if (Velocity = Critical_Velocity and Temperature = Critical_Temperature)
or else Altitude = Critical_Altitude
then
Ada.Text_IO.Put_Line ("Crash");
else
...
end if;
Related
I struggle to code an if statement in Pine.
The idea is that the stop-loss is tightened when a waring signal is triggered.
This tighter stop should replace the regular stop-loss and trailing-stop. I hoped is was a simple syntax error, but I can't seem to fix it by editing indents and spaces. Is there something more fundamental I'm overlooking?
Thanks for taking the time to read this!
longLossPerc = input(title="Long Stop Loss (%)", minval=0.0, step=0.1, defval=3.4) * 0.01
longTrailPerc = input(title="Long Trail Loss (%)", minval=0.0, step=0.1, defval=2.5) * 0.01
longTightStopPerc = input(title="Long K Loss (%)", minval=0.0, step=0.1, defval=0.6) * 0.01
longStopPrice = strategy.position_avg_price * (1 - longLossPerc)
longTrailPrice = strategy.position_avg_price * (1 - longTrailPerc)
longTightStop = high * (1 - longTightStopPerc)
stopValueLong = 0.0
if (Signal)
stopValueLong = longTightStop
else
stopValueLong = max(longStopPrice, longTailPrice)
else
0
You'll need something more or less of the following form, which supposes you have state variables to know the distinction between when you are in a trade or not:
var stopValueLong = 0.0
if (Signal)
stopValueLong := longTightStop
else if inTrade
stopValueLong := max(longStopPrice, longTailPrice)
else if closeTrade
stopValueLong := na
The := operator is important to assign values to your stopValueLong variable from with an if statement's local scopes. See:
https://www.tradingview.com/pine-script-docs/en/v4/language/Expressions_declarations_and_statements.html#variable-assignment
Does not display anything after data entry
program g
implicit none
real::q,n,s,z,q2,y,free_board,r,b,e,A,h,t
write(*,100)"pls insert discharge Q ="
read(*,*)q
write(*,100)"please insert Manning coefficient n ="
read(*,*)n
write(*,*)"please insert slope of the hydraulic channel ="
read(*,*)s
write(*,*)"please inset Z ="
read(*,*)z
write(*,*)"how much of b/y do you want?"
write(*,*)"if it not important right 2.5"
read(*,*)e
if(e<2.or.e>5)then
stop
end if
y=0
do
b=y*e
A=b+2*y*((1+Z**2)**(0.5))
R=((b+z*y)*y)/(b+(2*y*(1+z**2)**(0.5)))
h=(1/n)*(r**(2/3))*A*(s)**0.5
if( abs(h-q)<0.01) then
exit
end if
y=0.001+y
end do
free_board=0.2*y
h=free_board+y
t=b+2*y*z
write(*,100)"free board="
write(*,*) free_board
write(*,100)"y="
write(*,*)y
write(*,100)"b="
write(*,*)b
write(*,100)"T="
write(*,100)t
100 format(A)
end program g
this not work and not show anything after enter data
For sure the line write(*,100)t will give you a wrong output, since "t" it is a real, not a string. Please change it to write(*,*).
With all inputs equal to 1.0 and by putting e=2.5, I see these output (at screen):
free board = 3.7200041E-02
y = 0.1860002
b = 0.4650005
T = 0.8370009
If you don't see outputs, maybe you are choosing wrong "e" values (less than 2 or more than 5).
I'm writing a VBScript that searches the Active Directory for a computer object. If the object does not exist or exists and is in the correct OU, then it should run a separate script that creates/joins the computer to the AD.
