Function to recursively change base

**Klarth** · Feb 21 '07, 01:55 AM

Re: Function to recursively change base

If the problem is to do with the "grrrr .... is 0" appearing in the
output, it is because in your recursive function, it is outside of
your if statement. When the recursion comes back up, it is going to
execute it.

On Feb 21, 10:33 am, ComaGr...@gmail .com wrote:

I'm trying to "Write a function itob(n, s, b) that converts the
integer n into a base b character represeantation in the string s. In
particular, itob(n, s, 16) formats n as a hexadecimal integre in s.
itob must be recursive. " It's #3-5 in the C Programming language by
KandR. I also made a main to test it:
>
void itob(int n, char s[], int b);
main ()
{
/*void itob(int n, char s[], int b);*/
int b[]={2, 6, 8, 10, 16, 32};
int p[]={2, 6, 8, 10, 16, 32, 1239757};
char s[BUFSIZ];
>
int i, j;
>
for(i=0; i<7; i++){
for (j=0; j<6; j++){
itob(p[i], s, b[j]);
printf("s is %d\n",s);
printf("******* ******\n");
}
}
>
}
>
void itob(int n, char s[], int b)
{
int i, sign;
i=0;
printf("number is %d\n",n);
printf("base is %d\n",b);
>
if ((sign = n) <0 )
n = -n;
>
if (n/b>0)
itob(n/b,s,b);
s[i++]=n%b;
s[i]='\0';
printf("grrrr s[i++]is %d\n",s[i]);
>
}
>
But this is the sort of output I get (just a very small sample)
>
*************
number is 1239757
base is 16
number is 77484
base is 16
number is 4842
base is 16
number is 302
base is 16
number is 18
base is 16
number is 1
base is 16
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
s is 134510020
*************
number is 1239757
base is 32
number is 38742
base is 32
number is 1210
base is 32
number is 37
base is 32
number is 1
base is 32
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
s is 134510020
>
I just don't enjoy recursion :,[

**ComaGrrrl@gmail.com** · Feb 21 '07, 02:05 AM

Re: Function to recursively change base

On Feb 20, 8:42 pm, "Klarth" <kah....@gmail. comwrote:

If the problem is to do with the "grrrr .... is 0" appearing in the
output, it is because in your recursive function, it is outside of
your if statement. When the recursion comes back up, it is going to
execute it.
>
>

Sorry I should have been more specific. I'm confused because I keep
getting s is 134510020
and the grr thing was just to see what was going in (or not going in)
my array which is 0 for some reason when I'd like it to be number%base.

**user923005** · Feb 21 '07, 02:55 AM

Re: Function to recursively change base

On Feb 20, 5:33 pm, ComaGr...@gmail .com wrote:

I'm trying to "Write a function itob(n, s, b) that converts the
integer n into a base b character represeantation in the string s. In
particular, itob(n, s, 16) formats n as a hexadecimal integre in s.
itob must be recursive. " It's #3-5 in the C Programming language by
KandR. I also made a main to test it:
>
void itob(int n, char s[], int b);
main ()
{
/*void itob(int n, char s[], int b);*/
int b[]={2, 6, 8, 10, 16, 32};
int p[]={2, 6, 8, 10, 16, 32, 1239757};
char s[BUFSIZ];
>
int i, j;
>
for(i=0; i<7; i++){
for (j=0; j<6; j++){
itob(p[i], s, b[j]);
printf("s is %d\n",s);
printf("******* ******\n");
}
}
>
}
>
void itob(int n, char s[], int b)
{
int i, sign;
i=0;
printf("number is %d\n",n);
printf("base is %d\n",b);
>
if ((sign = n) <0 )
n = -n;
>
if (n/b>0)
itob(n/b,s,b);
s[i++]=n%b;
s[i]='\0';
printf("grrrr s[i++]is %d\n",s[i]);
>
}
>
But this is the sort of output I get (just a very small sample)
>
*************
number is 1239757
base is 16
number is 77484
base is 16
number is 4842
base is 16
number is 302
base is 16
number is 18
base is 16
number is 1
base is 16
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
s is 134510020
*************
number is 1239757
base is 32
number is 38742
base is 32
number is 1210
base is 32
number is 37
base is 32
number is 1
base is 32
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
grrrr s[i++]is 0
s is 134510020
>
I just don't enjoy recursion :,[

