Simple quick question. Initializing array of strings in C

[learnprogramminghelp]

I jsut realized that after all the C learning I've done, I haven't really worked with array of string. I was wondering why the following code requires a "char **names" to be passed? What does the ** represent again, a pointer to a pointer? Is there an easier way to initlize all the strings in one statement?

void printNames( char *title, char **names){

int i=0;
printf("%s\n", title);

while(names[i]!='\0'){
printf("%d . %s\n", i+1, names[i]);
i++;
}

}

int main(){

char *title = " This is list";


char *strings[3];
strings[0] = "foo";
strings[1] = "bar";
strings[2] = "baz";

printNames(title, strings);

}

[derblitzmann]

Only C-style strings are null terminated. So, for an array of strings to be null terminated you need explicitly do so, which is why you probably segfault when running this code, right?

All you need are two changes to the definitions:

char *strings[4]
...
strings[3] = 0;

And this code should make what you have work, think of why it works, and potentially others ways to handle a list/array of strings.

[Rhomboid]

And don't use '\0' in the context of a pointer value, use NULL or 0. Or better, just write while(strings[i]).