how to get different digits starting from the most significant digit in a number in c?

Question

I am solving a problem in which a positive integer is given and i have to display it in words.

For example if a number is 2134 the output should be "two one three four" . If i use the modulus operator and use the recursion technique, i get the digits starting from the least significant digit ie "four three one two"

I can also reverse the number and then use modulus operator but i am looking for a better method.

What can be a better approach of solving this problem?.What concept am i missing?

Answer 1

My simple answer:

void printNum(int x)
{
    static const char * const num[] = {
        "zero ", "one ", "two "  , "three ", "four ",
        "five ", "six ", "seven ", "eight ", "nine "
    };

    if (x < 10) {
        printf(num[x]);
        return;
    }
    printNum(x / 10);
    printNum(x % 10);
}

edit

After some more experimenting and tweaking, I came up with this version. I think this is the about the most "pure" recursive function I can make.

void printNum(int x)
{
    static const char * const num[] = {"zero ",  "one ", "two ", "three ",
                                       "four ",  "five ", "six ", "seven ",
                                       "eight ", "nine "};
    (x < 10)? printf(num[x]) : (printNum(x / 10), printNum(x % 10));
}

Answer 2

In case you're looking for recursive solution:

void num2word(int n) {
    if (n / 10 == 0) {
        // print the word for n
    }
    else {
        num2word(n / 10);
        // print the word for n % 10
    }
}

Answer 3

Going for a pure math answer, because I think that's what you're looking for:

#include <math.h>
#include <stdio.h>

main() {
    long long mynum = 2987612345;
    long long firstDigitValue;
    int       firstDigit;
    char *names[] = { "zero", "one", "two", "three", "four", "five",
                      "six", "seven", "eight", "nine" };

    while (mynum > 0) {
        firstDigit = (int) (mynum/pow(10,(int)log10(mynum)));
        firstDigitValue = firstDigit * pow(10,(int)log10(mynum));
        mynum -= firstDigitValue;
        printf("%s ", names[firstDigit]); 
    }
    printf("\n");
}

And then running it produces:

two nine eight seven six one two three four five 

It should work with any size number that a long long can handle (in other languages, or with a big number library in C, it can handle something arbitrarily large).

Now... I'm not sure the use of log10 and pow is the fastest. But it's the most fun :-)

Answer 4

If it is a null terminated char *, then why not just do this:

int i = 0;
while(char[i] != null){
    switch(char[i]):

         case '1': printf("One "); break;
         ....
         i++;

}

if its not a char *, turn it into one with char * temp = itoa(number);

Answer 5

1234 / 1000 => 1
1234 % 1000 => 234     234 / 100 => 2
1234 % 100  => 34      34  / 10  => 3
1234 % 10   => 4

You can pretty easely build a loop around that, but you may go even easer with itoa, sprintf, and std::to_string .