Good day, I'm doing some Codeforces exercises in my free time, and I had a problem to test if the user was a boy or a girl, well, my problem isn't that, i have just demonstrated the code.
While compiling my code in my computer ( I'm using version 3.0.4 for i386 ) i get no error, but codeforces gives me this error
program.pas(15,16) Error: Operator is not overloaded: "freq(Char;AnsiString):LongInt;" + "ShortInt"
program.pas(46,4) Fatal: There were 1 errors compiling module, stopping
The error wasn't clear enough to me, as the same script was perfectly compiled with my version.
The platform is using ( version 3.0.2 i386-Win32 ).
program A236;
uses wincrt, sysutils;
var
username : String;
function freq(char: char; username : String): Integer;
var
i: Integer;
begin
freq:= 0;
for i:= 1 to length(username) do
if char = username[i] then
freq:= freq + 1;
//writeln(freq);
end;
function OddUserName(username : String): Boolean;
var
i, counter: Integer;
begin
OddUserName:= false; // even
counter:= 0;
for i:= 1 to length(username) do
if freq(username[i], username) <> 1 then
delete(username, i, 1)
else
counter:= counter + 1;
if counter mod 2 <> 0 then
OddUserName:= true; // odd
//writeln(counter);
//writeln(OddUserName);
end;
begin
readln(username);
if not OddUserName(username) then
writeln('CHAT WITH HER!')
else
writeln('IGNORE HIM!');
//readkey();
end.
The error is supposed to be at this line probably :
function freq(character: char; username : String): Integer;
Thanks for everyone who helps.
Inside of a function, the function's name can be used as a substitute for using an explicit local variable or Result. freq() and OddUserName() are both doing that, but only freq() is using the function name as an operand on the right-hand side of an assignment. freq := freq + 1; should be a legal statement in modern Pascal compilers, see Why i can use function name in pascal as variable name without definition?.
However, it would seem the error message is suggesting that the failing compiler is treating freq in the statement freg + 1 as a function type and not as a local variable. That would explain why it is complaining about not being able to add a ShortInt with a function type.
So, you will have to use an explicit local variable instead, (or the special Result variable, if your compiler provides that), eg:
function freq(charToFind: char; username : String): Integer;
var
i, f: Integer;
begin
f := 0;
for i := 1 to Length(username) do
if charToFind = username[i] then
f := f + 1;
//writeln(f);
freq := f;
end;
function freq(charToFind: char; username : String): Integer;
var
i: Integer;
begin
Result := 0;
for i := 1 to Length(username) do
if charToFind = username[i] then
Result := Result + 1;
//writeln(f);
end;
I got a problem in my Lazarus project: everytime I want to use a function it throws the above error (External: SIGSEGV). I don't know what that means, but some debugging showed me, that this is the code, causing the error:
class function TUtils.AsStringArray(const Strs:TStrings): TStringArray;
var
s:string;
i:integer;
begin
SetLength(Result, Strs.Count);
i := 1;
for s in Strs do
begin
Result[i] := s;
i := i + 1;
end;
end;
And the definitions
TStringArray = array of string;
TUtils = class
public
[...]
class function AsStringArray(const Strs:TStrings): TStringArray; static;
end;
The exception occurs after i := i + 1;. I would be really thankful if you could help me!
Dynamic arrays such as TStringArray = array of string; are zero-based; your code uses it as 1-based and raises access violation.
You should replace i := 1; by i := 0;
To the second Problem, it is because you are accesing to the index i, wich at the start it is 1 that is why you have the problem, the range of the array is determined by "length - 1", so if your length is 1, then your range is 0. So to solve the problem in your for loop you have to put Result[i-1] := s; like this you acces the index you really want.
More of this on http://wiki.freepascal.org/Dynamic_array
A USB memory stick has two partitions - one read only and the other read-write.
My program runs from the read-only partition.
The volume labels for both partitions are fixed by the manufacturer:
MYDISK-RO and MYDISK-RW
When inserted in Windows, each partition (volume) gets a different drive letter. These drive letters are different on different computers depending on the configuration ie. the number of drive letters already allocated to disk drives.
My question is:
Which is the best (most efficient) way for the program to find the drive letter of the read-write partition, using the volume label?
It needs to work on Windows XP and up.
