My code works but keeps giving me access violation error.
" Access violation at address 00440690B in module. read of address 01F62C42."
what is wrong? and how can I make it work?
The second loop does nothing. please help!
Var
num1, num2, k : Integer;
LL : string;
begin
LL := ' ';
num1 := 4;
num2 := 4;
for k := 1 to 7 do
begin
LL[num1] := '*';
LL[num2] := '*';
redt.Lines.Add(LL);
num1 := num1 +1;
num2 := num2 -1;
end;
for k := 1 to 3 do
redt.Lines.Add(' * ');
end;
My code works.
No, it does not. You are accessing elements of LL that are out-of-bounds. In the final iteration of the first loop, num1 has value 10, and num2 has value -2. Both of these are out-of-bounds when used as indices for LL. Valid indices for LL are 1 to 7. So I guess that the first loop should run for 1 to 4.
If you would enable range checking in the compiler options, the compiler would be able to tell you this. I cannot stress enough the importance of using range checking. Use it, and let the compiler find your defects.
Related
Having a problem whenever i want when typing in 2 4 6 8 10 etc i want to have the answer to come out as Even but cant seem to to find the solutioin
if (Edit1.Text = '2' ) then
Edit2.Text := 'Even'
else
Edit2.Text := 'Odd'
Tryed to divide 2 but it always fails.
Convert the text to an integer and use the Odd function:
if Odd(StrToInt(Text)) then
// the value is odd
You need to first convert the text in the textbox to a numeric type like Integer, and then compare that value by modulus of 2 (the remainder of integer division):
var
value: Integer;
begin
...
value := StrToInt(Edit1.Text);
if ((value mod 2) = 0) then
Edit2.Text := 'Even'
else
Edit2.Text := 'Odd';
...
end;
There is nice and short explanation on mod operator here:
FreePascal Wiki on Mod:
mod (modulus) divides two numbers and returns only the remainder that
is a whole number. For instance, the expression a:= 13 mod 4; would
evaluate to 1 (a=1), while b := 12 mod 4; would evaluate to 0 (b=0).
I'm coding this function where if a string differs only by one character, returns the distinct characters position, if they're right the same is supposed to return -1 and -10 in the case they differ by more than 1 character.
Just for giving and example, '010' and '110' or '100' and '110' works good, returning 0 and 1 each...
However, when I try with '100' and '101'or with '110' and '111' I get a result of -1 when it should be 2! I've done the desktop testing and I can't just see the mistake.
function combine (m1, m2 : string) : integer;
var
dash : integer;
distinct : integer;
i : integer;
begin
distinct := 0;
dash := -1;
for i := 0 to Length(m1)-1 do
begin
if m1[i] <> m2[i] then
begin
distinct := distinct+1;
dash := i;
if distinct > 1 then
begin
result:= -10;
exit;
end;
end;
end;
result := dash;
end;
I'm always getting same length strings,
What am I doing wrong?
The main problem is that Delphi strings are 1-based. Your loop needs to run from 1 to Length(m1).
If you enabled range checking in the compiler options, then the compiler would have raised an error at runtime which would have led you to the fault. I cannot stress enough that you should enable range checking. It will lead to the compiler finding errors in your code.
Note also that this means that the returned values will also be 1-based. So an input of '100', '101' will give the result 3 since that is the index of the first difference.
You should also check that m1 and m2 are the same length. If not raise an exception.
Another tip. The idiomatic way to increment a variable by 1 is like so:
inc(distinct);
If you want to increment by a different value write:
inc(distinct, n);
So, I would write the function like this:
function combine(const m1, m2: string): integer;
var
i: integer;
dash: integer;
distinct: integer;
begin
if Length(m1)<>Length(m2) then begin
raise EAssertionFailed.Create('m1 and m2 must be the same length');
end;
distinct := 0;
dash := -1;
for i := 1 to Length(m1) do
begin
if m1[i] <> m2[i] then
begin
inc(distinct);
dash := i;
if distinct > 1 then
begin
result := -10;
exit;
end;
end;
end;
result := dash;
end;
I have a BIG problem here and do not even know how to start...
In short explanation, I need to know if a number is in a set of results from a random combination...
