Re: A Certain Size/sizeof

From: Gene Wirchenko (gwirchenkoEXCEPT_at_CAPITALSwencomine.com)
Date: 11/15/03 

Date: Fri, 14 Nov 2003 23:47:32 GMT

On Fri, 14 Nov 2003 18:19:28 -0500, Robert W Hand
<rwhand@NOSPAMoperamail.com> wrote:

[snip]

>count should have the number of rows in the array. Let's take it from
>the beginning. Note that I shall indicate the object array as arr to
>avoid hopeless confusion in the following note.
>
>arr is an array that has 17 columns. The number of rows is determined
>by the initializer. sizeof is an operator that works on the compiler
>side. It determines the size of the object operand in bytes.
>
>So sizeof arr will be the number of bytes that is in arr. Such a size
>will be implementation defined. Note that when the operand of sizeof
>is an array, the array does not decay into a pointer to some type.
>
>arr is an array of an array of 17 ints. When used as the operand of
 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>the indirection operator, arr decays into a pointer into an array of
>17 ints. So applying the indirection operator to arr results in an
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>array of 17 ints. When the sizeof operator is then applied, the
 ^^^^^^^^^^^^^^^^^
>resulting array will not decay further. Rather it will yield the size
>of each row in bytes.
>
>So dividing sizeof arr by sizeof *arr will be the number of rows in
>the object array, arr.

     That was a very clear explanation. I could follow it
step-by-step. The above were the missing bits (for me).

     Thank you.

><Observation>
>Reading the "arguments" of both sides, I really do not see any reason
>for further anger. It reads like a misunderstanding by both parties.
>Please kiss and make up. ;-)
></Observation>

Sincerely,

Gene Wirchenko

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