Rather than enumerate all drive letters and compare the volume label to the one that we need, I'm looking ideally for a single function call to Windows .. something like:
GetDriveLetterByVolumeName(AVolumeLabel: String);
or
GetVolumeInformation(AVolumeLabel: String);
Is there such a function or is enumerating the drive letters and comparing each volume label the only solution?
TIA.
Long long time ago I used this code (Was on Delphi7)
This procedure add in combobox all the root of all Removable drives found
Procedure TfMain.GetDiskDrives();
var
r: LongWord;
Drives: array[0..128] of char;
pDrive: pchar;
begin
Result := '';
r := GetLogicalDriveStrings(sizeof(Drives), Drives);
if r = 0 then exit;
if r > sizeof(Drives) then
raise Exception.Create(SysErrorMessage(ERROR_OUTOFMEMORY));
pDrive := Drives; // Point to the first drive
while pDrive^ <> #0 do begin
if GetDriveType(pDrive) = DRIVE_REMOVABLE then begin
cDrive.Items.Add(pDrive);
end;
inc(pDrive, 4); // Point to the next drive
end;
if cDrive.Items.Count=1 then cDrive.ItemIndex:=0;
end;
After that you can use the following function to get the volume name
function GetVolumeName(DriveLetter: Char): string;
var
dummy: DWORD;
buffer: array[0..MAX_PATH] of Char;
oldmode: LongInt;
begin
oldmode := SetErrorMode(SEM_FAILCRITICALERRORS);
try
GetVolumeInformation(PChar(DriveLetter + ':\'),
buffer,
SizeOf(buffer),
nil,
dummy,
dummy,
nil,
0);
Result := StrPas(buffer);
finally
SetErrorMode(oldmode);
end;
end;
I'm posting my code adapted from Gianluca Colombo's answer:
Tested and working with Delphi XE2 Update 4.1 on Windows 7 x64.
unit uDiskUtils;
interface
uses Windows, Classes, SysUtils;
Procedure GetDiskDrives(var ADriveList: TStrings);
function GetVolumeName(const ADriveLetter: Char): string;
function FindDiskDriveByVolumeName(const AVolumeName: String): Char;
implementation
Procedure GetDiskDrives(var ADriveList: TStrings);
var
r: LongWord;
Drives: array [0 .. 128] of Char;
pDrive: pchar;
begin
ADriveList.Clear;
r := GetLogicalDriveStrings(sizeof(Drives), Drives);
if r = 0 then
exit;
if r > sizeof(Drives) then
raise Exception.Create(SysErrorMessage(ERROR_OUTOFMEMORY));
pDrive := Drives; // Point to the first drive
while pDrive^ <> #0 do
begin
if GetDriveType(pDrive) = DRIVE_REMOVABLE then
begin
ADriveList.Add(pDrive);
end;
inc(pDrive, 4); // Point to the next drive
end;
end;
function GetVolumeName(const ADriveLetter: Char): string;
var
dummy: DWORD;
buffer: array [0 .. MAX_PATH] of Char;
oldmode: LongInt;
begin
oldmode := SetErrorMode(SEM_FAILCRITICALERRORS);
try
GetVolumeInformation(pchar(ADriveLetter + ':\'), buffer, sizeof(buffer), nil, dummy, dummy, nil, 0);
Result := StrPas(buffer);
finally
SetErrorMode(oldmode);
end;
end;
function FindDiskDriveByVolumeName(const AVolumeName: String): Char;
var
dl: TStringList;
c: Integer;
begin
Result := ' ';
dl := TStringList.Create;
try
GetDiskDrives(TStrings(dl));
for c := 0 to dl.Count - 1 do
if (AVolumeName = GetVolumeName(dl[c][1])) then
Result := dl[c][1];
finally
dl.Free;
end;
end;
end.
I was trying to speed up a certain routine in an application, and my profiler, AQTime, identified one method in particular as a bottleneck. The method has been with us for years, and is part of a "misc"-unit:
function cwLeftPad(aString:string; aCharCount:integer; aChar:char): string;
var
i,vLength:integer;
begin
Result := aString;
vLength := Length(aString);
for I := (vLength + 1) to aCharCount do
Result := aChar + Result;
end;
In the part of the program that I'm optimizing at the moment the method was called ~35k times, and it took a stunning 56% of the execution time!