Let me explain better: I created a random "number" with 3 integer chars from 1 to 8, like this:
procedure TForm1.btn1Click(Sender: TObject);
var
cTmp: Char;
sTmp: String[3];
begin
sTmp := '';
While (Length(sTmp) < 3) Do
Begin
Randomize;
cTmp := IntToStr(Random(7) + 1)[1];
If (Pos(cTmp, sTmp) = 0) Then
sTmp := sTmp + cTmp;
end;
edt1.Text := sTmp;
end;
Now I need to know is some other random number, let's say "324" (example), is in the set of results of that random combination.
Please, someone can help? A link to get the equations to solve this problem will be enough...
Ok, let me try to add some useful information:
Please, first check this link https://en.wikipedia.org/wiki/Combination
Once I get some number typed by user, in an editbox, I need to check if it is in the set of this random combination: S = (1..8) and k = 3
Tricky, hum?
Here is what I got. Maybe it be usefull for someone in the future. Thank you for all people that tried to help!
Function IsNumOnSet(const Min, Max, Num: Integer): Boolean;
var
X, Y, Z: Integer;
Begin
Result := False;
For X := Min to Max Do
For Y := Min to Max Do
For Z := Min to Max Do
If (X <> Y) and (X <> Z) and (Y <> Z) Then
If (X * 100 + Y * 10 + Z = Num) Then
Begin
Result := True;
Exit;
end;
end;
You want to test whether something is a combination. To do this you need to verify that the putative combination satisfies the following conditions:
Each element is in the range 1..N and
No element appears more than once.
So, implement it like this.
Declare an array of counts, say array [1..N] of Integer. If N varies at runtime you will need a dynamic array.
Initialise all members of the array to zero.
Loop through each element of the putative combination. Check that the element is in the range 1..N. And increment the count for that element.
If any element has a count greater than 1 then this is not a valid combination.
Now you can simplify by replacing the array of integers with an array of booleans but that should be self evident.
You have your generator. Once your value is built, do something like
function isValidCode( Digits : Array of Char; Value : String ) : Boolean;
var
nI : Integer;
begin
for nI := 0 to High(Digits) do
begin
result := Pos(Digits[nI], Value ) > 0;
if not result then break;
end;
end;
Call like this...
isValidCode(["3","2","4"], RandomValue);
Note : it works only because you have unique digits, the digit 3 is only once in you final number. For something more generic, you'll have to tweak this function. (testing "3","3","2" would return true but it would be false !)
UPDATED :
I dislike the nested loop ^^. Here is a function that return the nTh digit of an integer. It will return -1 if the digits do not exists. :
function TForm1.getDigits(value : integer; ndigits : Integer ) : Integer;
var
base : Integer;
begin
base := Round(IntPower( 10, ndigits-1 ));
result := Trunc( value / BASE ) mod 10;
end;
nDigits is the digits number from right to left starting at 1. It will return the value of the digit.
GetDigits( 234, 1) returns 4
GetDigits( 234, 2) returns 3
GetDigits( 234, 3) returns 2.
GetDigits( 234, 4) returns 0.
Now this last function checks if a value is a good combination, specifying the maxdigits you're looking for :
function isValidCombination( value : integer; MinVal, MaxVal : Integer; MaxDigits : Integer ) : Boolean;
var
Buff : Array[0..9] of Integer;
nI, digit: Integer;
begin
ZeroMemory( #Buff, 10*4);
// Store the count of digits for
for nI := 1 to MaxDigits do
begin
digit := getDigits(value, nI);
Buff[digit] := Buff[digit] + 1;
end;
// Check if the value is more than the number of digits.
if Value >= Round(IntPower( 10, MaxDigits )) then
begin
result := False;
exit;
end;
// Check if the value has less than MaxDigits.
if Value < Round(IntPower( 10, MaxDigits-1 )) then
begin
result := False;
exit;
end;
result := true;
for nI := 0 to 9 do
begin
// Exit if more than One occurence of digit.
result := Buff[nI] < 2 ;
if not result then break;
// Check if digit is present and valid.
result := (Buff[nI] = 0) or InRange( nI, MinVal, MaxVal );
if not result then break;
end;
end;
Question does not seem too vague to me,
Maybe a bit poorly stated.
