SysMemCpy and number of bytes

2020-02-27T14:27:48.359000Z

Great !
Thanks a lot !
Cheers.

SysMemCpy and number of bytes

2020-02-27T14:27:46.975000Z

Originally created by: scott_cunningham

The problem is byte alignment that happens behind the scenes. In your first example, a DWORD can only start on a even, even byte (such as position 0 or 4 or 8, etc). But you list a WORD first and it takes position 0 and 1. A DWORD cannot start at position 2, so the compiler moves it to position 4. In this case your structure is 8 bytes (6 +2 dead bytes).

Same idea is true for your second example (a WORD cannot start on an odd position).

To avoid this problem, you should always use SIZEOF(structure) to define the number of bytes to copy. Do this also for "normal" variables to avoid platform specific size problems (not every hardware system has a byte taking only one space...)

DUT2_BYTE_SIZE : __XWORD := SIZEOF(DUT2);

SysMemCpy and number of bytes

2020-02-27T14:27:46.314000Z

I have this DUT

TYPE DUT2 :
STRUCT
   word1:DWORD;
   dword1:DWORD;
END_STRUCT
END_TYPE

and this

TYPE DUT3 :
STRUCT
   by1:BYTE;
   wd1:WORD;
END_STRUCT
END_TYPE

my pou for test is

PROGRAM TEST
VAR
   Src: DUT2 ;
   Dest: DUT2 ;
   N_of_bytes: __XWORD := 6;
END_VAR

SysMemCpy(
    pDest :=ADR(Dest),  (* destination *)    
    pSrc := ADR(Src),  (* source*)    
    udiCount := N_of_bytes,  (* N of bytes to copy *)    
);

Here my first question:
i thinked one word + one doubleword = 6 byte but
Why if i set N_of_bytes=6 i can't copy the doubleword right? just using 8 ?

and my second question, using

PROGRAM TEST
VAR
   Src: DUT3 ;
   Dest: DUT3 ;
   N_of_bytes: __XWORD := 3;
END_VAR

i thinked one byte + one word = 3 byte but
Why if i set N_of_bytes=3 i can't copy the word right? just using 4 ?

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