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Spool

Joined: 08 Jan 2013
Posts: 151

Spool 12 Jan 2013, 16:55

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Last edited by Spool on 17 Mar 2013, 04:07; edited 2 times in total

12 Jan 2013, 16:55

HaHaAnonymous

Joined: 02 Dec 2012
Posts: 1178
Location: Unknown

HaHaAnonymous 12 Jan 2013, 17:11

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12 Jan 2013, 17:11

Spool

Joined: 08 Jan 2013
Posts: 151

Spool 12 Jan 2013, 17:18

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Last edited by Spool on 17 Mar 2013, 03:51; edited 1 time in total

12 Jan 2013, 17:18

HaHaAnonymous

Joined: 02 Dec 2012
Posts: 1178
Location: Unknown

HaHaAnonymous 12 Jan 2013, 17:35

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12 Jan 2013, 17:35

Bargest

Joined: 09 Feb 2012
Posts: 79
Location: Russia

Bargest 12 Jan 2013, 18:27

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.

12 Jan 2013, 18:27

JohnFound

Joined: 16 Jun 2003
Posts: 3499
Location: Bulgaria

JohnFound 12 Jan 2013, 20:18

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

12 Jan 2013, 20:18

Bargest

Joined: 09 Feb 2012
Posts: 79
Location: Russia

Bargest 12 Jan 2013, 20:24

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

12 Jan 2013, 20:24

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