From what I understand you want to check if a string is in a set of randomly generated characters.
Here is how that would work fastest, keep a sorted array of all letters and how many times you have each letter.
Subtract each letter from the target string
If any value in the sorted int array goes under 0 then that means the string can not be made from those characters.
I made it just work with case insensitive strings but it can easily be made to work with any string by making the alphabet array 255 characters long and not starting from A.
This will not allow you to use characters twice like the other example
so 'boom' is not in 'b' 'o' 'm'
Hope this helps you.
function TForm1.isWordInArray(word: string; arr: array of Char):Boolean;
var
alphabetCount: array[0..25] of Integer;
i, baseval, position : Integer;
s: String;
c: Char;
begin
for i := 0 to 25 do alphabetCount[i] := 0; // init alphabet
s := UpperCase(word); // make string uppercase
baseval := Ord('A'); // count A as the 0th letter
for i := 0 to Length(arr)-1 do begin // disect array and build alhabet
c := UpCase(arr[i]); // get current letter
inc(alphabetCount[(Ord(c)-baseval)]); // add 1 to the letter count for that letter
end;
for i := 1 to Length(s) do begin // disect string
c := s[i]; // get current letter
position := (Ord(c)-baseval);
if(alphabetCount[position]>0) then // if there is still latters of that kind left
dec(alphabetCount[position]) // delete 1 to the letter count for that letter
else begin // letternot there!, exit with a negative result
Result := False;
Exit;
end;
end;
Result := True; // all tests where passed, the string is in the array
end;
implemented like so:
if isWordInArray('Delphi',['d','l','e','P','i','h']) then Caption := 'Yup' else Caption := 'Nope'; //yup
if isWordInArray('boom',['b','o','m']) then Caption := 'Yup' else Caption := 'Nope'; //nope, a char can only be used once
Delphi rocks!
begin
Randomize; //only need to execute this once.
sTmp := '';
While (Length(sTmp) < 3) Do
Begin
cTmp := IntToStr(Random(7) + 1)[1]; // RANDOM(7) produces # from 0..6
// so result will be '1'..'7', not '8'
// Alternative: clmp := chr(48 + random(8));
If (Pos(cTmp, sTmp) = 0) Then
sTmp := sTmp + cTmp;
IF SLMP = '324' THEN
DOSOMETHING; // don't know what you actually want to do
// Perhaps SET SLMP=''; to make sure '324'
// isn't generated?
end;
edt1.Text := sTmp;
end;
For a registration code I want to convert an Int64 to base30 (30 so that only uppercase characters and excluding 0,O,I,1,etc.) and back.
This is not too difficult using functions like:
const
Base = 30;
Base30CharSet = '23456789ABCDEFGHJKLMNPRSTVWXYZ';
function ConvertIntToBase30(ANumber: Int64): string;
begin
if(ANumber = 0) then
Result := Copy(Base30CharSet, 1, 1)
else begin
Result := '';
while(ANumber <> 0) do begin
Result := Copy(Base30CharSet, (ANumber mod Base)+1, 1) + Result;
ANumber := ANumber div Base;
end;
end;
end;
function ConvertBase30ToInt(ANumber: string): Int64;
var
i: integer;
begin
Result := 0;
for i := 1 to Length(ANumber) do begin
Result := Result + (Pos(ANumber[i], Base30CharSet)-1);
if(i < Length(ANumber)) then
Result := Result * Base;
end;
end;
The snag is that I am interested in the Int64's bits, so I could be dealing with a number like $FFFFFFFFFFFFFFFF = -1.
To work around this I thought I would store and remove the sign (abs()) and include the sign as an extra character appended to the base30 result. The problem the occurs at the lower limit of Int64 as calling abs(-9223372036854775808) results in an overflow.
Does anyone have a solution or better algorithm to solve this problem?