ObjExist_CorrectOU_7 = Null
ObjExist_CorrectOU_10 = Null
If compare = True Then
Win7_OU = "OU=DisallowRDP,OU=64Bit,OU=Win8"
Win10_OU = "OU=DisallowRDP,OU=64Bit,OU=Win10"
For x = 16 To 46
If Asc(Mid(objRS.Fields("distinguishedName"), x, 1)) = Asc(Mid(Win7_OU, (x - 15), 1)) Then
ObjExist_CorrectOU_7 = True
Else
ObjExist_CorrectOU_7 = False
End If
Next
For y = 16 To 46
If Asc(Mid(objRS.Fields("distinguishedName"), y, 1)) = Asc(Mid(Win10_OU, (y - 15), 1)) Then
ObjExist_CorrectOU_10 = True
Else
ObjExist_CorrectOU_10 = False
End If
Next
End If
If ObjExist_CorrectOU_7 = True Then
WScript.Echo "TRUE"
End If
Dim objShell
Set objShell = WScript.CreateObject("WScript.Shell")
filename = "C:\programdata\dell\kace\k2000_deployment_info.conf"
Win7_Deployment = "deployment_name=Windows 7 x64 with SP1, join AD"
Win10_Deployment = "deployment_name=Development Windows 10 (x64), join AD"
Set fso = CreateObject("Scripting.FileSystemObject")
Set f = fso.OpenTextFile(filename)
Do While Not f.AtEndOfStream
If ((f.ReadLine = Win7_Deployment) Or ((f.ReadLine = Win7_Deployment) And (ObjExist_CorrectOU_7 = True))) Then
WScript.Echo "IT WORKED!"
'objShell.Run "JoinAD_Win7.vbs"
Exit Do
End If
On Error Resume Next
Loop
f.Close
Set g = fso.OpenTextFile(filename)
Do While Not f.AtEndOfStream
If ((g.ReadLine = Win10_Deployment) Or ((g.ReadLine = Win10_Deployment) And (ObjExist_CorrectOU_10 = True))) Then
'objShell.Run "JoinAD_Win10.vbs"
WScript.Echo "IT WORKED AGAIN!"
Exit Do
End If
On Error Resume Next
Loop
g.Close
Set objShell = Nothing
The problem I'm running into is that the two If..Then statements execute every time, even though I know the conditions are absolutely NOT being met.
Does it have to do with my use of Or and And?
Your question does not satisfy Minimal, Complete, and Verifiable example criteria. However, at first sight: read On Error Statement and ReadLine Method and Working with Files documentation.
Do While Not f.AtEndOfStream
''' ↓ this `ReadLine` reads every uneven line i.e. the 1st, 3rd, 5th, …
If ((f.ReadLine = Win7_Deployment) Or ((f.ReadLine = Win7_Deployment) And (ObjExist_CorrectOU_7 = True))) Then
''' this one reads every even line ↑ i.e. the 2nd, 4th, 6th, …
WScript.Echo "IT WORKED!"
'objShell.Run "JoinAD_Win7.vbs"
Exit Do
End If
On Error Resume Next ' this causes that script continues on line next to IF … THEN
' in case of uneven records in file.
' i.e. runtimme error "Input past end of file"
Loop
Use something like
Do While Not f.AtEndOfStream
sReadLine = f.ReadLine
If ((sReadLine = Win7_Deployment) Or ((sReadLine = Win7_Deployment) And (ObjExist_CorrectOU_7 = True))) Then
WScript.Echo "IT WORKED!"
'objShell.Run "JoinAD_Win7.vbs"
Exit Do
End If
''' get rid of `On Error Resume Next` statement at all
Loop
And what about Do While Not f.AtEndOfStream followed up by g.ReadLine? Use either f or g (the same TextStream object in both)…
When querying a model, with model_completion set to false in Z3_model_eval: What will Z3 return to signify that an interpretation is a "don't care"?
In case someone wants to suggest that: I guess it's probably not the return value of the function Z3_model_eval, since the z3++.h file (the C++ API) contains the lines:
Z3_bool status = Z3_model_eval(ctx(), m_model, n, model_completion, &r);
check_error();
if (status == Z3_FALSE)
throw exception("failed to evaluate expression");
In general: How does Z3 signify that a certain constant is a "don't care" in a model?
