K&R2 Exercise 4-12

**Richard Heathfield** · Jun 27 '08, 07:27 PM

Re: K&R2 Exercise 4-12

Lax said:

Here is my 1st attempt at the solution for K&R2 Exercise 4-12:
>
void itoa(int n, char *s)
{
>
if (n < 0)
{
*s++ = '-';
n = -n;
}
>
if( n/10 0 )
itoa( n/10, s );
>
*s++ = n%10 + '0';
*s = 0;
}
>
>
I realize why my code is not working, it's because the increments
don't carry forward when the recursive calls "unwind."

In a real itoa, you'd want to take a maximum length, to reduce the risk of
buffer overruns. The following fixes your problem, I think (although I
haven't actually tested it).

void itoasub(int n, char **s)
{
if(n / 10 0)
{
itoasub(n / 10, s);
}
**s = n % 10 + '0';
++*s;
}
void itoa(int n, char *s)
{
if(n < 0)
{
*s++ = '-';
n = -n; /* beware INT_MIN! */
}
itoasub(n, &s);
*s = 0;
}

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

**Bartc** · Jun 27 '08, 07:27 PM

Re: K&R2 Exercise 4-12

"Lax" <Lax.Clarke@gma il.comwrote in message
news:1737a679-a647-40be-a053-47a96987c45b@k3 7g2000hsf.googl egroups.com...

Here is my 1st attempt at the solution for K&R2 Exercise 4-12:
>
void itoa(int n, char *s)
{
>
if (n < 0)
{
*s++ = '-';
n = -n;
}
>
if( n/10 0 )
itoa( n/10, s );
>
*s++ = n%10 + '0';
*s = 0;
}

Try this; it uses a local variable, and calls strlen a lot, so is a bit
slower:

void itoa(int n, char *s)
{int len;

*s=0;

if (n < 0)
{
*s++ = '-';
n = -n;
}

if( n>=10 )
itoa( n/10, s );
len=strlen(s);
s[len]= n%10 + '0';
s[len+1]=0;
}

--
Bart

**Ben Bacarisse** · Jun 27 '08, 07:27 PM

Re: K&R2 Exercise 4-12

Lax <Lax.Clarke@gma il.comwrites:

Here is my 1st attempt at the solution for K&R2 Exercise 4-12:

I can solve this this way:
>
void itoa2(int n, char *s)
{
static char *p = 0;
>
if(p == 0)
p = s;
>
if (n < 0)
{
*p++ = '-';
n = -n;

Small point: in C -n can overflow. I am sure that K&R don't intend
that you deal with this here, but a real itoa would have to.

}
>
if( n/10 0 )
itoa2( n/10, p );
>
*p++ = n%10 + '0';
*p = 0;
}
>
>
This keeps the p pointer up-to-date while the recursive calls unwind.
>
My quesiton is, does anyone know of a solution that doesn't require
the introduction of a static duration variable? Something like a small
modifcation to my first attempt?

Well, not a tweak but a general idea: recursion works particularly
well with functions that have simple value parameters and return useful
results. In fact, that pattern can be grown into a whole methodology
of programming.

Since the interface of itoa is fixed, you need a helper function. The
trick is thinking up the most helpful function you can imagine. Often
that function can then be written in terms of itself and the desired
function becomes a simple call of it with a little extra
house-keeping. For example:

Given 'char *aux_itoa(int n, char *s);' that takes a positive 'n',
puts the character representation of 'n' into 's' and returns a pointer
to the end of that representation, could you write both itoa and also
aux_itoa?

--
Ben.

**Martin** · Jun 27 '08, 07:28 PM

Re: K&R2 Exercise 4-12

On Sun, 13 Apr 2008 08:14:18 +0100, Lax <Lax.Clarke@gma il.comwrote:

Here is my 1st attempt at the solution for K&R2 Exercise 4-12:
>
void itoa(int n, char *s)
{
>
if (n < 0)
{
*s++ = '-';
n = -n;
}
>
if( n/10 0 )
itoa( n/10, s );
>
*s++ = n%10 + '0';
*s = 0;
}

Any comments on Tondo & Gimpel's solution, in THE C ANSWER BOOK?

#include <math.h>

/* itoa: convert n to characters in s; recursive */
void itoa(int n, char s[])
{
static int i;

if ( n / 10 )
itoa(n / 10, s);
else {
i = 0;
if (n < 0)
s[i++] = '-';
}
s[i++] = abs(n) % 10 + '0';
s[i] = '0';
}

--
Martin

**Flash Gordon** · Jun 27 '08, 07:28 PM

Re: K&R2 Exercise 4-12

Martin wrote, On 15/04/08 13:44:

<snip>

Any comments on Tondo & Gimpel's solution, in THE C ANSWER BOOK?

