C Get size of Array of array of chars

Question

i want to get the number of elements at Text array the answer should be 2

char Text[5][10] = {
    "Big12345",
    "Big54321",
};

i want to a code to count number of elements in array of chars

Answer 1

You are mistaken. The number of elements in the array is 5. Two elements have non-empty strings and three elements have empty strings. But In fact an empty string can be placed anywhere in the array. For example

char Text[5][10] = 
{
    "Big12345",
    "",
    "Big54321",
};

This declaration is equivalent to

char Text[5][10] = 
{
    "Big12345",
    "",
    "Big54321",
    "",
    ""
};

You could write a function that determines how many elements contain non-empty strings. For example

#include <stdio.h>

size_t count_non_empty( size_t m, size_t n, char s[][n] )
{
    size_t count = 0;

    for ( size_t i = 0; i < m; i++ )
    {
        count += s[i][0] != '\0';
    }

    return count;
}

int main(void) 
{
    char Text[5][10] = 
    {
        "Big12345",
        "",
        "Big54321",
    };

    printf( "There are %zu non-empty elements\n", count_non_empty( 5, 10, Text ) );

    return 0;
}

The program output is

There are 2 non-empty elements

Answer 2

In this particular case, anything after the initializers will be 0, so:

size_t counter = 0;

while ( Text[counter][0] != 0 )
  counter++;

But, in general, C doesn't give you a good way of doing this. You either have to track the number of elements being used separately, or you have to use a sentinel value in the array.

Answer 3

Use the following to find the number of elements in your array which have a string assigned to them:

#include <stdio.h>
#include <string.h>

int main()
  {
  char Text[5][10] = {"Big12345",
                      "Big54321",};
  int i, n;

  for(i = 0, n = 0 ; i < 5 ; i++)
    if(strlen(Text[i]) > 0)
      n += 1;

  printf("%d elements have a length > 0\n", n);

  return 0;
  }