Why this code is not getting valid for 10 ?

Question

#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main() {
char s[50];
int n;
fgets(s,50,stdin);
n=strlen(s);
for (int i=0;i<n;i++)
{ if(isdigit(s[i])&&(isdigit(s[i+1])!='0'))
{ switch(s[i])
{ case '1': printf("one");break;
case '2': printf("two");break;
case '3': printf("three");
break;
case '4': printf("four");break;
case '5': printf("five");break;
case '6': printf("six");break;
case '7': printf("seven");break;
case '8': printf("eight");break;
case '9': printf("nine");break;
// case '10':printf("ten");break;
default: printf("zero");break;}}
else if((isdigit(s[i])=='1')&&(isdigit(s[i+1])=='0'))
{
printf("ten");
}
else
{
printf ("%c",s[i]);
}
}
return 0;

}

Answer 1

Because you are often comparing a character to the return value of isdigit() (which is either zero or a positive number), in a context where you only want to compare the characters.
For example, to check if the next character is a zero,

s[ i + 1 ] == '0' would be sufficient.

Also, keep in mind that ten is the only numeral with two digits. What happens if you increment once like you would normally do after printing "ten"?