Increment array to a certain number

Question

I have an integer array (representing a 4-digit number) that I need to increment so that each integer never goes higher than 3. Basically, it needs to print every 4-digit number that does not have 4 or higher in it. Here's the output I'm expecting compared to the actual output:

Expected: 0000 0001 0002 0003 0010 0011 0012 0013 0020 0021 0022 .... 3333
Received: 0000 1000 2000 3000 3100 3200 3300 3310 3320 3330 3331 3332

I know my algorithm's messed up but I don't know what to do to it:

int i, c[4];

memset(c, 0, sizeof(c));
i = 0;
while (1) {
    testprint(c);
    c[i]++;
    if (c[i] == 3)
        i++;
    if (i == 3)
        break;
}

All testprint does is display every digit in the array. So how should I change my code to correctly increment the array? And do I even need to use an array? How would I do this without one?

Your code wouldn't print any 4 's ... are you sure that is your real output? — M.M
– M.M, Commented Oct 12, 2016 at 21:54
If you want to print every 4 digit number, you're going to need to include some more logic. Currently you cycle the first digit then the second etc., this won't get numbers where the first digit is lower than the second digit for example. — M.M
– M.M, Commented Oct 12, 2016 at 21:55
@M.M sorry about the 4s, it was from a previous version... — MD XF
– MD XF, Commented Oct 12, 2016 at 21:56

chqrlie · Accepted Answer · 2017-06-10 11:23:39Z

2

You can use a function to convert a number into its representation in base 4 (ie: using only digits 0, 1, 2 and 3):

#include <stdio.h>

char *itoa4(unsigned value, char *dest, int digits) {
    dest[digits] = '\0';
    while (digits-- > 0) {
        dest[digits] = '0' + value % 4;
        value /= 4;
    }
    return dest;
}

int main(void) {
    char buf[5];
    for (int i = 0; i < 256; i++) {
        printf("%s\n", itoa4(i, buf, 4));
    }
    return 0;
}

edited Jun 10, 2017 at 11:23

answered Oct 12, 2016 at 22:21

chqrlie

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MD XF · Accepted Answer · 2016-10-13 19:33:27Z

0

All right, I found the answer with help from @chqrlie:

int *ito4(int value, int dest[])
{
    int i = 4;
    while (i-- > 0) {
        dest[i] = value % 4;
        value /= 4;
    }
    return dest;
}

int main(void)
{
    int i, arr[4];
    for (i = 0; i < 256; i++)
        print_array(ito4(i, buf), 4);
    return 0;
}

answered Oct 13, 2016 at 19:33

MD XF

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6 Comments

chqrlie Over a year ago

Do you mind accepting my answer since you copied it?

MD XF Over a year ago

@chqrlie I did not copy your answer. Your answer reminded me of something I saw in a C Programming book. I found it in there and modified it to fix my problem

chqrlie Over a year ago

my bad, the function name and layout looked strikingly similar to my code, but indeed they are different.

Weather Vane Over a year ago

@chqrlie can the OP really accept own answer to own question?

MD XF Over a year ago

@WeatherVane ...yes?

|

Jack · Accepted Answer · 2016-10-12 22:04:48Z

-1

You don't need to use an array at all for this purpose. How you choose to represent a value and the value itself are two different things.

The same value, for example 10, can be represented in any different base you wish, it's 10 in base 10, 0xA in base 16, 12 in base 8 and so on.

At the same time incrementing a value by 1 works under every base. Assuming this you can easily just format an unsigned integer to print it as a base4 number.

Mind that with (value >> (i*2)) & 0x3 you extract the i-th digit, then everything becomes trivial.

answered Oct 12, 2016 at 22:04

Jack

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1 Comment

MD XF Over a year ago

How do I implement this?

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