Author: Daniel Shawul
Date: 03:34:18 07/21/04
Go up one level in this thread
On July 21, 2004 at 06:27:06, Paul Clarke wrote:
>On July 21, 2004 at 06:16:10, Daniel Shawul wrote:
>
>>
>>Isn't the size of a struct the sum of the sizes of its members.
>>
>>but for this struct sizeof(HASH_E) gives me 16 when i expect 14
>>
>>struct HASH_E
>>{
>> HASHKEY checksum;
>> char from;
>> char to;
>> short eval;
>> unsigned char depth;
>> unsigned char entry_threat_promote_seq;
>>};
>
>Structures are padded so that their size is a multiple of the largest alignment
>size required by any of the members. In this case that would be the HASHKEY
>which, depending on your architecture, the compiler will probably want to align
>on a four- or eight-byte boundary, so the compiler adds a couple of extra bytes
>to make the structure a multiple of four or eight bytes. The idea is that, if
>you have an array of these structs, all the fields in all the elements of the
>array will have the required alignment.
Thanks very much.
I changed the "struct member alignment" to 1 byte. Previously it was
8 byte. And now it displays the correct size 14.
Is there any efficiecy loss by changing the alignment?? If there is any
i don't want to do that?
daniel
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