Author: Daniel Shawul
Date: 05:30:33 07/21/04
Go up one level in this thread
On July 21, 2004 at 06:42:23, Aivaras Juzvikas wrote:
>On July 21, 2004 at 06:34:18, Daniel Shawul wrote:
>
>>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
>
>how did u change it?
follow the link if you are using MSVC.
project -> settings -> c/c++ -> catagory [code generation] -> struct member
alignment.
daniel
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