Unique behaviour of incremental operators in recursion

See <url: http://www.parashift.com/c++-f aq-lite/how-to-post.html#faq-5
..2>.

Ramashish

* Bangalore:[color=blue]
> In reversing string usring recursive function, i found problems in
> using incremental postfix , and incremental prefix
>
>
> 1 void rev(char*);
> 2 void main()[/color]

Incorrect.
<url: http://www.parashift.com/c++-faq-lite/newbie.html#faq-29.3>
<url: http://www.research.att.com/~bs/bs_faq2.html#void-main>


[color=blue]
> 3 {
> 4 char s[]="STRING";
> 5 rev(s);
> 6 }
> 7 void rev(char* s)
> 8 {
> 9 if(*s != '\0')
> 10 {
> 11 rev(s++);[/color]

<url: http://home.no.net/dubjai/win32cpptut/html/w32cpptut_01_02_11.html>

[color=blue]
> 12 printf("%c",*s);[/color]

As a newbie, use C++ iostreams.

[color=blue]
> 13 }
> 14 else
> 15 return;[/color]

Indentation.
[color=blue]
> 16 }
>
> If run the above code, excution will be stuck in infinite loop(It will
> be in first address of the string).[/color]

See above, or look up the definition of postfix ++ in your textbook.

--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

You cant change the address of array variable s, bcoz STRING directly
stores in variable s (it's done with the statement s[]="STRING").
i think it may possible through char *s="STRING". in this case s
contains the address (memory location of string i.e "STRING" )

prefix and postfix operators cant be applied on array variables.

try it.

>>Bangalore wrote: excution will be stuck in infinite loop
See the code.[color=blue][color=green]
>> rev(s++);
>> printf("%c",*s);[/color][/color]
You are using postfix operator , so First it will call rev(s) , and
then it will increment s by one. So each time it is l calling only
rev(s) recursively infinite time.[color=blue][color=green]
>> if I chage postfix increment operator (line 11) to prefix increment
>>operator.
>>In the output it won't print first character.[/color][/color]
if u do[color=blue][color=green]
>> rev(++s);
>> printf("%c",*s);[/color][/color]
then it is already incrementing s, so you will next character printed.
Just , print first and call rev() later, you will get what u want.
[color=blue][color=green]
>> printf("%c",*s);
>> rev(++s);[/color][/color]
[color=blue][color=green]
>> If I replace s++ with s+1 ( line 11 ), I am getting correct output.[/color][/color]
No doubt, since s +1 means simply add 1 to s and then pass to rev()
function. It doesn't change the value of s itself.[color=blue][color=green]
>> Plz clarify me, .[/color][/color]
Use a good book on C/C++.
writing ++s, means
s = s+1
and then use s.
writing s++ means
first use s and then s = s+1.
test this
int a=0, b =10;
a = ++b;
cout<<a<<b<<endl;
a =0;
b = 10;
a = b++;
cout<<a<<b<<endl;

upashu2 wrote:[color=blue][color=green][color=darkred]
> >>Bangalore wrote: excution will be stuck in infinite loop[/color][/color]
> See the code.[color=green][color=darkred]
> >> rev(s++);
> >> printf("%c",*s);[/color][/color]
> You are using postfix operator , so First it will call rev(s) , and
> then it will increment s by one. So each time it is l calling only
> rev(s) recursively infinite time.[/color]


No, consider this line more closely:

rev(s++);

what actually happens is this: s is incremented *before* the call to
rev, not afterwards. And s is not actually passed to rev (because it
has already been incremented). Now it may appear that the compiler is
in a bit of bind. What does it pass to rev? Fortunately the compiler
had the foresight to make a copy of s before it was incremented, so it
passes this copy of s in the function call to rev.

In C++, all expressions, even those with postincrement operations,
which appear as parameters in a function call must be completely
evaluated before the function call can be made.

Greg