i have a interger
val = 99999;
and i want to add the content of val i.e., 9+9+9+9+9 = 45;
how to do that....
need urgent help
val = 99999;
and i want to add the content of val i.e., 9+9+9+9+9 = 45;
how to do that....
need urgent help
P
Pranav
take a temporary variable to store sum and initialize it to zero,
Now by recursively taking modulus and dividing the number you can each
digit separated and you can go on adding the digit..,
can u explain with code...
i didn't understand.....
N
Nate Eldredge
@$|-|. DUBEY said:
Hint:
val % 10 is the rightmost digit of val.
val / 10 shifts all digits to the right.
N
Nick Keighley
not only do you want help with your homework but you want the code as
well!
Another approach would be to use sprintf()
--
Nick Keighley
Yes:
Error on line 0: Lazy programmer.
(comp.lang.c++)
i got it you saved my day........
this group rocks.!!!!!!!!!!!!!!!1
P
Pranav
Thats good Always try to solve on your own or by following an
algorithm never expect a valid code.., It will help you a lot,
K
Keith Thompson
@$|-|. DUBEY said:
It's difficult to imagine how adding the digits of the decimal
representation of an integer could be "urgent". Unless, of course,
you're asking us to do your homework for you.
N
Nick Keighley
perhaps he's a number theorist or something. Tho' a 5
digit number seems kind of small. Ah! perhaps "interger"
isn't meant to be "integer" but indicates some other
mathematical object with more complex properties
(tho' Plain Old Integers a pretty complicated).
We need more info about these inergers of which you
speak.
good grief!
--
Nick Keighley
Many astrologers think that this concentration on [the sun-sign
column] has
done untold damage to serious astrology.
The Independent
L
lovecreatesbeauty
Hope it doesn't invoke any UB
/* given eg 54321, return 1 + 2 + 3 + 4 + 5. */
int my_sum(int n)
{
int sum = 0;
for (; n; n /= 10)
sum += n % 10;
return sum;
}
D
Doug Miller
... and it's due today.
S
ssksakthi -image processing
main()
{
int a,b,sum=0;
printf("enter the number\n");
scanf("%d", &a);
while(a>0)
{
a=a%10;
sum=sum+a;
a=a/10;
}
printf("%d", sum);
getch();
}
H
Hyuga
Doesn't compile on my system:
~$ gcc -Wall sumdigits.c -o sumdigits
sumdigits.c:2: warning: return type defaults to ‘int’
sumdigits.c: In function ‘main’:
sumdigits.c:4: warning: implicit declaration of function ‘printf’
sumdigits.c:4: warning: incompatible implicit declaration of built-in
function ‘printf’
sumdigits.c:5: warning: implicit declaration of function ‘scanf’
sumdigits.c:5: warning: incompatible implicit declaration of built-in
function ‘scanf’
sumdigits.c:15: warning: implicit declaration of function ‘getch’
sumdigits.c:3: warning: unused variable ‘b’
sumdigits.c:16: warning: control reaches end of non-void function
/tmp/ccalL9CH.o: In function `main':
sumdigits.c
.text+0xb1): undefined reference to `getch'collect2: ld returned 1 exit status
Which I suppose is fine, because the OP can hardly learn anything if
you do his homework for him. Or maybe that was your point?
Hyuga
B
Ben Bacarisse
Hyuga said:
There are so many way this program is wrong (not least that the
algorithm is all wrong) that I'd like to believe it was all intended.
N
Nate Eldredge
Eric Sosman said:
You, sir, are a very bad person.
C
CBFalconer
Keith said:
Another possibility - he is trying to develop a check-digit (or a
check-digit checking) system. Not likely, but feasible.
U
user923005
Or even pure-math research:
The distribution of the sum-of-digits function (1998)
by Michael Drmota, Johannes Gajdosik
J. Theor. Nombres Bordx
http://www.geometrie.tuwien.ac.at/drmota/distdig.ps.gz
U
user923005
Hope it doesn't invoke any UB
/* given eg 54321, return 1 + 2 + 3 + 4 + 5. */
int my_sum(int n)
{
int sum = 0;
for (; n; n /= 10)
sum += n % 10;
return sum;
}
Since you have so neatly destroyed any hope of education for the O.P.,
let's complete the job:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
char *revstr(char *s)
{
const char *start = s;
char *end = s;
char c;
while (*s++) {;
}
s -= 2;
while (end < s) {
c = *end;
*end++ = *s;
*s-- = c;
}
return (start);
}
// The digit sum of N can be defined as the sum of the digits of N
// See: http://mathworld.wolfram.com/DigitSum.html
unsigned long digit_sum(unsigned long long n)
{
unsigned long sum;
for (sum = 0; n; n /= 10)
sum += n % 10;
return sum;
}
// The digit root of N is the digit sum repeated until we have a
single digit
unsigned long digit_root(unsigned long long n)
{
while (n > 9) {
n = (unsigned long long) digit_sum(n);
}
return (unsigned long) n;
}
// What strange magic is this?
unsigned long dr(unsigned long long n)
{
if (n)
return 1 + (n - 1) % 9;
return 0;
}
// What is so special about these?
unsigned long long magic_numbers[] =
{1, 81, 1458, 1729};
int main(void)
{
const size_t s = sizeof magic_numbers / sizeof
magic_numbers[0];
size_t index;
for (index = 0; index < s; index++) {
const unsigned long l = digit_sum(magic_numbers[index]);
unsigned long lrev;
char revnum[12];
sprintf(revnum, "%lu", l);
lrev = (unsigned long) atol(revstr(revnum));
printf("%lu * %lu = %lu\n", l, lrev, l * lrev);
}
for (index = 9999999; index <= 10000099; index++) {
printf("digit root %lu is %lu\n", (unsigned long) index,
digit_root(index));
printf("dr %lu is %lu\n", (unsigned long) index,
dr(index));
}
return 0;
}