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int array address, sizeof

RicardasSim | PRO | 07/06/19 08:51:47 AM UTC | 0 ⭐ | 644 👁️ | Never ⏰ | []
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#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
 
int main()
{
 
    int arr[10]={0};
   
    printf("Address:\n\n");
    
    printf("arr: %p\n", (void*) arr); 
    printf("&arr: %p\n", (void*) &arr);
    printf("&arr[0]: %p\n", (void*) &arr[0]);
    printf("&arr[1]: %p\n", (void*) &arr[1]);
    
    printf("\nsizeof:\n\n");
    
    printf("arr: %zu\n", sizeof arr); 
    printf("&arr: %zu\n", sizeof &arr);
    printf("&arr[0]: %zu\n", sizeof &arr[0]);
    printf("&arr[1]: %zu\n", sizeof &arr[1]);
    printf("*arr: %zu\n", sizeof *arr);
    printf("arr[0]: %zu\n", sizeof arr[0]);
    
    return 0;
}
 
/*
output:
 
Address:
 
arr: 0x7fffe67bf280
&arr: 0x7fffe67bf280
&arr[0]: 0x7fffe67bf280
&arr[1]: 0x7fffe67bf284
 
sizeof:
 
arr: 40
&arr: 8
&arr[0]: 8
&arr[1]: 8
*arr: 4
arr[0]: 4
 
 
*/

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