The way to deal with it is having a character to indicate it is a negative number so that you can decode back. For negative number, just flip the bit from 1 to 0 and remove the sign bit before encoding and when decode, do a flip back and add the sign bit. Below is working codes
function InvertIntOff(const ANumberL, ANumberH: Integer): Int64;
asm
XOR EAX,$FFFFFFFF
XOR EDX,$FFFFFFFF
end;
function InvertIntOn(const ANumberL, ANumberH: Integer): Int64;
asm
XOR EAX,$FFFFFFFF
XOR EDX,$FFFFFFFF
OR EDX,$80000000
end;
function ConvertIntToBase(ANumber: Int64): string;
const
CBaseMap: array[0..31] of Char = (
'2','3','4','5','6','7','8','9', //0-7
'A','B','C','D','E','F','G','H', //8-15
'J','K','L','M','N', //16-20
'P','Q','R','S','T','U','V','X','W','Y','Z'); //21-31
var
I: Integer;
begin
SetLength(Result, 15);
I := 0;
if ANumber < 0 then
begin
Inc(I);
Result[I] := '1';
ANumber := InvertIntOff(ANumber and $FFFFFFFF, (ANumber and $FFFFFFFF00000000) shr 32);
end;
while ANumber <> 0 do
begin
Inc(I);
Result[I] := CBaseMap[ANumber and $1F];
ANumber := ANumber shr 5;
end;
SetLength(Result, I);
end;
function ConvertBaseToInt(const ABase: string): Int64;
var
I, Index: Integer;
N: Int64;
begin
Result := 0;
if Length(ABase) > 0 then
begin
if ABase[1] = '1' then
Index := 2
else
Index := 1;
for I := Index to Length(ABase) do
begin
case ABase[I] of
'2'..'9':
N := Ord(ABase[I]) - Ord('2');
'A'..'H':
N := Ord(ABase[I]) - Ord('A') + 8;
'J'..'N':
N := Ord(ABase[I]) - Ord('J') + 16;
'P'..'Z':
N := Ord(ABase[I]) - Ord('P') + 21;
else
raise Exception.Create('error');
end;
if I > Index then
Result := Result or (N shl ((I - Index) * 5))
else
Result := N;
end;
if ABase[1] = '1' then
Result := InvertIntOn(Result and $FFFFFFFF, (Result and $FFFFFFFF00000000) shr 32);
end;
end;
procedure TestBase32;
var
S: string;
begin
S := ConvertIntToBase(-1);
ShowMessage(S + ' / ' + IntToStr(ConvertBaseToInt(S)) + ' ? -1');
S := ConvertIntToBase(-31);
ShowMessage(S + ' / ' + IntToStr(ConvertBaseToInt(S)) + ' ? -31');
S := ConvertIntToBase(1);
ShowMessage(S + ' / ' + IntToStr(ConvertBaseToInt(S)) + ' ? 1');
S := ConvertIntToBase(123456789);
ShowMessage(S + ' / ' + IntToStr(ConvertBaseToInt(S)) + ' ? 123456789');
S := ConvertIntToBase(-123456789);
ShowMessage(S + ' / ' + IntToStr(ConvertBaseToInt(S)) + ' ? -123456789');
end;
I think you are almost there by considering abs()...
But rather than using abs() why not simply ignore the sign for processing the value of the Int64 itself ? As far as I can tell, you are in fact already doing this so only one minor addition is needed to the encoding routine:
if aNumber < 0 then
// negative
else
// positive;
The only problem then is the LOSS of sign information in the resulting Base30 string. So treat that as a separate problem to be solved using the new information gained from the aNumber < 0 test...
I see you have excluded all chars that could be confused for 0 or 1 but have also excluded 0 and 1 themselves. You could therefore use 0 and 1 to indicate positive or negative (or vice versa).
Depending on the purpose of these routines, the placement of the 0/1 in the result could be entirely arbitrary (if you wished to obfuscate things and make the placement of the 0/1 random rather than a consistent lead/trail character).
When encoding simply drop a sign indicator into the result string at random, and when decoding handle the 0/1 character whenever as the sign marker it is encountered, but skipped for the purposes of decoding the value.
Of course, if obfuscation is not an issue then simply consistently pre or post fix the sign indicator.
