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Spool



Joined: 08 Jan 2013
Posts: 154
Spool
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Last edited by Spool on 17 Mar 2013, 04:07; edited 2 times in total
Post 12 Jan 2013, 16:55
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HaHaAnonymous



Joined: 02 Dec 2012
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Post 12 Jan 2013, 17:11
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Spool



Joined: 08 Jan 2013
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Spool
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Last edited by Spool on 17 Mar 2013, 03:51; edited 1 time in total
Post 12 Jan 2013, 17:18
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HaHaAnonymous



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Post 12 Jan 2013, 17:35
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Bargest



Joined: 09 Feb 2012
Posts: 79
Location: Russia
Bargest
Memory operations are much slower than register ones.
Code:
mov eax, [a] ; 1
inc eax
mov [a], eax ; 2
    

2 memory operations.
Code:
inc dword [a] ; 1
    

1 memory operation.
But if we have more complex calculations, there will be more memory operations in the second case.
Post 12 Jan 2013, 18:27
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JohnFound



Joined: 16 Jun 2003
Posts: 3502
Location: Bulgaria
JohnFound
Bargest, "inc [memory]" takes 2 memory operations as well. Because the CPU can not do arithmetic directly in the memory. The value have to be read, processed and then write. Of course the cash memory can affect this process as well. (although I also think "inc [memory] faster" because at least only one instruction have to be read from the memory, decoded and executed. Well, from the cash actually. Smile
Post 12 Jan 2013, 20:18
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Bargest



Joined: 09 Feb 2012
Posts: 79
Location: Russia
Bargest
Quote:
Bargest, "inc [memory]" takes 2 memory operations as well.

I know this. But:
1) It is one operation, so just one opcode is peeked out from buffer and decoded;
2) I think in this instruction address is set on address bus just for one time (it is one of the longest parts of memory operations).

So I consider it is 1.5 memory operation.Very Happy

And cache improves speed of first case as well.Smile
Post 12 Jan 2013, 20:24
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