I'm very new to Pascal and still learning much. I have to write a code that :
Takes input of a string
Split the string into two characters each (Snippet)
Use the Snippet to get an index from an array
Transpose the Snippet to a certain value
If Index + Transpose is larger than the length of the Array, return nothing
If not, append the transposed Snippet to a result string
Return the transposed string
I can only write 1 through 3, the rest is still a blur for me. Helps are appreciated.
(And I also want to improve it without many for loops. Any thoughts?)
program TransposeString;
var
melody : Array[1..24] of String[2] = ('c.', 'c#', 'd.', 'd#', 'e.', 'f.', 'f#', 'g.', 'g#', 'a.', 'a#', 'b.', 'C.', 'C#', 'D.', 'D#', 'E.', 'F.', 'F#', 'G.', 'G#', 'A.', 'A#', 'B.');
songstring, transposedstring : String;
transposevalue : byte;
function Transpose(song : String; transposevalue : byte): String;
var
songsnippet : String[2];
iter_song, iter_index, index : byte;
begin
for iter_song := 1 to length(song) do
begin
if iter_song mod 2 = 0 then continue;
songsnippet := song[iter_song] + song[iter_song + 1]; //Split the string into 2 characters each
for iter_index := 1 to 24 do
begin
if melody[iter_index] = songsnippet then
begin
index := iter_index; //Get Index
break;
end;
end;
//Check Transpose + Index
//Transpose Snippet
//Append Snippet to Result String
end;
end;
begin
readln(songstring);
readln(transposevalue);
transposedstring := transpose(songstring, transposevalue);
writeln(transposedstring);
end.
As a starter for you to work from, rather than just spoon-feeding an answer:
You have the index of the snippet (note) in index. Assuming the notes are in order you need to return the note from the array positions above it, so
result := result + melody[iter_index + transposevalue];
You need to check the length of the array before trying to read from it, otherwise it'll crash (step 5). This is just an if statement.
I wouldn't worry too much about for loops - 2 deep nesting isn't that bad. If you wanted to split it out a bit then GetTransposedNote(const note:string): string; could be split out as a new function.
Things you may want to think about are:
What if you can't find the note in the array?
Do you want to be case-sensitive
What if the input string has an odd number of characters?
You are most of the way there already, though.
Related
i have a function that puts a number in a var (var1). if var1 = 3 then create 3 new variables and assign a value for later send that values in an out type. I don't know how can i do, my code is something like this where "pos" acts like an INSTR() that i know its each 13 chars. And "newvar||var1" is because i want my vars names be newvar1, newvar2, newvar3, etc.:
FOR n IN 1..var1
LOOP
pos NUMBER(4):= 0;
newvar||var1 NUMBER(3);
newvar||var1 := SUBSTR(TO_NUMBER(numberInsideString), 1, 1+pos);
pos := pos+13;
END LOOP;
My question is, how a person who really know PLSQL would do that, im learning PLSQL please help. thank you.
What you want to try to do suggests that you need a one dimensional array in order repeatedly to assign new values for each iteration within the loop. One of the types in OWA package such as vc_arr(composed of VARCHAR2 data type values) is handy to define an array.
Coming to the code, you can start with moving the variables into the declaration section, and write such a simple code as a sample exercise
SET serveroutput ON
DECLARE
pos INT := 0;
numberInsideString VARCHAR2(100):='as12df34sdg345apl8976gkh4f11öjhöh678u7jgvj';
newvar OWA.vc_arr;
BEGIN
FOR n IN 1..3
LOOP
newvar(n) := SUBSTR(numberInsideString, 1, 1+pos);
pos := pos+5;
Dbms_Output.Put_Line(newvar(n));
END LOOP;
END;
/
a
as12df
as12df34sdg
I need to construct a matrix; a number of columns and rows are also in the first row of the matrix, I'll make an example so its more clearer.
4 3
1 2 3
5 6 7
9 10 8
1 11 13
Where m=4 (number of rows) and n=3 (number of columns)
This is an example of a text file. Is something like this even possible?
