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    <p>Hey ! I recognize a famous homework I had in first year of my school too (@Epitech students : don't copy/paste the following code, try to come up with your own solution, it's for your own good ^^)</p> <p>The solution to your problem is to perform the problem in a recursive way :</p> <pre><code>void my_putnbr_base(int num, int base) { int start; int remainder; remainder = num % base; start = (num - remainder) / base; if (start != 0) my_putnbr_base(start, base); if (remainder &gt;= 10) printf("%c", remainder + 55); else printf("%d", remainder); } </code></pre> <p>Does your homework specifies that it should only work with positives numbers ? If not, it's easy to include the negative numbers handling :</p> <pre><code>void my_putnbr_base(int num, int base) { int start; int remainder; if (num &lt; 0) { putchar('-'); my_putnbr_base(-num, base); } else { remainder = num % base; start = (num - remainder) / base; if (start != 0) my_putnbr_base(start, base); if (remainder &gt;= 10) printf("%c", remainder + 55); else printf("%d", remainder); } } </code></pre> <p>@arno : that's true, because the exemple code is using ASCII table. If we want something trully flexible we need the base in parameter. For example :</p> <pre><code>&gt;&gt; my_putnbr_base(4242, "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ") 39U &gt;&gt; my_putnbr_base(42, "0123456789ABCDEF") 2A </code></pre> <p>this implements the example :</p> <pre><code>void my_putnbr_base(int num, char *base) { int start; int remainder; int len; len = strlen(base); if (num &lt; 0) { putchar('-'); my_putnbr_base(-num, base); } else { remainder = num % len; start = (num - remainder) / len; if (start != 0) my_putnbr_base(start, base); printf("%c", base[remainder]); } } </code></pre> <p>I hope it solves your problem !</p> <p>edit: I didn't read correctly ^^ You are not allowed to use functions, so recursion is out of the question... Here is an interative way, you can put this in a main(). You can improve this code by adding the negative numbers handling and flexible bases, as I showed you :)</p> <pre><code>int my_putnbr_base_it(int num, int base) { unsigned int quotient = 1; unsigned int remainder; while ((num / quotient) &gt;= base) quotient *= base; while (quotient) { if ((remainder = (num / quotient) % base) &lt; 10) printf("%d", remainder); else printf("%c", 55 + remainder); quotient /= base; } return (0); } </code></pre> <p>Hope it solves everything now !</p>
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    1. POPrint the digits of a number in reverse order without arrays or functions
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    1. USThomas Joulin
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