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PIC18F47J53 CCS C Compiler 5.090 bit shift

 
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Marco27293



Joined: 09 May 2020
Posts: 58

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PIC18F47J53 CCS C Compiler 5.090 bit shift
PostPosted: Tue Jul 21, 2020 9:24 am     Reply with quote

Hi,

Is there any diference between:

Code:
for(k=0;k<(Hex_File_Info[Trunks_number]<<1);k+=2)                                       
   {
      err+=M24M02_random_read_2byte_Loader(1,0x208+(k<<1),&trunc_address[k]);
      err+=M24M02_random_read_2byte_Loader(1,0x20A+(k<<1),&trunc_address[k+1]);
   }

//and

for(k=0;k<(Hex_File_Info[Trunks_number]*2);k+=2)                                       
   {
      err+=M24M02_random_read_2byte_Loader(1,0x208+(k*2),&trunc_address[k]);
      err+=M24M02_random_read_2byte_Loader(1,0x20A+(k*2),&trunc_address[k+1]);
   }


k RAM variable value behaviour is the same or not?
temtronic



Joined: 01 Jul 2010
Posts: 8220
Location: Greensville,Ontario

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PostPosted: Tue Jul 21, 2020 10:40 am     Reply with quote

The easiest way to see is cut a small program with both versions, compile it, then dump the listing and look at the code that was generated.

The compiler is smart and probably knows <<1 is same as *2 BUT you really should look and see

Jay
Ttelmah



Joined: 11 Mar 2010
Posts: 17478

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PostPosted: Tue Jul 21, 2020 11:49 am     Reply with quote

It depends on the size and type of 'k'.
Bit rotation is only guaranteed to be a *2, on an positive value, with
less than one minus the number of bits held by the number of bits
storable by the value.

So if you have a signed int16, if the rotation goes out of the fifteenth bit,
the equivalence is not guaranteed. Similarly rotation of a -ve number
does not have this guaranteed equivalence.
jeremiah



Joined: 20 Jul 2010
Posts: 1248

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PostPosted: Tue Jul 21, 2020 1:50 pm     Reply with quote

Also keep in mind that per the C standard, left shifts of signed integers are potentially undefined (Section 6.5.7/4 of the standard), so make sure any data that you left shift is not signed.
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