defining and iterating through array of strings in c

Question

How can I define an array of string in c then iterate with a loop through the items in the array?

So far I have

char myStrings[][10] = { "one", "two", "three", "four", "five" };
// do I need to specify the "10" maximum length?
// also does it automatically create a null ending character after the string?

int i = 0;
for( i = 0; i < ; i++)
{
// I need to pass each string to  a function which accepts
// const char *
}

Answer 1

When you declare a char sequence with "", null terminator is added.

char myStrings[][10] = { "one", "two", "three", "four", "five" };

for (size_t i = 0; i < sizeof(myStrings) / sizeof(myStrings[0]); i++)
{
    fooThatReceivesOneString(myStrings[i]);
}

---

Edit - sizeof()

sizeof() returns the size of a variable. It doesn't matter if the variable is an int, array or 2d array.

For example, see the following program

#include <stdio.h>
int main() {
    char myStrings[][10] = { "one", "two", "three", "four", "five" };
    printf("sizeof(myStrings):    %zu\n", sizeof(myStrings));
    printf("sizeof(myStrings[0]): %zu\n", sizeof(myStrings[0]));
    return 0;
}

Which outputs (on my machine):

sizeof(myStrings):    50
sizeof(myStrings[0]): 10

Since each element in the array has the same size, we can divide the sizeof an array with the sizeof the first element to get the amount of elements in the array.

Answer 2

void loopftn (void)
{
  char *numbers[] = {"One", "Two", "Three", ""}, **n;

  n = numbers;
  while (*n != "") {
    printf ("%s\n",  *n++);
  }
  return;
}

1. You don't need to have a maximum length.
2. All strings in C are null-terminated.