It's easy to see that it's a horrible way to left-pad a string, so I replaced it with
function cwLeftPad(const aString:string; aCharCount:integer; aChar:char): string;
begin
Result := StringOfChar(aChar, aCharCount-length(aString))+aString;
end;
which gave a significant boost. Total running time went from 10,2 sec to 5,4 sec. Awesome! But, cwLeftPad still accounts for about 13% of the total running time. Is there an easy way to optimize this method further?
Your new function involves three strings, the input, the result from StringOfChar, and the function result. One of them gets destroyed when your function returns. You could do it in two, with nothing getting destroyed or re-allocated.
Allocate a string of the total required length.
Fill the first portion of it with your padding character.
Fill the rest of it with the input string.
Here's an example:
function cwLeftPad(const aString: AnsiString; aCharCount: Integer; aChar: AnsiChar): AnsiString;
var
PadCount: Integer;
begin
PadCount := ACharCount - Length(AString);
if PadCount > 0 then begin
SetLength(Result, ACharCount);
FillChar(Result[1], PadCount, AChar);
Move(AString[1], Result[PadCount + 1], Length(AString));
end else
Result := AString;
end;
I don't know whether Delphi 2009 and later provide a double-byte Char-based equivalent of FillChar, and if they do, I don't know what it's called, so I have changed the signature of the function to explicitly use AnsiString. If you need WideString or UnicodeString, you'll have to find the FillChar replacement that handles two-byte characters. (FillChar has a confusing name as of Delphi 2009 since it doesn't handle full-sized Char values.)
Another thing to consider is whether you really need to call that function so often in the first place. The fastest code is the code that never runs.
Another thought - if this is Delphi 2009 or 2010, disable "String format checking" in Project, Options, Delphi Compiler, Compiling, Code Generation.
StringOfChar is very fast and I doubt you can improve this code a lot. Still, try this one, maybe it's faster:
function cwLeftPad(aString:string; aCharCount:integer; aChar:char): string;
var
i,vLength:integer;
origSize: integer;
begin
Result := aString;
origSize := Length(Result);
if aCharCount <= origSize then
Exit;
SetLength(Result, aCharCount);
Move(Result[1], Result[aCharCount-origSize+1], origSize * SizeOf(char));
for i := 1 to aCharCount - origSize do
Result[i] := aChar;
end;
EDIT: I did some testing and my function is slower than your improved cwLeftPad. But I found something else - there's no way your CPU needs 5 seconds to execute 35k cwLeftPad functions except if you're running on PC XT or formatting gigabyte strings.
I tested with this simple code
for i := 1 to 35000 do begin
a := 'abcd1234';
b := cwLeftPad(a, 73, '.');
end;
and I got 255 milliseconds for your original cwLeftPad, 8 milliseconds for your improved cwLeftPad and 16 milliseconds for my version.
You call StringOfChar every time now. Of course this method checks if it has something to do and jumps out if length is small enough, but maybe the call to StringOfChar is time consuming, because internally it does another call before jumping out.
So my first idea would be to jump out by myself if there is nothing to do:
function cwLeftPad(const aString: string; aCharCount: Integer; aChar: Char;): string;
var
l_restLength: Integer;
begin
Result := aString;
l_restLength := aCharCount - Length(aString);
if (l_restLength < 1) then
exit;
Result := StringOfChar(aChar, l_restLength) + aString;
end;
You can speed up this routine even more by using lookup array.
Of course it depends on your requirements. If you don't mind wasting some memory...
I guess that the function is called 35 k times but it has not 35000 different padding lengths and many different chars.
So if you know (or you are able to estimate in some quick way) the range of paddings and the padding chars you could build an two-dimensional array which include those parameters.
For the sake of simplicity I assume that you have 10 different padding lengths and you are padding with one character - '.', so in example it will be one-dimensional array.
You implement it like this:
type
TPaddingArray = array of String;
var
PaddingArray: TPaddingArray;
TestString: String;
function cwLeftPad4(const aString:string; const aCharCount:integer; const aChar:char; var anArray: TPaddingArray ): string;
begin
Result := anArray[aCharCount-length(aString)] + aString;
end;
begin
//fill up the array
SetLength(StrArray, 10);
PaddingArray[0] := '';
PaddingArray[1] := '.';
PaddingArray[2] := '..';
PaddingArray[3] := '...';
PaddingArray[4] := '....';
PaddingArray[5] := '.....';
PaddingArray[6] := '......';
PaddingArray[7] := '.......';
PaddingArray[8] := '........';
PaddingArray[9] := '.........';
//and you call it..
