• RenatoL <rexlen gmail.com> Dec 24 2011
• "Mr. Anonymous" <mailnew4ster gmail.com> Dec 24 2011
• "Regan Heath" <regan netmail.co.nz> Dec 28 2011

RenatoL <rexlen gmail.com> writes:

snippet:


int[] arr1 = [1,2,3,4,5];
int[5] arr2 = [1,2,3,4,5];
writeln(arr1.sizeof);
writeln(arr2.sizeof);

Output:
8
20

"0 is ok to me but why "8"??

"Mr. Anonymous" <mailnew4ster gmail.com> writes:

On 24.12.2011 18:46, RenatoL wrote:
 snippet:


 int[] arr1 = [1,2,3,4,5];
 int[5] arr2 = [1,2,3,4,5];
 writeln(arr1.sizeof);
 writeln(arr2.sizeof);

 Output:
 8
 20

 "0 is ok to me but why "8"??


8 is the size of the int[] type, which contains two pointers (or a 
pointer and a size).
To get 20, you can use:
arr1[0].sizeof * arr1.length

"Regan Heath" <regan netmail.co.nz> writes:

On Sat, 24 Dec 2011 16:46:18 -0000, RenatoL <rexlen gmail.com> wrote:

 snippet:


 int[] arr1 = [1,2,3,4,5];
 int[5] arr2 = [1,2,3,4,5];
 writeln(arr1.sizeof);
 writeln(arr2.sizeof);

 Output:
 8
 20

 "0 is ok to me but why "8"??


It's a quirk of D that int[] is a reference type, so you get the size of  
the reference (as My Anonymous said, a length and pointer) whereas int[5]  
is a value type, so you get the size of the value.

It's the same as the following C..

int *arr1;
int arr2[5];

printf("%d\n", sizeof(arr1));
printf("%d\n", sizeof(arr2));

R

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