Test if an integer is a palindrome in c language

[Wabz]

Hello mates,

Does anyone know how to write a function that tests if an integer is a
palindrome in C language?

 

[Kevin Handy]

Does anyone know how to write a function that tests if an integer is a
palindrome in C language?

Yes. I'm quite sure that someone does.

 

[Wabz]

Yes. I'm quite sure that someone does.

Could this someone possibly help me write the said program?

 

[Flash Gordon]

Wabz wrote, On 14/08/07 18:42:
Please don't quote things you are not commenting on, such as the
annoying text appended to Kevin's post.

Could this someone possibly help me write the said program?

Yes, I'm sure one of the people meeting those requirements could help if
they were so desired. However, knowledgeable people are unlikely to do
your homework for you when you have not even made an attempt. So if you
want help you should attempt it and then post your attempt here with any
questions you have. The more effort you put in the more people will be
prepared to help you.

 

[osmium]

Could this someone possibly help me write the said program?

Start by extracting the individual digits of the integer. You may find the
modulo operator, %, helpful. There is another way, involving sprintf() in
<stdio.h>, but if this is a school assignment, the instructor probably
expects the modulo way.

 

[Fred Kleinschmidt]

Hello mates,

Does anyone know how to write a function that tests if an integer is a
palindrome in C language?

What base? Eleven is a palindrome when written in base 10,
but not when written in base 8.
Create a buffer large enough to hold the characters to display the
largest integer. Use sprintf to fill it. Check whether palindrome.

 

[Army1987]

Hello mates,

Does anyone know how to write a function that tests if an integer is a
palindrome in C language?

I assume you mean palindrome in base 10.
sprintf() it to a string, and then check whether the string is
palindrome:
len = strlen(str);
for (i = 0; i < len/2; i++) {
if (str != str[len - i - 1])
break; /*it is not palindome*/
}

 

[Kevin Handy]

Could this someone possibly help me write the said program?

Should have asked that in the first place.

Many ways of doing it. You need to decide
what number system you want to
use: binary, hex, octal, decimal, etc.

1. Convert to string in that format, and compare
front end of the resulting string to the back end.

2. A number can be a palindrome in one base, while
not being one in another. If you can choose your
base for a specific number N, you can pick
base (N-1), which gives you the result (11),
which is always a palindrome. So your program
could then be hard coded to always print "true".

3. Shift digits of N into a new value, in reverse
order (% helps here). Compare N and reverse N.

....

 

[user923005]

I assume you mean palindrome in base 10.

Otherwise (for instance) FFFF is a palindrome, but 65535 will say
'nope'

sprintf() it to a string, and then check whether the string is
palindrome:
len = strlen(str);
for (i = 0; i < len/2; i++) {
if (str != str[len - i - 1])
break; /*it is not palindome*/

}

It might be interesting to try every base from 2 to 36.
Given that condition, what percentage of integers are palindromes?

 

[user923005]

I assume you mean palindrome in base 10.

Otherwise (for instance) FFFF is a palindrome, but 65535 will say
'nope'

sprintf() it to a string, and then check whether the string is
palindrome:
len = strlen(str);
for (i = 0; i < len/2; i++) {
if (str != str[len - i - 1])
break; /*it is not palindome*/

}

It might be interesting to try every base from 2 to 36.
Given that condition, what percentage of integers are palindromes?

It seems that what numbers are not palindromes in some base might be a
more interesting question. Of course if we include bases up to that
number, then the answer is obviously 'none'.

#include <stdlib.h>
#include <string.h>
/* ltostr is from snippets (I fixed the broken bit). */
char *ltostr(long long num, char *string, size_t max_chars, unsigned
base)
{
char remainder;
int sign = 0;
if (base < 2 || base > 36)
return ((void *) 0);
if (num < 0) {
sign = 1;
num = -num;
}
if (num == 0) /* bugbug:drc formerly wrong result here... */
return "0";
string[--max_chars] = '\0';
for (max_chars--; max_chars > sign && num != 0; max_chars--) {
remainder = (char) (num % base);
if (remainder < 9)
string[max_chars] = remainder + '0';
else
string[max_chars] = remainder - 10 + 'A';
num /= base;
}
if (sign)
string[--max_chars] = '-';
if (max_chars > 0)
memset(string, ' ', max_chars + 1);
return string + max_chars;
}

int ispal(const char *start)
{
const char *end = start + strlen(start) - 1;
while (end > start) {
if (*start != *end) {
return 0;
}
start++;
end--;
}
return 1;
}