TestString := cwLeftPad4('Some string', 20, '.', PaddingArray);
end;
Here are benchmark results:
Time1 - oryginal cwLeftPad : 27,0043604142394 ms.
Time2 - your modyfication cwLeftPad : 9,25971967336897 ms.
Time3 - Rob Kennedy's version : 7,64538131122457 ms.
Time4 - cwLeftPad4 : 6,6417059620664 ms.
Updated benchmarks:
Time1 - oryginal cwLeftPad : 26,8360194218451 ms.
Time2 - your modyfication cwLeftPad : 9,69653117046119 ms.
Time3 - Rob Kennedy's version : 7,71149259179622 ms.
Time4 - cwLeftPad4 : 6,58248533610693 ms.
Time5 - JosephStyons's version : 8,76641780969192 ms.
The question is: is it worth the hassle?;-)
It's possible that it may be quicker to use StringOfChar to allocate an entirely new string the length of string and padding and then use move to copy the existing text over the back of it.
My thinking is that you create two new strings above (one with FillChar and one with the plus). This requires two memory allocates and constructions of the string pseudo-object. This will be slow. It may be quicker to waste a few CPU cycles doing some redundant filling to avoid the extra memory operations.
It may be even quicker if you allocated the memory space then did a FillChar and a Move, but the extra fn call may slow that down.
These things are often trial-and-error!
You can get dramatically better performance if you pre-allocate the string.
function cwLeftPadMine
{$IFDEF VER210} //delphi 2010
(aString: ansistring; aCharCount: integer; aChar: ansichar): ansistring;
{$ELSE}
(aString: string; aCharCount: integer; aChar: char): string;
{$ENDIF}
var
i,n,padCount: integer;
begin
padCount := aCharCount - Length(aString);
if padCount > 0 then begin
//go ahead and set Result to what it's final length will be
SetLength(Result,aCharCount);
//pre-fill with our pad character
FillChar(Result[1],aCharCount,aChar);
//begin after the padding should stop, and restore the original to the end
n := 1;
for i := padCount+1 to aCharCount do begin
Result[i] := aString[n];
end;
end
else begin
Result := aString;
end;
end;
And here is a template that is useful for doing comparisons:
procedure TForm1.btnPadTestClick(Sender: TObject);
const
c_EvalCount = 5000; //how many times will we run the test?
c_PadHowMany = 1000; //how many characters will we pad
c_PadChar = 'x'; //what is our pad character?
var
startTime, endTime, freq: Int64;
i: integer;
secondsTaken: double;
padIt: string;
begin
//store the input locally
padIt := edtPadInput.Text;
//display the results on the screen for reference
//(but we aren't testing performance, yet)
edtPadOutput.Text := cwLeftPad(padIt,c_PadHowMany,c_PadChar);
//get the frequency interval of the OS timer
QueryPerformanceFrequency(freq);
//get the time before our test begins
QueryPerformanceCounter(startTime);
//repeat the test as many times as we like
for i := 0 to c_EvalCount - 1 do begin
cwLeftPad(padIt,c_PadHowMany,c_PadChar);
end;
//get the time after the tests are done
QueryPerformanceCounter(endTime);
//translate internal time to # of seconds and display evals / second
secondsTaken := (endTime - startTime) / freq;
if secondsTaken > 0 then begin
ShowMessage('Eval/sec = ' + FormatFloat('#,###,###,###,##0',
(c_EvalCount/secondsTaken)));
end
else begin
ShowMessage('No time has passed');
end;
end;
Using that benchmark template, I get the following results:
The original: 5,000 / second
Your first revision: 2.4 million / second
My version: 3.9 million / second
Rob Kennedy's version: 3.9 million / second
This is my solution. I use StringOfChar instead of FillChar because it can handle unicode strings/characters:
function PadLeft(const Str: string; Ch: Char; Count: Integer): string;
begin
if Length(Str) < Count then
begin
Result := StringOfChar(Ch, Count);
Move(Str[1], Result[Count - Length(Str) + 1], Length(Str) * SizeOf(Char));
end
else Result := Str;
end;
function PadRight(const Str: string; Ch: Char; Count: Integer): string;
begin
if Length(Str) < Count then
begin
Result := StringOfChar(Ch, Count);
Move(Str[1], Result[1], Length(Str) * SizeOf(Char));
end
else Result := Str;
end;
It's a bit faster if you store the length of the original string in a variable:
function PadLeft(const Str: string; Ch: Char; Count: Integer): string;
var
Len: Integer;
begin
Len := Length(Str);
if Len < Count then
begin
Result := StringOfChar(Ch, Count);
Move(Str[1], Result[Count - Len + 1], Len * SizeOf(Char));
end
else Result := Str;
end;
function PadRight(const Str: string; Ch: Char; Count: Integer): string;
var
Len: Integer;
begin
Len := Length(Str);
if Len < Count then
begin
Result := StringOfChar(Ch, Count);
Move(Str[1], Result[1], Len * SizeOf(Char));
end
else Result := Str;
end;
I want to develop a setup package for conditionally upgrading an existing package. I want to check the existing software version against to-be-installed version. In order to do that, I have to compare the version strings.
