19 Nov
2002
19 Nov
'02
8:42 a.m.
HI friends
I was experimenting with a common c factorial problem
when I got a totally unexpected result.
here is the program
# include <stdio.h>
main(){
int fact;
int x;
fact=factorial(6);
printf ("%i\n",fff());
}
int factorial(int x){
if(x>1) return(x*factorial(x-1));
}
int fff () {
;
}
Naturally I was expecting 1 as the answer. But I am getting the answer 720
which is 6!.
If I comment the 5th line fact=factorial(6); I am getting the expected result
"1".
The normal explanation for this is memory leak. Can anyone please tell me what
is happening and are you too getting the same result.
I am using redhat linux 7.3 with gcc 2.96
regards
Abhijeet D Mhatre
19 Nov
19 Nov
9:33 a.m.
On Dec 18, 2002 at 08:29, Abhijeet D Mhatre wrote:
Naturally I was expecting 1 as the answer. But I am
getting the answer 720
which is 6!.
t.c:2: warning: return-type defaults to `int'
t.c: In function `main':
t.c:5: warning: implicit declaration of function `factorial'
t.c:6: warning: implicit declaration of function `fff'
t.c:4: warning: unused variable `x'
t.c:7: warning: control reaches end of non-void function
t.c: In function `factorial':
t.c:12: warning: control reaches end of non-void function
t.c: In function `fff':
t.c:16: warning: control reaches end of non-void function
[satyap]~$ ./a.out
720
[satyap]~$ gcc -v
Reading specs from /usr/lib/gcc-lib/i386-linux/2.95.4/specs
gcc version 2.95.4 20011002 (Debian prerelease)
[satyap]~$
--
Satya. <URL:http://satya.virtualave.net/> I need a job!
Perl, Apache, Linux, C. http://www-scf.usc.edu/~phanse/resume.pdf
But I forgot all about the Amnesia Conference!!
10:14 a.m.
On Wed, 18 Dec 2002, Abhijeet D Mhatre wrote:
I was experimenting with a common c factorial problem
when I got a totally unexpected result.
unexpected?
# include <stdio.h>
main(){
return type defaults to int
int fact;
int x;
fact=factorial(6);
printf ("%i\n",fff());
so where is the statement that returns an int?
}
int factorial(int x){
if(x>1) return(x*factorial(x-1));
what happens if x<=1? what is the return value then?
}
int fff () {
;
}
and what is this supposed to return? the printf will print the return
value of this function, but this function doesn't return anything.
Naturally I was expecting 1 as the answer. But I am
getting the answer
720 which is 6!.
pray tell us why you expected this so naturally.
Philip
--
You scratch my tape, and I'll scratch yours.
10:23 a.m.
On Wed, 18 Dec 2002 08:29:38 +0530
Abhijeet D Mhatre wrote:
# include <stdio.h>
main(){
int fact;
int x;
fact=factorial(6);
printf ("%i\n",fff());
}
int factorial(int x){
if(x>1) return(x*factorial(x-1));
}
int fff () {
;
}
Naturally I was expecting 1 as the answer. But I am getting the
answer 720 which is 6!.
Why were you expecting 1? Could you really expect any particular
value from a call to fff()?
--
Tahir Hashmi (VSE, NCST)
http://staff.ncst.ernet.in/tahir
tahir AT ncst DOT ernet DOT in
We, the rest of humanity, wish GNU luck and Godspeed
10:28 a.m.
On Wed, Dec 18, 2002 at 08:29:38AM +0530, Abhijeet D Mhatre wrote:
****************************************
* LUG meet on 12 Jan. 2003 @ VJTI
****************************************
int fact;
int x;
fact=factorial(6);
printf ("%i\n",fff());
}
int factorial(int x){
if(x>1) return(x*factorial(x-1));
}
int fff () {
;
}
Something very strage, I did the following (inserted wait(1),
thats it)
# include <stdio.h>
main(){
int fact;
int x;
fact=factorial(6);
wait(1);
printf ("%i\n",fff());
}
int factorial(int x){
if(x>1) return(x*factorial(x-1));
}
int fff () {
;
}
And the output I get is (-1).
11:29 a.m.
