Getting unexpected output for string array

Question

a = 1;
b = 9;
char s[9][5]={"one","two","three","four","five","six","seven","eight","nine"};
for(int i=a;i<=b;i++)
{
  if(i<10)
     printf("%s\n",s[i-1]);
  else
    {
      if(i%2==1)
         printf("odd\n");
      else
         printf("even\n");
    }  
}

expected:

one
two
three
four
five
six
seven
eight
nine

got:

one
two
threefour
four
five
six
seveneightnine
eightnine
nine

Answer 1

Not all elements of this array

char s[9][5]={"one","two","three","four","five","six","seven","eight","nine"};

contain a string. The type of elements is char[5]. So for example the string literal "three" is not completely contained in the third element of the array because there is no space to store the terminating zero character '\0' of the string literal and the conversion specifier %s is designed to output characters until the terminating zero character '\0' is encountered. This is the reason of the appearance of such an output like

threefour
seveneightnine
eightnine

So either you need to increase the size of elements of the array like

char s[9][6]= { /*...*/ };

or to use the following format string in the call of printf

printf("%.*s\n", 5, s[i-1]);

Pay attention to that this if statement

  if(i<10)

does not make a great sense because i is always less than 10.