If you have correctly presented it, then it is not very good.

#include <math.h>

This header is not needed.

/* itoa: convert n to characters in s; recursive */
void itoa(int n, char s[])
{
static int i;
>
if ( n / 10 )
itoa(n / 10, s);
else {
i = 0;
if (n < 0)
s[i++] = '-';
}
s[i++] = abs(n) % 10 + '0';

For abs you should include stdlib.h

s[i] = '0';
}

Now try running it with the following main function calling it...

int main(void)
{
char s1[]="abcdef",s2[]="ghijk";
itoa(1,s1);
itoa(2,s2);
printf("1='%s' 2='%s'\n",s1,s2 );
return 0;
}

Then try reading the code and see if you can find the error.
--
Flash Gordon

**Dann Corbit** · Jun 27 '08, 07:28 PM

Re: K&R2 Exercise 4-12

"Flash Gordon" <spam@flash-gordon.me.ukwro te in message
news:oe7fd5xqel .ln2@news.flash-gordon.me.uk...

Martin wrote, On 15/04/08 13:44:
>
<snip>
>

>Any comments on Tondo & Gimpel's solution, in THE C ANSWER BOOK?

>
If you have correctly presented it, then it is not very good.
>

>#include <math.h>

>
This header is not needed.
>

>/* itoa: convert n to characters in s; recursive */
>void itoa(int n, char s[])
>{
> static int i;
>>
> if ( n / 10 )
> itoa(n / 10, s);
> else {
> i = 0;
> if (n < 0)
> s[i++] = '-';
> }
> s[i++] = abs(n) % 10 + '0';

>
For abs you should include stdlib.h
>

> s[i] = '0';
>}

>
Now try running it with the following main function calling it...
>
int main(void)
{
char s1[]="abcdef",s2[]="ghijk";
itoa(1,s1);
itoa(2,s2);
printf("1='%s' 2='%s'\n",s1,s2 );
return 0;
}
>
Then try reading the code and see if you can find the error.

It has to be a typo (obviously). But even after the obvious fix, there is
still a problem with INT_MIN:

#include <stdlib.h>
/* my_itoa: convert n to characters in s; recursive */
void my_itoa(int n, char s[])
{
static int i;

if (n / 10)
my_itoa(n / 10, s);
else {
i = 0;
if (n < 0)
s[i++] = '-';
}
s[i++] = abs(n) % 10 + '0';
s[i] = 0;
}

#include <stdio.h>
#include <string.h>
#include <limits.h>
int main(void)
{
char s1[80] = "abcdef";
char s2[80] = "abcdef";
my_itoa(INT_MAX , s1);
my_itoa(INT_MIN , s2);
printf("INT_MAX ='%s' INT_MIN='%s'\n" , s1, s2);
return 0;
}

** Posted from http://www.teranews.com **

**Peter Nilsson** · Jun 27 '08, 07:28 PM

Re: K&R2 Exercise 4-12

Lax wrote:

Here is my 1st attempt at the solution for K&R2 Exercise 4-12:
>
void itoa(int n, char *s)

<snip>

My quesiton is, does anyone know of a solution that doesn't require
the introduction of a static duration variable? Something like a small
modifcation to my first attempt?

Google comp.lang.c for itoa. This is not the first time the function
has been discussed.

--
Peter

**Bartc** · Jun 27 '08, 07:28 PM

Re: K&R2 Exercise 4-12

Martin wrote:

On Sun, 13 Apr 2008 08:14:18 +0100, Lax <Lax.Clarke@gma il.comwrote:

>Here is my 1st attempt at the solution for K&R2 Exercise 4-12:
>>
>void itoa(int n, char *s)
>{
>>
> if (n < 0)
> {
> *s++ = '-';
> n = -n;
> }
>>
> if( n/10 0 )
> itoa( n/10, s );
>>
> *s++ = n%10 + '0';
> *s = 0;
>}

>
Any comments on Tondo & Gimpel's solution, in THE C ANSWER BOOK?
>
#include <math.h>
>
/* itoa: convert n to characters in s; recursive */
void itoa(int n, char s[])
{
static int i;

The OP wanted something without a 'static duration variable'.

-- Bartc

**user923005** · Jun 27 '08, 07:29 PM

Re: K&R2 Exercise 4-12

On Apr 16, 3:14 am, "Bartc" <b...@freeuk.co mwrote:

Martin wrote:

On Sun, 13 Apr 2008 08:14:18 +0100, Lax <Lax.Cla...@gma il.comwrote:

Here is my 1st attempt at the solution for K&R2 Exercise 4-12:

>

void itoa(int n, char *s)
{

>

if (n < 0)
{
*s++ = '-';
n = -n;
}

>

if( n/10 0 )
itoa( n/10, s );

>

*s++ = n%10 + '0';
*s = 0;
}

>

Any comments on Tondo & Gimpel's solution, in THE C ANSWER BOOK?

