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I'm new to programming, and I'm stuck at a problem. I want my program to identify the separate digits in a given number, like if I input 4692, it should identify the digits and print 4 6 9 2. And yeah, without using arrays.

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7 Answers 7

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A perfect recursion problem to tackle if you're new to programming...

4692/1000 = 4

4692%1000 = 692

692/100 = 6

692%100 = 92

92/10 = 9

92%10 = 2

You should get the idea for the loop you should use now, so that it works for any number. :)

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3 Comments

4692%1000 = 692, not 2. 4692/1000 = 4, not 469.
Yeah, +1 your answer is good because it leads to the answer rather than just solving the homework, but please correct your numbers!
The problem with typing too fast, is I type faster than I think sometimes. ;) Corrected!
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Haven't written C code in year, but this should work.

int i = 12345; while( i > 0 ){ int nextVal = i % 10; printf( "%d", nextVal ); i = i / 10; }
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3 Comments

This is completely wrong. You print the value of i 5 times: "123451234123121". Even if you print nextVal instead of i, you're doing it in the wrong order and will print "54321".
You should store the result to a string and reverse it before printing. You can also used a stack to reverse
@Babak Naffas's answer is a rather simple yet typical solution. But if you want more speed you can try double dabble to split the number into BCDs. After that it's easy to take each number out and print.
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Simple and nice

void PrintDigits(const long n) { int m = -1; int i = 1; while(true) { m = (n%(10*i))/i; i*= 10; cout << m << endl; if (0 == n/i) break; } }
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Works only for non negative numbers.
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Another approach is to have two loops.

1) First loop: Reverse the number.

int j = 0; while( i ) { j *= 10; j += i % 10; i /= 10; }

2) Second loop: Print the numbers from right to left.

while( j ) { std::cout << j % 10 << ' '; j /= 10; }

This is assuming you want the digits printed from right to left. I noticed there are several solutions here that do not have this assumption. If not, then just the second loop would suffice.

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Note part 1 breaks if i ends with zeros, because then j would stay at zero at the beginning and multiplying it by 10 would not have any effect on it.
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I think the idea is to have non reapeating digits printed (otherwise it would be too simple)... well, you can keep track of the already printed integers without having an array encoding them in another integer.

some pseudo C, to give you a clue:

int encoding = 0; int d; while (keep_looking()) { d = get_digit(); if (encoding/(2**d)%2 == 0) { print(d); encoding += 2**d; } }
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3 Comments

What is d**2? I think you mean d*d.
If I remember right, ** is Fortran's exponentiation operator. C doesn't have an exponentiation operator. You may replace 2**d with 1 << d.
that's why I said "pseudo-c", and I didn't want to expose a novice to bit shifting and masking operations (%2 is the same as &1 too), that's why I limited myself to arithmetic operations.
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Here is a simple solution if you want to just print the digits from the number.

#include <stdio.h> /** printdigits */ void printDigits(int num) { char buff[128] = ""; sprintf(buff, "%d ", num); int i = 0; while (buff[i] != '\0') { printf("%c ", buff[i]); i++; } printf("\n"); } /* main function */ int main(int argc, char** argv) { int digits = 4321; printDigits(digits); return 0; }
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You don't need 128 characters to print an integer. Also, why initialize buff if you're just going to overwrite it with sprintf() later? Also, why do you have char *p if you never use it?
ya, earlier I was planning to iterate using pointer so I have char *p. Regarding size, he did not mentioned length of numbers so to be on safe size, used 128.
An int can only hold numbers up to 4 billion on most platforms, so you should be safe with 10 digits (11 characters with the nul-terminator). If you're worried about 64-bits, that's still only around 20 to 25 characters.
He said "without using arrays"
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Is it correct 

int main() { int number; cin>>number; int nod=0; int same=number; while(same){ same/=10; nod++; } while(nod--){ cout<<(int)number/(int)pow10(nod)%10<<"\t"; } return 0; }
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Dont understand the logic completely