Lazarus readln doesn't read the variable

Question

I would like to ask for some help as i'm not able to identify what i did wrong in the following code as in the procedure , it doesn't read the first unit of the record , but after that it reads the uj.nev too. It only misses it for the first time which is what i can't understand ,and not only on the school computer but mine too. I would appreciate it if you could explain to me what i did wrong. Thanks in advance! (the program supposed to read records nev , varos , fizetes , and sort them into ascending order, and then write it out into a text file)

program adatbazis;
uses crt,Sysutils;

const
  C_FNAME = 'adatbazis.txt';
var
  a,n,j,l:integer;
  tfout: Textfile;

type
  Tember=record
    nev:string;
    varos:string;
    fizetes:longint;
  end;

procedure beiras(var uj:Tember);
begin
  writeln('nev',j,':');
  readln(uj.nev);
  writeln('varos',j,':');
  readln(uj.varos);
  writeln('fizetes',j,':');
  readln(uj.fizetes);
end;

var
  i,k:integer;
  seged:Tember;
  tomb: array[1..20] of Tember;
begin
  write('n :');
  read(n);
  for j:= 1 to n do begin
    beiras(tomb[j]);
  end;
  writeln('Mi szerint legyen rendezve?');
  repeat
    writeln(' Nev:1 , Fizetes:2 , varos:3');
    readln(l);
  until l<> 1 or 2 or 3;
  if l=1 then  begin
    For j:= 1 to n-1 do begin
      for k:= 2 to n do begin
        if tomb[j].nev>tomb[k].nev then
        begin
          seged:=tomb[j];
          tomb[j]:=tomb[k];
          tomb[k]:=seged;
        end;
      end;
    end;
  end;
  if l=2 then begin
    For j:= 1 to n-1 do begin
      for k:= 2 to n do begin
        if tomb[j].fizetes>tomb[k].fizetes then
        begin
          seged:=tomb[j];
          tomb[j]:=tomb[k];
          tomb[k]:=seged;
        end;
      end;
    end;
  end;
  if l=3 then  begin
    For j:= 1 to n-1 do begin
      for k:= 2 to n do begin
        if tomb[j].varos>tomb[k].varos then
        begin
          seged:=tomb[j];
          tomb[j]:=tomb[k];
          tomb[k]:=seged;
        end;
      end;
    end;
  end;
  AssignFile(tfout, C_FNAME);
  rewrite(tfout);
  for i:=1 to n do begin
    writeln(tfout,tomb[i].nev,'  ',tomb[i].varos,'  ',tomb[i].fizetes);
  end;
  CloseFile(tfout);
  clrscr;
  writeln('done');
  readln;
end.

Answer 1

I noticed that the essential difference between the FPC Read() and ReadLn functions when used with standard input is not very well documented (in the Delphi documentation it can be read). Therefore, the answer to your question is the following.

According to the FPC documentation, CRLF, CR and LF, all are recognized as End-Of-Line (EOL) characters, so in the following EOL stands for any of those.

When calling Read(s) where s: string; the program waits for user input until an EOL is added to the input buffer. The input characters up until (but excluding) the EOL are removed from the buffer and passed to s. The EOL is left in the input buffer.

Next call to Read(s) will immediately recognize the EOL in the input buffer, and will not wait for user input. The EOL continues to remain in the buffer, and therefore no subsequent Read(s) will wait for user input.

The first call to ReadLn(s) after a Read(s) will also immediately detect the EOL in the buffer and not wait for user input.

When calling ReadLn(s) the program waits (as in previous example) for user input until an EOL is added to the input buffer. The input characters up until (but excluding) the EOL are removed from the buffer and passed to s. Then the EOL is removed from the buffer and thrown away.

Next call to ReadLn(s) will again stop and wait for user input.

In your code you are calling Read(n) in the very beginning of the execution. Then, in beiras() you are calling ReadLn() which detects the EOL in the buffer and therefore returns immediately, but also removes the EOL from the buffer. Subsequent calls to ReadLn() will now wait for user input.

The cure is iow to change the Read()n to a ReadLn(n) in the beginning of your program.