Well i think this is related to stack and how the data
is stored
by the OS in stack.
I had put sleep before calling printf statement and
the return result was 0. Or if we give a specific
return then that value appears in the printf.
Well if anyone knows about this then do enlighten.
Ciao
Tapesh
--- Tahir Hashmi <code_martial(a)softhome.net> wrote:
****************************************
* LUG meet on 12 Jan. 2003 @ VJTI
****************************************
On Wed, 18 Dec 2002 08:29:38 +0530
Abhijeet D Mhatre wrote:
__________________________________________________
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# include <stdio.h>
main(){
int fact;
int x;
fact=factorial(6);
printf ("%i\n",fff());
}
int factorial(int x){
if(x>1) return(x*factorial(x-1));
}
int fff () {
;
}
Naturally I was expecting 1 as the answer. But I
am getting the
answer 720 which is 6!.
Why were you expecting 1? Could you really expect
any particular
value from a call to fff()?
12:44 p.m.
On Tue, 17 Dec 2002 21:52:51 -0800 (PST)
Tapeshwar Nath wrote:
Well i think this is related to stack and how the data
is stored by the OS in stack.
The return value for that function would be undefined. It
would vary with different combinations of operating systems and
compilers. Yes, it's got to do with how the compiler would
manage the stack and what policies it has for otherwise undefined
behavior. I got 0 with gcc 3.2, sleep() or no sleep().
the return result was 0. Or if we give a specific
return then that value appears in the printf.
Of course, it would.
A quick look for functions and return values through
_The_C_Programming_Language_ 2/e by Kernighan and Ritchie should
leave no scope for doubts.
--
Tahir Hashmi (VSE, NCST)
http://staff.ncst.ernet.in/tahir
tahir AT ncst DOT ernet DOT in
We, the rest of humanity, wish GNU luck and Godspeed
18 Dec
18 Dec
5:19 p.m.
> int fact;
> int x;
> fact=factorial(6);
> printf ("%i\n",fff());
> }
>
>
> int factorial(int x){
> if(x>1) return(x*factorial(x-1));
> }
>
> int fff () {
> ;
> }
>
OK I give up, what do you mean by the fff() function?
Something very strage, I did the following (inserted wait(1),
thats it)
# include <stdio.h>
main(){
int fact;
int x;
fact=factorial(6);
wait(1);
printf ("%i\n",fff());
}
You've defined fff() as int, and it _doesn't_ return anything, plus you are
printing the value it returns (the printf statement)!
If you are really interested in the factorial of 6, I guess you should be
printing the return value of factorial(6).
That is your printf statement should look something like this:
printf("%d\n", factorial(6)); or simply printf("%d\n", fact);
Also, the int x declaration in main() is superfluous - you can do without
it; and as a good programming practice never choose variable names that
collide. Your main() has an x, and so does your factorial() - they are
different and don't mean the same x.
Warm wishes,
Amol Hatwar
26 Dec
26 Dec
12:47 p.m.
Abhijeet D Mhatre <abhijeetmhatre(a)hotpop.com> writes:
****************************************
* LUG meet on 12 Jan. 2003 @ VJTI
****************************************
HI friends
I was experimenting with a common c factorial problem
when I got a totally unexpected result.
here is the program
# include <stdio.h>
main(){
int fact;
int x;
fact=factorial(6);
printf ("%i\n",fff());
}
int factorial(int x){
if(x>1) return(x*factorial(x-1));
}
int fff () {
;
}
Naturally I was expecting 1 as the answer. But I am getting the answer 720
which is 6!.
The function fff() is supposed to return an integer according to definition. But it does
not. The return of the previous function which is on the stack is taken instead. If you
insert a 'return int;' then the value of int is printed.
A good experiment anyway. This is the reason why I love C and so many hate it. '-)
If I comment the 5th line fact=factorial(6); I am getting the expected result
"1".
But I am wondering why it should print "1"... & why is it expected.
quasi
7653
days inactive
7690
days old
8 comments
8 participants
participants (8)
-
Abhijeet D Mhatre -
Amish Munshi -
Amol Hatwar -
Philip S Tellis -
q u a s i -
Satya -
Tahir Hashmi -
Tapeshwar Nath