#include <stdio.h>
static char string[50] = {0};
int main(void)
{
long long index;
unsigned base;
for (index = 0; index < 4000000000; index++) {
for (base = 2; base < 37; base++) {
long long val = index;
char *p = ltostr(val, string, sizeof string,
base);
if (isspace(*p))
p++;
if (ispal(p)) {
printf("%llu = %s is a palindrome in base %u\n", val,
p, base);
}
}
}
return 0;
}

 

[user923005]

/* Original ltostr() was hosed. Fixed version: */
static const char B36TAB[] =
{
'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'
};
#include <stdlib.h>
#include <string.h>
/* ltostr is from snippets (I fixed the broken bit). */
char *ltostr(long long num, char *string, size_t max_chars, unsigned
base)
{
char remainder;
int sign = 0;
if (base < 2 || base > 36)
return ((void *) 0);
if (num < 0) {
sign = 1;
num = -num;
}
if (num == 0) /* bugbug:drc formerly wrong result
here... */
return "0";
string[--max_chars] = '\0';
for (max_chars--; max_chars > sign && num != 0; max_chars--) {
remainder = (char) (num % base);
string[max_chars] = B36TAB[remainder];
num /= base;
}
if (sign)
string[--max_chars] = '-';
if (max_chars > 0)
memset(string, ' ', max_chars + 1);
return string + max_chars;
}

int ispal(const char *start)
{
const char *end = start + strlen(start) - 1;
while (end > start) {
if (*start != *end) {
return 0;
}
start++;
end--;
}
return 1;
}

#include <stdio.h>
static char string[50] =
{0};
int main(void)
{
long long index;
unsigned base;
for (index = 0; index < 4000000000; index++) {
for (base = 2; base < 37; base++) {
long long val = index;
char *p = ltostr(val, string, sizeof string,
base);
if (isspace(*p))
p++;
if (ispal(p)) {
printf("%llu = %s is a palindrome in base %u\n", val,
p, base);
}
}
}
return 0;
}
/*
Sample output:
....
2290134 = E656E is a palindrome in base 20
2290141 = 84548 is a palindrome in base 23
2290156 = 1IEI1 is a palindrome in base 35
2290187 = 9H1H9 is a palindrome in base 22
2290220 = 36Q63 is a palindrome in base 29
2290252 = 11022100122011 is a palindrome in base 3
2290293 = 50805 is a palindrome in base 26
2290322 = 2OOO2 is a palindrome in base 30
2290353 = 925529 is a palindrome in base 12
2290378 = 4270724 is a palindrome in base 9
2290391 = HAHAH is a palindrome in base 19
2290460 = BG6GB is a palindrome in base 21
2290499 = 2ERE2 is a palindrome in base 31
2290501 = 1D3D1 is a palindrome in base 36
2290530 = 5LEL5 is a palindrome in base 25
2290534 = E666E is a palindrome in base 20
2290627 = 3K9K3 is a palindrome in base 28
2290641 = 1000101111001111010001 is a palindrome in base 2
2290658 = 20233033202 is a palindrome in base 4
2290670 = 84648 is a palindrome in base 23
2290671 = 9H2H9 is a palindrome in base 22
2290686 = 6LGL6 is a palindrome in base 24
2290735 = 13EE31 is a palindrome in base 18
2290738 = 48A84 is a palindrome in base 27
2290752 = HAIAH is a palindrome in base 19
2290850 = 25T52 is a palindrome in base 32
2290853 = 1O9O1 is a palindrome in base 34
2290901 = BG7GB is a palindrome in base 21
2290922 = 2290922 is a palindrome in base 10
....
*/

 

[Ben Bacarisse]

/* Original ltostr() was hosed. Fixed version: */

<string-based palindrome tester snipped>

I like this way:

int pal_aux(int n, int r, int b)
{
return n == 0 ? r : pal_aux(n / b, r * b + n % b, b);
}

int is_pal_in_base(int n, int b)
{
return pal(n, 0, b) == n;
}

(not well tested, but it *looks* sound!)

 

[user923005]

<string-based palindrome tester snipped>

I like this way:

int pal_aux(int n, int r, int b)
{
return n == 0 ? r : pal_aux(n / b, r * b + n % b, b);
}

int is_pal_in_base(int n, int b)
{
return pal(n, 0, b) == n;

/*Did you mean:*/
return pal_aux(n, 0, b) == n;

}

(not well tested, but it *looks* sound!)

