Question

How to get the size of an array of pointers in C

Answer 1

#include <stdio.h>
   
int main() {
    int array[] = {8, 3, 17, 6, 90};
    
    int* arrayp[] = {&array[0], &array[1], &array[2], &array[3], &array[4]};

    printf("%zu", sizeof(arrayp) / sizeof(arrayp[0]));

    return 0;
}


        
/*
run:
      
5
         
*/

 

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