For true don't-cares, the model will not assign any value. Consequently, calls to eval or Z3_model_eval with model_completion set to false will keep the original constants untouched and only replace those for which a model value is assigned (and it will potentially simplify the expression). Here's an example:
context c;
expr e = c.int_const("x");
solver s(c);
s.add(e == e);
model m = s.get_model();
std::cout << m.eval(e, false) << std::endl;
std::cout << m.eval(e, true) << std::endl;
Note that the first line of output prints x, i.e., the original expression is untouched, while the call to eval with model_completion set to true will print 0.
This is not homework, but an old exam question. I am curious to see the answer.
We are given an alphabet S={0,1,2,3,4,5,6,7,8,9,+}. Define the language L as the set of strings w from this alphabet such that w is in L if:
a) w is a number such as 42 or w is the (finite) sum of numbers such as 34 + 16 or 34 + 2 + 10
and
b) The number represented by w is divisible by 3.
Write a regular expression (and a DFA) for L.
This should work:
^(?:0|(?:(?:[369]|[147](?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]0*(?:\+?(?:0\
+)*[369]0*)*\+?(?:0\+)*[258])*(?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]|0*(?:
\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147])|[
258](?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0
\+)*[147])*(?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]|0*(?:\+?(?:0\+)*[369]0*)
*\+?(?:0\+)*[258]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]))0*)+)(?:\+(?:0|(?:(?
:[369]|[147](?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]0*(?:\+?(?:0\+)*[369]0*)
*\+?(?:0\+)*[258])*(?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]|0*(?:\+?(?:0\+)*
[369]0*)*\+?(?:0\+)*[147]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147])|[258](?:0*(?
:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147])*
(?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]|0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)
*[258]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]))0*)+))*$
It works by having three states representing the sum of the digits so far modulo 3. It disallows leading zeros on numbers, and plus signs at the start and end of the string, as well as two consecutive plus signs.
Generation of regular expression and test bed:
a = r'0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*'
b = r'a[147]'
c = r'a[258]'
r1 = '[369]|[147](?:bc)*(?:c|bb)|[258](?:cb)*(?:b|cc)'
r2 = '(?:0|(?:(?:' + r1 + ')0*)+)'
r3 = '^' + r2 + r'(?:\+' + r2 + ')*$'
r = r3.replace('b', b).replace('c', c).replace('a', a)
print r
# Test on 10000 examples.
import random, re
random.seed(1)
r = re.compile(r)
for _ in range(10000):
x = ''.join(random.choice('0123456789+') for j in range(random.randint(1,50)))
if re.search(r'(?:\+|^)(?:\+|0[0-9])|\+$', x):
valid = False
else:
valid = eval(x) % 3 == 0
result = re.match(r, x) is not None
if result != valid:
print 'Failed for ' + x
Note that my memory of DFA syntax is woefully out of date, so my answer is undoubtedly a little broken. Hopefully this gives you a general idea. I've chosen to ignore + completely. As AmirW states, abc+def and abcdef are the same for divisibility purposes.
Accept state is C.
A=1,4,7,BB,AC,CA
B=2,5,8,AA,BC,CB
C=0,3,6,9,AB,BA,CC
Notice that the above language uses all 9 possible ABC pairings. It will always end at either A,B,or C, and the fact that every variable use is paired means that each iteration of processing will shorten the string of variables.
Example:
1490 = AACC = BCC = BC = B (Fail)
1491 = AACA = BCA = BA = C (Success)
Not a full solution, just an idea:
(B) alone: The "plus" signs don't matter here. abc + def is the same as abcdef for the sake of divisibility by 3. For the latter case, there is a regexp here: http://blog.vkistudios.com/index.cfm/2008/12/30/Regular-Expression-to-determine-if-a-base-10-number-is-divisible-by-3
to combine this with requirement (A), we can take the solution of (B) and modify it:
First read character must be in 0..9 (not a plus)
Input must not end with a plus, so: Duplicate each state (will use S for the original state and S' for the duplicate to distinguish between them). If we're in state S and we read a plus we'll move to S'.
When reading a number we'll go to the new state as if we were in S. S' states cannot accept (another) plus.
Also, S' is not "accept state" even if S is. (because input must not end with a plus).