Thread: how do I find the length of and array using sizeof()

Damm I forgot this I knew this a week ago but it slipped out.

Code:

#include<stdio.h>

main() 
{
       int x;
       char array[100][50] =
       {
            {"element 1"},
            {"element 2"},
            {"element 3"},
            {"element 4"}
       };
        
       x = sizeof(array)/sizeof(array[0]);
       printf("the size of the array is %d",x);
       getchar();
}

I am not getting the right results it should be 4 not 100 I forgot this I knew this a week ago. I guess the old saying is true in programming if you don't use it you lose it

is there a short cut notation in c to initialize the rest of the elements to the null character?

Thanks for the reply by the way I really appreciate it

char array[100] = {0};

you use this short hand notation to explicitly set the first element to zero and implicitly set all the other elements to zero
when you first initialize an array


I just had to get that into in my head

Code:

     int i;
  
      char array[100] = {0}; // explicitly set the first element to zero and implicitly set all the other elements to zero 

      array[0] = 1;
      array[1] = 2;
      
      for(i =0; array[i] != 0; i++)
      {
      }      
      
      printf("the size of the array is %i",i);

thanks all later gents I like this forum by the way.

This does only work if the actual content of an array element isn't 0, which is absolutely possible.