title: How to implement a^b without `pow`?
body: <p>I need to write a function to compute a^b but I am not allowed to use <code>pow</code>. Any ideas? I am lost. 
  It looks like problem is in main now...
  Somewhere it gets that vys is what i characterise it. So if i set that vys=1 in main i get 1 in output.. </p>
  
  <pre><code>#include &lt;stdio.h&gt;
  #include &lt;time.h&gt;
  #include &lt;math.h&gt;
  #include &lt;unistd.h&gt;
  
  void multiplied(int b, int n)
  {
    int i=1, vys=1;
    while (i&lt;=n)
    {
      vys *=b;
  
      i++;
    }
    return vys;
  }
  
  
  main(void)
  {
    int b=0, n=0, vys=1;
    printf("Give numbers b and n but they must be in interval &lt;0,10&gt;!\n");
    scanf("%d %d", &amp;b, &amp;n);
    if ((b &lt; 0 || b&gt;10) || (n&lt;0 || n&gt;10))
    {
      printf("Numbers are not in interval &lt;0,10&gt;!\n");
    }
    else
    {
      printf("Number is in interval so i continue...\n");
      sleep(2);
     vys= multiplied(&amp;b, &amp;n);
      printf("%d", vys);
  }
  </code></pre>