Program Feb;
const
max=100;
type
Matrix=array[1..max,1..max] of integer;
var datoteka:text;
m,n:integer;
counter:integer;
begin
assign(datoteka,'datoteka.txt');
reset(datoteka);
while not eoln(datoteka) do
begin
read(datoteka, m);
read(datoteka, n);
end;
repeat
read eoln(n)
until eof(datoteka)
write (m,n);
end.
My code isn't a big help, cause I don't know how to write it.
First, have a look at the code I wrote to do the task, and then look at my explanation below.
program Matrixtest;
uses
sysutils;
var
NoOfCols,
NoOfRows : Integer;
Source : TextFile;
Matrix : array of array of integer;
FileName : String;
Row,
Col : Integer; // for-loop iterators to access a single cell of the matrix
Value : Integer;
begin
// First, construct the name of the file defining the matrix
// This assumes that the file is in the same folder as this app
FileName := ExtractFilePath(ParamStr(0)) + 'MatrixDef.Txt';
writeln(FileName); // echo it back to the screen so we can see it
// Next, open the file
Assign(Source, FileName);
Reset(Source);
read(Source, NoOfRows, NoOfCols);
writeln('Cols: ', NoOfCols, 'Rows: ', NoOfRows);
SetLength(Matrix, NoOfCols, NoOfRows);
readln(source); // move to next line in file
// Next, read the array data
for Row := 1 to NoOfRows do begin
for Col := 1 to NoOfCols do begin
read(Source, Value);
Matrix[Col - 1, Row - 1] := Value;
end;
end;
// Display the array contents
for Row := 1 to NoOfRows do begin
for Col := 1 to NoOfCols do begin
writeln('Row: ', Row, ' contents', Matrix[Col - 1, Row - 1]);
end;
end;
Close(Source); // We're done with the file, so close it to release OS resources
readln; // this waits until you press a key, so you can read what's been displayed
end.
In your program, you can use a two-dimensional array to represent your matrix. Free Pascal supports multi-dimensional arrays; see https://wiki.lazarus.freepascal.org/Multidimensional_arrays for more information.
This is a complex task, so it helps to know how to do more basic things like reading an array of a size known at compile-time from a text file.
The wrinkle in this task is that you are supposed to read the dimensions (numbers of rows and columns) of the matrix at run-time from the file which contains the matrix's contents.
One inefficient way to do this would be to declare the matrix array with huge dimensions, larger than anything you would expect in practice, using the type of array declaration in the Wiki page linked above.
A better way is to use dynamic arrays, whose dimensions you can set at run-time. To use this, you need to know:
How to declare a dynamic array in Free Pascal
How to set the dimensions of the array at run-time, once you've picked them up from your matrix-definition file (hint: SetLength is the way to do this)
The fact that a Free Pascal dynamic array is zero-based
The easiest way of managing zero-based arrays is to write your code (in terms of Row and Column variables) as if the matrix were declared as array[1..NoOfRows, 1..NoOfColumns] and subtract one from the array indexes only when you actually access the array, as in:
Row := 3;
Column := 4;
Value := Matrix[Row - 1, Column - 1];
I am getting this error when compiling my code: "cars2.pp(3,8) Fatal: Syntax error, "=" expected but ":" found"
Here's my code:
program vehInfo;
type
wheels: array [1 .. 6] of integer;
purchaseYear: array [1919 .. 2051] of integer;
style = (sports, SUV, minivan, motorcycle, sedan, exotic);
pwrSrc = (electric, hybrid, gas, diesel);
vehicle = record
wheel : wheels;
buyDate : purchaseYear;
styles : style;
source : pwrSrc;
end;
var
myVehicle: vehicle;
listOfCars: file of vehicle;
begin
assign(listOfCars, 'hwkcarsfile.txt');
reset(listOfCars);
read(listOfCars, myVehicle);
writeln('wheel type: ' , myVehicle.wheel);
writeln('year purchased: ' , myVehicle.buyDate);
writeln('style: ' , myVehicle.styles);
writeln('power source: ' , myVehicle.source)
close(listOfCars);
end.