>

#include <math.h>

>

/* itoa: convert n to characters in s; recursive */
void itoa(int n, char s[])
{
static int i;

>
The OP wanted something without a 'static duration variable'.

As a sensible solution, snprintf() springs to mind. A simpler way to
remove the static and remain recursive is to have one top level
function that calls the recursive function and passes in an
initialized counter.

**Willem** · Jun 27 '08, 07:29 PM

Re: K&R2 Exercise 4-12

Lax wrote:
) Here is my 1st attempt at the solution for K&R2 Exercise 4-12:
)
) void itoa(int n, char *s)
) {
)
) if (n < 0)
) {
) *s++ = '-';
) n = -n;
) }
)
) if( n/10 0 )
) itoa( n/10, s );
)
) *s++ = n%10 + '0';
) *s = 0;
) }
)
)
) I realize why my code is not working, it's because the increments
) don't carry forward when the recursive calls "unwind."
)
) I can solve this this way:
)
)
) void itoa2(int n, char *s)
) {
) static char *p = 0;
)
) if(p == 0)
) p = s;
)
) if (n < 0)
) {
) *p++ = '-';
) n = -n;
) }
)
) if( n/10 0 )
) itoa2( n/10, p );
)
) *p++ = n%10 + '0';
) *p = 0;
) }
)
)
) This keeps the p pointer up-to-date while the recursive calls unwind.
)
) My quesiton is, does anyone know of a solution that doesn't require
) the introduction of a static duration variable? Something like a small
) modifcation to my first attempt?

A simple way would be to have two functions and pass a pointer to pointer:

void itoa_r(int n, char **s)
{
if( n/10 0 )
itoa2( n/10, s );
(*s)++ = n%10 + '0';
}

void itoa(int n, char *s)
{
if (n < 0) {
*s++ = '-';
n = -n;
}
itoa_r(n, &s);
*s = 0;
}

SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

**Martin** · Jun 27 '08, 07:31 PM

Re: K&R2 Exercise 4-12

On Tue, 15 Apr 2008 23:04:49 +0100, Dann Corbit <dcorbit@connx. comwrote:

It has to be a typo (obviously).

Indeed. I missed out the backslash, so the incorrect line:

s[i] = '0';

should be:

s[i] = '\0';

But Dann's

s[i] = 0;

will do!

That was the only typing mistake I made. T&G's response does include
<math.hin its solution.

But even after the obvious fix, there is still a problem with INT_MIN:
>
... itoa ...
>
#include <stdio.h>
#include <string.h>
#include <limits.h>
int main(void)
{
char s1[80] = "abcdef";
char s2[80] = "abcdef";
my_itoa(INT_MAX , s1);
my_itoa(INT_MIN , s2);
printf("INT_MAX ='%s' INT_MIN='%s'\n" , s1, s2);
return 0;
}

I get:

INT_MAX='214748 3647' INT_MIN='-214748364('

which is puzzling.

--
Martin

**Martin** · Jun 27 '08, 07:31 PM

Re: K&R2 Exercise 4-12

On Wed, 16 Apr 2008 11:14:06 +0100, Bartc <bc@freeuk.comw rote:

The OP wanted something without a 'static duration variable'.

Yes. T&G's solution is answering K&R2's question (or trying to anyway).

--
Martin

**Ben Bacarisse** · Jun 27 '08, 07:31 PM

Re: K&R2 Exercise 4-12

Martin <m@b.cwrites:

I get:
>
INT_MAX='214748 3647' INT_MIN='-214748364('
>
which is puzzling.

There is a bug (two, really). The solution should include <stdlib.h>
not <math.hand abs(n) is undefined when n == INT_MIN on some
systems. I think this version is correct (but I prefer my other
posted solution!):

#include <stdlib.h>

void itoa(int n, char s[])
{
static int i;

if ( n / 10 )
itoa(n / 10, s);
else {
i = 0;
if (n < 0)
s[i++] = '-';
}
s[i++] = abs(n % 10) + '0'; /* Note: not abs(n) % 10. */
s[i] = '\0';
}

--
Ben.

**CBFalconer** · Jun 27 '08, 07:31 PM

Re: K&R2 Exercise 4-12

Martin wrote:

>

.... snip ...

>
I get:
>
INT_MAX='214748 3647' INT_MIN='-214748364('
>
which is puzzling.

Sounds like a bug in limits.h. The entry for INT_MIN should be:

#define INT_MIN = (-2137482647 - 1)

and probably is:

#define INT_MIN = (-2137482648)

because the expression "-number" loads number and then negates it.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.

** Posted from http://www.teranews.com **

K&R2 Exercise 4-12

K&R2 Exercise 4-12

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