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
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.
chars, terminated by acharwith value zero. For an array to hold such a sequence, it needs space for the terminating zero. If there is no zero, it is not a string, and treating it like a string makes no sense.char *s[] = { "one", "two", "etc..." };That way you don't need to think about how long the strings are, or even how many are in the array. If you need to know the number of strings in the array, usesize_t count = sizeof(s) / sizeof(s[0]);