Certainly much brighter than converting to character.
But it makes it harder to see the answers.

 

[Ben Bacarisse]

/*Did you mean:*/
return pal_aux(n, 0, b) == n;

Yes. Thanks for spotting it that.

One day I'll remember just to paste the code, and to "tidy it up" for
posting. (I added the _aux because I thought "pal" rather too sort a
name, even for a small helper function.)

 

[CBFalconer]

Could this someone possibly help me write the said program?

I suggest you first find him and then ask him (or her). An
alternative would be to do your own homework.

 

[pete]

Hello mates,

Does anyone know how to write a function that tests if an integer is a
palindrome in C language?

/* BEGIN new.c */

#include <stdio.h>
#include <string.h>
#include <limits.h>

int integer_is_palindrome(long unsigned integer);
char *str_rev(char *s);

int main(void)
{
int integer;

for (integer = 0; 1000 > integer; ++integer) {
if (integer_is_palindrome(integer)) {
printf("%d\n", integer);
}
}
return 0;
}

int integer_is_palindrome(long unsigned integer)
{
char lutoa_buff[(sizeof(long) * CHAR_BIT) / 3 + 1];
char copy[sizeof lutoa_buff];

sprintf(lutoa_buff, "%lu", integer);
strcpy(copy, lutoa_buff);
str_rev(copy);
return strcmp(copy, lutoa_buff) == 0;
}

char *str_rev(char *s)
{
char *const p = s;
char *t = p;
char swap;

if (*t != '\0' && *++t != '\0') {
t += strlen(t + 1);
do {
swap = *t;
*t-- = *s;
*s++ = swap;
} while (t > s);
}
return p;
}

/* END new.c */

 

[Richard]

Hello mates,

Does anyone know how to write a function that tests if an integer is a
palindrome in C language?

/* BEGIN new.c */

#include <stdio.h>
#include <string.h>
#include <limits.h>

int integer_is_palindrome(long unsigned integer);
char *str_rev(char *s);

int main(void)
{
int integer;

for (integer = 0; 1000 > integer; ++integer) {
if (integer_is_palindrome(integer)) {
printf("%d\n", integer);
}
}
return 0;
}

int integer_is_palindrome(long unsigned integer)
{
char lutoa_buff[(sizeof(long) * CHAR_BIT) / 3 + 1];
char copy[sizeof lutoa_buff];

sprintf(lutoa_buff, "%lu", integer);
strcpy(copy, lutoa_buff);
str_rev(copy);
return strcmp(copy, lutoa_buff) == 0;
}

char *str_rev(char *s)
{
char *const p = s;
char *t = p;
char swap;

if (*t != '\0' && *++t != '\0') {
t += strlen(t + 1);
do {
swap = *t;
*t-- = *s;
*s++ = swap;
} while (t > s);
}
return p;
}

/* END new.c */

slow?

something like this for the palindrome check on the string. Untested and
pretty much pseudo code.

len=strlen(p);
endp=p+len-1;
while(endp>p)
if ((*endp--)!=(*p++))
return 0;
return 1;



--

 

[Ben Pfaff]

It might be interesting to try every base from 2 to 36.
Given that condition, what percentage of integers are palindromes?

I don't have an answer, but a few moments of arithmetic led me to
discover that in base B there are pow(B, -N/2) palindromic numbers
of length N, for even N. Somehow this is pleasing, even though
it's almost trivial.

 

[Harald van =?UTF-8?B?RMSzaw==?=]

I don't have an answer, but a few moments of arithmetic led me to
discover that in base B there are pow(B, -N/2) palindromic numbers
of length N, for even N. Somehow this is pleasing, even though
it's almost trivial.

What you calculated is the ratio of palindromic numbers in all numbers of
length N, not the amount of palindromic numbers of length N. In base 10,
there are not 0.1 but 9 palindromic numbers of length 2. I'm not sure
whether you meant to post an expression that would return 9, or whether you
meant to describe your expression as returning the ratio.

 

[Ben Pfaff]

What you calculated is the ratio of palindromic numbers in all numbers of
length N, not the amount of palindromic numbers of length N. In base 10,
there are not 0.1 but 9 palindromic numbers of length 2. I'm not sure
whether you meant to post an expression that would return 9, or whether you
meant to describe your expression as returning the ratio.

I meant the ratio. Oops.