Find the number of strings in an array of strings in C

Question

char* names[]={"A", "B", "C"}; 

Is there a way to find the number of strings in the array? For example, in this case it should output: 3.

Answer 1

In this case you can divide the total size by the size of the first element:

num = sizeof(names) / sizeof(names[0]); 

Careful though, this works with arrays. It won't work with pointers.

Answer 2

For an array, which the examples in the bounty are, doing sizeof(names)/sizeof(names[0]) is sufficient to determine the length of the array.

The fact that the strings in the examples are of different length does not matter. names is an array of char *, so the total size of the array in bytes is the number of elements in array times the size of each element (i.e. a char *). Each of those pointers could point to a string of any length, or to NULL. Doesn't matter.

Test program:

#include<stdio.h>  int main(void) {     char* names1[]={"A", "B", "C"}; // Three elements     char* names2[]={"A", "", "C"}; // Three elements     char* names3[]={"", "A", "C", ""}; // Four elements     char* names4[]={"John", "Paul", "George", "Ringo"}; // Four elements     char* names5[]={"", "B", NULL, NULL, "E"}; // Five elements      printf("len 1 = %zu\n",sizeof(names1)/sizeof(names1[0]));     printf("len 2 = %zu\n",sizeof(names2)/sizeof(names2[0]));     printf("len 3 = %zu\n",sizeof(names3)/sizeof(names3[0]));     printf("len 4 = %zu\n",sizeof(names4)/sizeof(names4[0]));     printf("len 5 = %zu\n",sizeof(names5)/sizeof(names5[0])); } 

Output:

len 1 = 3 len 2 = 3 len 3 = 4 len 4 = 4 len 5 = 5 

EDIT:

To clarify, this only works if you've defined an array, i.e. char *names[] or char names[][], and you're in the scope where the array was defined. If it's defined as char **names then you have a pointer which functions as an array and the above technique won't work. Similarly if char *names[] is a function parameter, in which case the array decays to the address of the first element.