I am new to Pascal, any help would be appreciated, thank you.
It is quite simple: type uses =, while variable declarations use :.
So:
type
wheels = 1..6; // not an array, but a subrange type!
purchaseYear = 1919..2051; // not an array, but a subrange type!
style = (sports, SUV, minivan, motorcycle, sedan, exotic);
pwrSrc = (electric, hybrid, gas, diesel);
vehicle = record
wheel: wheels; { a field of a record is a variable }
buyDate: purchaseYear;
styles: style;
source: pwrSrc;
end;
...
var
myVehicle: vehicle;
listOfCars: file of vehicle;
Subrange types are ordinal types (in this case, both are integers), but within the given range. Any value outside the range is illegal. You don't want to have arrays of numbers, you only want the number of wheels and the year (a number too) the vehicle was purchased. You don't need 133 different dates, do you?
It was not me who wrote this code, it was the previous programmer. However, I noticed he didn't provide a decryption algorithm, rendering the encryption useless.
How can I decrypt this?
function Encrypt(jstr: String): String;
var
I: Integer;
A: Real;
begin
if Length(jstr) = 0 Then begin
Result := '';
Exit;
end;
A := 0;
for I := 0 To Length(jstr) do
A := A + (Ord(jstr[I]) * Pos(jstr[I],jstr)) / 33;
Result := FormatFloat('0000000000.0000000000',A);
if Pos(',',Result) > 0 then begin
Insert('.',Result,Pos(',',Result));
Delete(Result,Pos(',',Result),1);
end;
end;
Thanks!
It looks like a one way hash and hence is not reversible. For example, is the string is very big the result is still a string representation of a float.
That function cannot be reversed. Since it takes input of arbitrary length and returns output of finite length, simple information theory tells you the futility of attempting to write a general inverse. Even for shorter input strings it seems to me that different input strings can result in the same encrypted string.
Even as a hash this function seems very brittle to me due to the bizarre use of floating point code. If I were you I would replace this function with something more fit for purpose.
Finally, I recommend that you undertake a review of all code produced by this developer. The low quality of this code and algorithm suggests to me that everything that this developer touched is liable to have defects.
I have a question about algorithm:
How to find all characters in a string whose appearance is greater than a specific number, say 2 for example efficiently?
Regards.
Counting sort will be extremely efficient for one-byte encodings, border case is two-byte encodings. For wider encodings it is not so efficient, but counting array may be replaced with hash table.
EDIT: By the way, that is too general solution, doing only counting phase and outputting results on the fly will be more than enough.
s= #your string
h=Hash.new(0)
s.each_char {|c| h[c]+=1 }
h.select {|char,count| count>2}
var word = "......";
var chars = word.GroupBy(w => w).Where(g => g.Count > 2).Select(g => new { character = g.Key, count = g.Count });
Couldn't resist to try this out.
Keep an internal array of 256 elements for each (ASCII) character.
Loop once over the input string.
Increment the count for given character using the ordinal value of the character as a direct access into the internal array.
Delphi implementation
Type
TCharCounter = class(TObject)
private
FCounts: array[0..255] of byte;
public
constructor Create(const Value: string);
function Count(const AChar: Char): Integer;
end;
{ TCharCounter }
constructor TCharCounter.Create(const Value: string);
var
I: Integer;
begin
inherited Create;
for I := 1 to Length(Value) do
Inc(FCounts[Ord(Value[I])]);
end;
function TCharCounter.Count(const AChar: Char): Integer;
begin
Result := FCounts[Ord(AChar)];
end;
I would sort the string, then just walk through it and keep a running tally for each letter. The last is just O(n) so it'll be as efficient as your sort.
the easier way is to use an array: occurrence[256], initialize them all with 0's
and for every char in string, occurrence[(int)char]++.
And then you just scan the occurrence to find the occurrence of characters satisfying your criterion.