How can I convert the string value to a numerical value in a Inno setup script?
RegQueryStringValue(HKEY_LOCAL_MACHINE, 'Software\Blah blah', 'Version', version)
version = 'V1.R2.12';
numVersion := ??string_to_numerical_value??(version);
This is a little more tricky, as you would want to handle versions like 'V1.R2.12' and 'V0.R15.42' correctly - with the simple conversion in the other answer you would get 1212 and 1542, which would not compare the way you would expect.
You need to decide how big each part of the version number can be, and multiply the parts by that value to get a correct end number. Something like this:
[Code]
function string_to_numerical_value(AString: string; AMaxVersion: LongWord): LongWord;
var
InsidePart: boolean;
NewPart: LongWord;
CharIndex: integer;
c: char;
begin
Result := 0;
InsidePart := FALSE;
// this assumes decimal version numbers !!!
for CharIndex := 1 to Length(AString) do begin
c := AString[CharIndex];
if (c >= '0') and (c <= '9') then begin
// new digit found
if not InsidePart then begin
Result := Result * AMaxVersion + NewPart;
NewPart := 0;
InsidePart := TRUE;
end;
NewPart := NewPart * 10 + Ord(c) - Ord('0');
end else
InsidePart := FALSE;
end;
// if last char was a digit the last part hasn't been added yet
if InsidePart then
Result := Result * AMaxVersion + NewPart;
end;
You can test this with the following code:
function InitializeSetup(): Boolean;
begin
if string_to_numerical_value('V1.R2.12', 1) < string_to_numerical_value('V0.R15.42', 1) then
MsgBox('Version ''V1.R2.12'' is not as recent as version ''V0.R15.42'' (false)', mbConfirmation, MB_OK);
if string_to_numerical_value('V1.R2.12', 100) > string_to_numerical_value('V0.R15.42', 100) then
MsgBox('Version ''V1.R2.12'' is more recent than version ''V0.R15.42'' (true)', mbConfirmation, MB_OK);
Result := FALSE;
end;
Whether you pass 10, 100 or 1000 for AMaxVersion depends on the number and range of your version number parts. Note that you must not overflow the LongWord result variable, which has a maximum value of 2^32 - 1.
I haven't tried that (and my Pascal knowledge is a bit rusty), but something like the following should work:
function NumericVersion(s: String): Integer;
var
i: Integer;
s1: String;
begin
s1 := '';
for i := 0 to Length(s)-1 do
if (s[i] >= '0') and (s[i] <= '9') then
s1 := s1 + s[i];
Result := StrToIntDef(s1, 0);
end;
Please not that you'll have to play with the start and end value for i as I'm not sure whether it is zero-based or not (s[0] may contain the length of the string if it is a "Pascal String").
I've implemented two version strings (actually one string and one dword value) in the registry to overcome complexity.
displayversion="v1.r1.0"
version="10100" (=1*10^4 + 1*10^2 + 0*10^0)
That's simple. Though not an answer to this question, however one might think the other way around when faced with complexity, which could be avoided in a simpler way.