Why sizeof(array) and sizeof(&array[0]) gives different results?

Question

#include 
int main(void){
    char array[20];

    printf( \"\\nSize of array is %d\\n\", sizeof(array) );  //outputs 20
    printf(\"\\nSize of &         


        

Answer 1

&array[0] returns a pointer to the first index of array. Using the sizeof operator on a pointer will yield the size of a pointer (depends on ABI), in your case, it's 4.