Reverse strings in an array

Question

procedure ReverseArray(var A : array of string);
var I,J,L : integer;
begin
  for I := Low(A) to High(A) do
  begin
    L := length(A[I]);
    for J := L downto 1 do M := M + A[I];
  end;
  writeln(M);
end;


begin
  for I := 1 to 4 do readln(T[I]);
  ReverseArray(T);
  sleep(40000);
end.

What I'm trying to do here basically is reverse every string in the array but I'm unable to do it , what the code above do is basically repeat the words depends on their length (I write 'bob' in the array , the procedure will give me 'bob' three times because the length is 3) ... not sure why it's not working properly and what I'm missing

Are you trying to reverse the **order** of the strings within the array? Are you trying to reverse the **characters** within each string? Both? What exactly are you trying to accomplish? Your description is a little vague. — Remy Lebeau, Jun 08 '16 at 23:18
Delphi has a [`ReverseString()`](http://docwiki.embarcadero.com/Libraries/en/System.StrUtils.ReverseString) function in the `StrUtils` unit. — Remy Lebeau, Jun 09 '16 at 05:21
It's not a palindrome . I got it working with the answer above :) thanks guys — user2779177, Jun 09 '16 at 15:13

score 3 · Answer 1 · answered Jun 09 '16 at 05:27

Delphi has a ReverseString() function in the StrUtils unit.

uses
  StrUtils;

type
  TStrArray = array of string;

procedure ReverseArray(var A : TStrArray);
var
  I: integer;
begin
  for I := Low(A) to High(A) do
    A[I] := ReverseString(A[I]);
end;

var
  T: TStrArray;
  I: Integer
begin
  SetLength(T, 4);
  for I := 1 to 4 do Readln(T[I]);
  ReverseArray(T);
  ...
end.

Johan · Accepted Answer · 2016-06-09T04:02:49.463

A string is an array of char with some extra bells and whistles added.
So an array of string is a lot like an array of array of char.
If you want to reverse the string, you'll have to access every char and reverse it.

procedure ReverseArray(var A : array of string);
var 
  i,j,Len : integer;
  B: string;
begin
  for i := Low(A) to High(A) do begin
    Len := length(A[i]);
    SetLength(B, Len);  //Make B the same length as A[i].
    //B[Len] = A[i][1]; B[Len-1]:= A[i][2] etc...
    for j := Len downto 1 do B[j]:= A[i][(Len-J)+1];
    //Store the reversed string back in the array.
    A[i]:= B; 
    //Because A is a var parameter it will be returned.
    //Writeln(B); //Write B for debugging purposes.
  end;
end;


var
  i: integer;
  Strings: array [0..3] of string;
begin
  for i := 0 to 3 do readln(Strings[i]);
  ReverseArray(Strings);
  for i := 0 to 3 do writeln(Strings[i]);
  WriteLn('Done, press a key...'); 
  ReadLn;
end.

Some tips:

Do not use global variables like M but declare a local variable instead.
Don't do AStr:= AStr + AChar in a loop, if you can avoid it. If you know how long the result is going to be use the SetLength trick as shown in the code. It's generates much faster code.
Instead of a Sleep you can use a ReadLn to halt a console app. It will continue as soon as you press a key.
Don't put the writeln in your working routine.
Note the first element in a string is 1, but the first element in a array is 0 (unless otherwise defined); Dynamic arrays always start counting from zero.
Note that array of string in a parameter definition is an open array; a different thing from a dynamic array.
Single uppercase identifiers like T, K, etc are usually used for generic types, you shouldn't use them for normal variables; Use a descriptive name instead.

score 1 · Answer 3 · answered Jun 09 '16 at 04:50

Come on! 'bob' is one of those words you shouldn't try to test a reverse routine. But the problem goes beyond that.

Your problem is in here

for J := L downto 1 do 
  M := M + A[I];

You are trying to add the whole string to the M variable instead of the character you are trying to access. So, it should be

for J := L downto 1 do 
   M := M + A[I][J];

Also you need to set M := '' inside the first loop where it will have nothing when you start accumulating characters in to it.

Third, move the writing part, WriteLn(M), inside the first loop where you get a nice, separated outputs.

Putting together, it is going to be:

for I := Low(A) to High(A) do
begin
    L := length(A[I]);

    M := '';        
    for J := L downto 1 do 
       M := M + A[I][J];

    writeln(M);
end;

score 1 · Answer 4 · answered Jun 09 '16 at 08:28

My preferred solution for this is

type
  TStringModifier = function(const s: string): string;

procedure ModifyEachOf( var aValues: array of string; aModifier: TStringModifier );
var
  lIdx: Integer;
begin
  for lIdx := Low(aValues) to High(aValues) do
    aValues[lIdx] := aModifier( aValues[lIdx] );
end;

and it ends up with

var
  MyStrings: array[1..3] of string;
begin
  MyStrings[1] := '123';
  MyStrings[2] := '456';
  MyStrings[3] := '789';

  ModifyEachOf( MyStrings, SysUtils.ReverseString );
end;

score 1 · Answer 5 · answered Jul 26 '18 at 13:31

1

uses
   System.SysUtils, System.StrUtils;

var
   Forwards, backwards : string;

begin
    forwards:= 'abcd';
    backwards:= ReverseString(forwards);
    Writeln(backwards);
    Readln;
end;

// dcba

answered Jul 26 '18 at 13:31

user7732148

11
2

2

Please give more explanation even though code is self-explanatory. – tetri Jul 26 '18 at 14:45

Reverse strings in an array

5 Answers5