I have ran in to a problem while writing a program for school that converts a string like abc to bcd, a becomes b and b becomes c and you can see the rest.
For i = 0 To length - 1
If (Asc(justatext.Substring(i, 1)) >= 65 And Asc(justatext.Substring(i, 1)) <= 90) Then
Asc(justatext.Substring(i, 1) = (Asc(justatext.Substring(i, 1) + 1)))
answer &= justatext.Substring(i, 1)
End If
Next
This is in a function and I return the value answer, but I always get an invalid cast exception. Is there a way you can do this with ansi codes?.
your problem can be found in the brackets, you have quite a lot of them, and I think you confused yourself with them.
I've tripped down your code and removed the brackets that aren't needed:
For i = 0 To justatext.Length - 1
If Asc(justatext.Substring(i, 1)) >= 65 And Asc(justatext.Substring(i, 1)) <= 90 Then
answer &= Chr(Asc(justatext.Substring(i, 1)) + 1)
End If
Next
Watch out: this code will only work for capital letters..
Related
I have come across a issue while working in VBA . I'm supposed to write program that is Numerical integration of trapeze method (I'm not sure if It is how it's called in English) of function 100*x^99 lower limit = 0 upper limit = 1 . Cells (j,5) contains numbers (10,30,100,300,1000,3000,10000) - amount of point splits . Code seems to work but given wrong results , for amount of splits it should be around
10 - 5.000295129200607
30 - 1.786588019299606
100 - 1.0812206997600746
300 - 1.0091505687770146
1000 - 1.0008248693208752
3000 - 1.0000916650530287
10000 - 1.000008249986933
Function F(x)
F = 100 * (x ^ 99)
End Function
Sub calka()
Dim n As Single
Dim xp As Single
Dim dx As Single
Dim xk As Single
Dim ip As Single
Dim pole As Single
xp = 0
xk = 1
For j = 5 To 11
n = Cells(j, 5)
dx = (xk - xp) / n
pole = 0
For i = 1 To n - 1
pole = pole + F(xp + i * dx)
Next i
pole = pole + ((F(xp) + F(xk)) / 2)
pole = pole * dx
Worksheets("Arkusz1").Cells(j, 7) = pole
Next j
End Sub
I tried to implement same code in java and c++ and it worked flawlessly but VBA always gives me wrong results , I'm not sure if it's rounds at some point and I can disable in settings or my code is just not written right .
Apologies for low clarity It's hard for me to translate mathematic to English.
Use Doubles rather than Singles
http://www.techrepublic.com/article/comparing-double-vs-single-data-types-in-vb6/
I've encountered a problem with a program I'm writing for school. I need to verify credit card numbers using the Luhn Algorithm, however I'm having some difficulty in getting the logic of the algorithm to work correctly. I believe I know where the problem is, but I'm unable to fix it.
I believe the problem is here:
For i = 0 To cardInput.Text.Length - 2 Step -2
Dim x = (i * 2)
If x > 9 Then
x = x - 9
End If
oddTotal += x
Next
'Sum of undoubled digits
For i = 0 To cardLength - 1 Step -2
evenTotal += i
Next
total = oddTotal + evenTotal
checkSum = total
infoOutput.Items.Add("CheckDigit: " & checkDigit)
infoOutput.Items.Add("CheckSum :" & checkSum)
'Verify that the card is valid by the Mod 10 (Lund algoritm)
If checkSum = checkDigit Or checkSum = 0 Then
valid = True
Else
valid = False
End If
If it's needed, the rest of my project can be seen here
My code doesn't seem to start at the last digit and take every other digit back to the beginning to be doubled. Is the Step -2 operator incorrect here? What am I doing wrong?
There are several problems here. Particularly:
If you want a loop to count backwards, you have to start at the higher index and end at the lower one. So:
For i = cardInput.Text.Length - 2 To 0 Step -2
Then, instead of using i directly, you should use the i-ith digit:
Dim x = Val(cardInput.Text(i))
The same applies to your sum of evens.
If you want to check if the last digit is zero, use the Mod operator:
valid = (checkSum Mod 10 = 0)
I saw this pseudo-code on another stackoverflow question found here Split a string to a string of valid words using Dynamic Programming.
The problem is a dynamic programming question to see if an input string can be split into words from a dictionary.
The third line, means to set an array b of size [N+1] to all false values? I'm pretty sure about that. But what I am really not sure about is the fifth line. Is that a for-loop or what? I feel like pseudo-code saying 'for i in range' would only have 2 values. What is that line saying?
def try_to_split(doc):
N = len(doc)
b = [False] * (N + 1)
b[N] = True
for i in range(N - 1, -1, -1):
for word starting at position i:
if b[i + len(word)]:
b[i] = True
break
return b
It's confusing syntax, and I'm pretty sure there's a mistake. It should be:
for i in range(N - 1, 0, -1) //0, not -1
which I believe means
for i from (N - 1) downto 0 //-1 was the step, like i-- or i -= 1
This makes sense with the algorithm, as it simply starts at the end of the string, and solves each trailing substring until it gets to the beginning. If b[0] is true at the end, then the input string can be split into words from the dictionary. for word starting at position i just checks all words in the dictionary to see if they start at that position.