I wrote a function that does something similar. But I decided no
sense hauling that bulky string all over the place. And I restricted
myself to using only putchar() as a library function call within the
bit of stuff that does the actual work. Of course, it would not
satisfy an assignment that asked you to stuff the answer into a
string. P.S. If you use 64 bit integers, then your string must be a
minimum of 65 characters in length. But I am not sure I see the
efficacy of making it BUFSIZ characters. So I have an assignment for
you. Comment this code and tell me what each line of it is doing.
After you have done that, it will be easy for you to write code that
does exactly what you want.

#include <stdio.h>
#include <assert.h>
#include <time.h>

static const char symbol_table[] =
{
'0', '1', '2', '3', '4', '5', '6', '7',
'8', '9', 'A', 'B', 'C', 'D', 'E', 'F',
'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N',
'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V',
'W', 'X', 'Y', 'Z',
};

void emit_digits(uns igned long long whomper, unsigned base)
{
unsigned long long digit = whomper % base;
assert(base <= 36); /* Else, you're going to need a new symbol
set */
assert(base >= 2); /* Else you'll have a heck of a long wait */

whomper /= base;
if (whomper)
emit_digits(who mper, base);
putchar(symbol_ table[digit]);
}

void change_base(uns igned long long whomper, unsigned base)
{
emit_digits(who mper, base);
putchar('\n');
}

#ifdef UNIT_TEST
int main(void)
{
unsigned long long two_to_the_31 = 2147483648;
unsigned long long some_other_thin g;
unsigned base;
size_t i;
unsigned j;
change_base(two _to_the_31, 2);
change_base(two _to_the_31, 16);
change_base(two _to_the_31, 32);
putchar('\n');
srand((unsigned ) time(NULL));
for (i = 1; i < 10; i++) {
some_other_thin g = rand();
some_other_thin g *= rand();
for (j = 2; j <= 36; j++) {
change_base(som e_other_thing, j);
}
putchar('\n');
}
putchar('\n');
return 0;
}
#endif

**Michael** · Feb 21 '07, 07:45 AM

Re: Function to recursively change base

On Feb 20, 5:33 pm, ComaGr...@gmail .com wrote:

I'm trying to "Write a function itob(n, s, b) that converts the
integer n into a base b character represeantation in the string s. In
particular, itob(n, s, 16) formats n as a hexadecimal integre in s.
itob must be recursive. " It's #3-5 in the C Programming language by
KandR. I also made a main to test it:
>
void itob(int n, char s[], int b);
main ()
{
/*void itob(int n, char s[], int b);*/
int b[]={2, 6, 8, 10, 16, 32};
int p[]={2, 6, 8, 10, 16, 32, 1239757};
char s[BUFSIZ];

Comment the following out

>
int i, j;
>
for(i=0; i<7; i++){
for (j=0; j<6; j++){
itob(p[i], s, b[j]);
printf("s is %d\n",s);
printf("******* ******\n");
}
}
>

Replace with:
itob(3, s, 16);
printf("s is %d\n", s);

}
>
void itob(int n, char s[], int b)
{
int i, sign;
i=0;
printf("number is %d\n",n);
printf("base is %d\n",b);
>
if ((sign = n) <0 )
n = -n;
>
if (n/b>0)
itob(n/b,s,b);
s[i++]=n%b;
s[i]='\0';

Replace

printf("grrrr s[i++]is %d\n",s[i]);

with

printf("grrrr i is %d and s[i]is %d\n", i, s[i]);

>
}
>

Then figure out what should be happening, and what actually is. (You
may need some more printfs, or you may need a debugger, or whatever.)
Fix problems.

Then replace
itob(3, s, 16);
with
itob(12, s, 16);

Figure out what should be happening, and what actually is. Fix
problems.

Then with
itob(20, s, 16);
Figure out what should be happening, and what actually is. Fix
problems.

At this point, comment back in your original code, and it should
work. If not, find the smallest number that doesn't work and put in a
single line as I've shown, then track it down.

Michael