You could even simply choose to use '1' to indicate negative and the LACK of a '1' to indicate/assume positive (this would simplify the zero value case a little I think)
The easy answer is to turn range checking off, even just for the method that you're calling abs in.
If you don't care about an extra char or two you could split the int64 into words or dwords and string those together. I would be more tempted to go to base32 and use bit shifts for speed and ease of use. Then your encoding becomes
Base32CharSet[(ANumber shr 5) % 32]
and a similar pos() based approach for the decode.
I'm defecting from C# to Delphi 2009, I'm liking it so far very much.
I wrote a binary search procedure, which works fine. I added a simple if-else statement at the end of my proc and it just doesn't fire! I can't see anything wrong with it and am embarrassed to have to say I am stuck. Please help!
procedure BinSearch;
var
min,max,mid, x: integer;
A : array[0..4] of integer;
rslt : integer;
begin
writeln('binary search');
A[0] := 34; A[1] := 65; A[2] := 98; A[3] := 123; A[4] := 176;
listarray(a);
x := 62;
min := 0;
max := 4;
repeat
begin
mid := (min + max) div 2;
if x > A[mid] then
min := mid + 1
else
max := mid - 1;
end;
until (A[mid] = x) or (min > max);
writeln(mid);
writeln(a[mid]);
if A[mid] = x then
rslt := mid
else
rslt := not mid;
if 54 = 65 then
rslt := mid
else
rslt := not mid;
end;
It's the if A[mid] = x then one that won't fire. when debugging neither true or false branches fire, the debugger just skips straight over them. Also the if 54 = 65 then which is just a test does the same.
The if inside my repeat loop works fine though.
If I copy the problem if statement into a mini test proc, and then call the proc it works, so it makes me think it's something else in the proc like a missing ; causing something strange to happen but I cannot see it. Please help!
The Delphi compiler is pretty smart, and it will happily remove unused code. When I compile your code I get compiler hints saying "Value assigned to 'rslt' never used". Since the value is never used, the compiler just skips over those statements.
If you add a Writeln(rslt); to the end of your procedure, you will find that the debugger will now trace through your if statement.
It could be that the debugger is just skipping over those statements even though they are actually running. Make sure that all of the options are turned on in the debugging options. In Delphi 7, they are under Project\Options under the Compiler tab.
The "Begin" statement just after the "Repeat" statement shouldn't be there. A "Repeat" doesn't use a begin. I would remove it just to be sure that it doesn't cause any problems.
"rslt" is not used. Therefore Delphi optimizes it out.
Obviously, you want to return your result. So change your declaration to:
procedure BinSearch(var rslt: integer);
or better, make it a function:
function BinSearch: integer;
and at the end put in:
Result := rslt;
Do either of the above, and you'll find that those statements are no longer skipped over because rslt is now being used.
But, you'll find you will have a problem with your statement:
rslt := not mid;
because mid is an integer. I'm not sure what you want to return here, but I know you don't want the "not" operator to be applied to "mid".
Look at this code that I got from wikibooks. It might help you figure it out.
(* Returns index of requested value in an integer array that has been sorted
in ascending order -- otherwise returns -1 if requested value does not exist. *)
function BinarySearch(const DataSortedAscending: array of Integer;
const ElementValueWanted: Integer): Integer;
var
MinIndex, MaxIndex: Integer;
{ When optimizing remove these variables: }
MedianIndex, MedianValue: Integer;
begin
MinIndex := Low(DataSortedAscending);
MaxIndex := High(DataSortedAscending);
while MinIndex <= MaxIndex do begin
MedianIndex := (MinIndex + MaxIndex) div 2; (* If you're going to change
the data type here e.g. Integer to SmallInt consider the possibility of
an overflow. All it needs to go bad is MinIndex=(High(MinIndex) div 2),
MaxIndex = Succ(MinIndex). *)
MedianValue := DataSortedAscending[MedianIndex];
if ElementValueWanted < MedianValue then
MaxIndex := Pred(MedianIndex)
else if ElementValueWanted = MedianValue then begin
Result := MedianIndex;
Exit; (* Successful exit. *)
end else
MinIndex := Succ(MedianIndex);
end;
Result := -1; (* We couldn't find it. *)
end;