If one wants to be able to reconstruct a solution, they can change b to an int array, initialize to 0s, and change the if to this:
if b[i + len(word)] != 0
b[i] = i + len(word) //len(word) works too
break
I'm working on some Hoare logic and I am wondering whether my approach is the right one.
I have the following program P:
s = 0
i = 1
while (i <= n) {
s = s + i
i = i + 1
}
It should satisfy the hoare triple {n >= 0}P{s = n*(n+1)/2} (so it just takes the sum). Now, initially I had |s = i*(i-1)/2| as my invariant, which works fine. However, I had a problem from going to the end of my loop, to my desired postcondition. Because for the impliciation
|s = i*(i-1)/2 & i > n|
=>
| s = n * (n+1) / 2 |
to hold, I need to prove that i is n+1, and not just any i bigger than n. So what I thought of is to add a (i <= n + 1) to the invariant, so that it becomes :
|s = i * (i-1)/2 & i <= n+1|
Then I can prove the program so I think it should be correct.
Nonetheless, I find the invariant to be a bit, less "invariantly" :). And not like anything I've seen in the course or in the exercises so far, so I was wondering if there was a more elegant solution here?
So what I thought of is to add a (i <= n + 1) to the invariant, so that it becomes :
|s = i * (i-1)/2 & i <= n+1|
Nonetheless, I find the invariant to be a bit, less "invariantly" :). And not like anything I've seen in the course or in the exercises so far, so I was wondering if there was a more elegant solution here?
Nope, given the way the code is written, that's exactly the way to go. (I can tell from experience since I've been teaching Hoare logic during several semesters in two different courses and since it's part of my graduate studies.)
Using i <= n is common practice when programming. In your particular program, you could just as well have written i != n+1 instead, in which case your first invariant (which indeed looks cleaner) would have sufficed since you get
| s=i*(i-1)/2 & i=n+1 |
=>
| s=n*(n+1)/2 |
which evidently holds.
There is another way to reason,given a more appropriate invariant (and other code)...searh n for final value of i...
I : s = i*(i+1)/2 and 0 <= i <=n
B : i < n
Now,evidently you have for post condition:
I and i >= n => s = i*(i+1)/2 and i=n => s = n*(n+1)/2
The code now becomes
s = 0
i = 0
while (i < n) {
s = s + (i+1)
i = i + 1
}
The invariant holds at init and keeps after each loop,since rewriting I as 2s=i*(i+1) we have to proof
I and i<n => 2(s + (i+1)) = (i+1)*(i+2)
2(s + (i+1) )=
2s + 2(i+1) =
i*(i+1) + 2(i+1)= (since I holds)
(i+1)(i+2)
Qed.
I got this question in a programming test. Do you think this question is even correct? Look at the answer choices. (2^x means 2 raised to x)
Consider the following pseudocode.
x := 1;
i := 1;
while (x >= 1000)
begin
x := 2^x;
i := i + 1;
end;
What is the value of i at the end of the pseudocode?
a) 4
b) 5
c) 6
d) 7
e) 8
I am sure that the value of i will 1. I told the examiner of the discrepancy and he advised me the leave the question unanswered if I felt it was incorrect. What else could I have done?
1
X < 1000, so it doesn't enter the while.
Or there is an error in the Question (and X should be <= 1000 and not >=1000)
If it's <= 1000 it should be 5:
2 - 4 - 16 - 65K
2 - 3 - 4 - 5
This question tests two things:
can you read code
can you communicate / interact
Since you asked about the discrepancy, you showed 1. to be true. I'm not so sure if you passed 2, it depends too much on the situation / expectations.
I believe I would have left a note on the answer sheet stating 'none of the given'.
Not an easy situation!
As written, the answer would be 1.
Had the test on the while been reversed (i.e. x < 1000), then the series is:
At the end of each loop iteration
i = 2, x = 2
i = 3, x = 2^2 = 4
i = 4, x = 2^4 = 16
i = 5, x = 2^16 = 65,536
So i would be 5
If I where you, I would say that it is none of the above and basically say that since x is less than 1000 when the while loop starts, the value of i is never modified. I think that it is a bad thing to leave blank answers in a quiz, it is always better to write something which you think is relevant. If you think that there is a mistake, you can always state an assumption before stating your answer, so in this case you can either say that the while loop never works, or else, you explicitly state an assumption, in this case, it would be something like "Assuming that there is a mistake in the question, and that "while (x >= 1000)" should in fact be "while (x <= 1000)"...
Then, you proceed with your working.