扭转了我想用递归函数反转字符串中的递归的方式

Question

一个字符串，这里是我的代码：

Program InvertTheString; 
var n:integer;word:string; 

Function Invert (N:integer; word:string) : string; 
begin 
    if N=1 then 
     Invert:=word[N]+Invert 
    Else 
     Invert:=Invert(N-1,word); 
end; 

BEGIN 
readln(word); 
n:=length(word); 
writeln (Invert(N,word)); 

writeln;write('Press Enter To Exit...'); 
readln; 
END. 

但它不工作，哪里是拨错？

Answer 1

Function Invert (N:integer; word:string) : string; 
begin 
    if N=0 then 
     Invert:='' 
    Else 
     Invert:= word[N] + Invert(N-1,word); 
end; 

Answer 2

我不做帕斯卡，但是典型的（简单）递归相反，在伪代码，看起来像：

function reverse(word) 
    if empty(word) return "" 
    else return reverse(withoutFirstLetter(word)) + firstLetterOf(word) 

Answer 3

Function Invert (ch:string) : string; 
begin 
if ch='' then 
Invert:='' 
else 
{get the last letter of the string + eliminate it and execute the function} 
Invert:=copy(ch,length(ch),1)+Invert(copy(ch,1,length(ch)-1)); 
end;