Android Question Very fast reading a binary file

[AlpVir]

The following code reads a file 2 bytes at a time and stores the result in an array. It 's all right, but ... there is some other method of reading faster ?
Thanks in advance

Dim raf As RandomAccessFile
    Dim H      As Int
    Dim i As Int
    Dim Buffer(1201*1201) As Int  '  very big array
   
    raf.Initialize2 (File.DirRootExternal  , "DEM/3/N44E007.hgt", True, True)  ' LittleEndian
    Do While raf.CurrentPosition < raf.Size
        B1=raf.ReadUnsignedByte(raf.CurrentPosition )
        B2=raf.ReadUnsignedByte(raf.CurrentPosition )
        H=B1*256+B2
        Buffer(N)=H
        Log ("B1=" & B1 & "  B2=" & B2 & "  H=" & H)
        N=N+1
        If N>10 Then Exit   ' debug
    Loop
    '--- read the buffer
    Log ("read the buffer")
    For i=0 To 9
        Log(Buffer(i))
    Next   
    raf.Close

 

[thedesolatesoul]

It will be faster to read all (or some) of the bytes using ReadBytes and then later converting them to unsigned and doing operations on it.

 

[Informatix]

To read and write bytes, forget the RandomAccessFile library. It is very very slow. Read or write your byte array with a simple InputStream/OutputStream.

 

[AlpVir]

Reading with InputStreamToBytes lasts 0.8 sec.
The For-Next loop to rework the result lasts for 43 seconds.
There is some internal function that make the calculations more quickly?
I also feel InputStream but now I feel I have the impression that the sticking point is precisely the elaboration of the results, not the reading.

Dim b() As Byte = Bit.InputStreamToBytes(File.OpenInput(File.DirRootExternal  , "DEM/3/N44E007.hgt"))
For i=0 To b.Length -1 Step 2
    Buffer(N)=b(i)*256+b(i+1)
    N=N+1  
Next

 

[AlpVir]

Slightly faster: reading file 0.192 sec; data processing 43.53 seconds
You can do better ?
Tanks !

Dim InputStream1 As InputStream
    InputStream1 = File.OpenInput (File.DirRootExternal, "DEM/3/N44E007.hgt")
    L=File.Size (File.DirRootExternal, "DEM/3/N44E007.hgt")
    Dim b(L) As Byte
    count = InputStream1.ReadBytes (b, 0, b.length-1)
    For i=0 To b.Length -1 Step 2
       Buffer(N)=b(i)*256+b(i+1)
       N=N+1  
    Next

 

[RandomCoder]

Would using a bit shift instruction be quicker than mathematically multiplying by 256 I wonder?
B4A Help docs, bit

Regards,
RandomCoder

 

[thedesolatesoul]

No idea. Try

Buffer(N) = Bit.Or(Bit.ShiftLeft(b(i),8), b(i+1))

i remember B4A internally cnoverts most things to double so i wonder if this is casting overhead.

 

[Daestrum]

You could use nio to read the byte buffer as Int(s), but it would read 4 bytes per Int, maybe read it as Short(s) if you just want 2 byte values.

 

[Erel]

There is no double conversion in this case.

Are you testing the speed in Release mode? What is the file size?

 

[AlpVir]

The file is 2818 KB
I've always used the debug mode for this short piece of code.
Moving on to a real device in release mode I got the following times
Solution Buffer (N) = b (i) * 256 + b (i + 1) 0.7 sec
Solution Buffer (N) = Bit.OR (Bit.ShiftLeft (b (i), 8), b (i + 1)) 0.9 sec

These are times that I think are acceptable, at least in this stage of development of the app that reads and interprets the DEM files
Thanks to all of the aid

 

[thedesolatesoul]

I do however finding interesting that a Shift+OR is slower than multiply+add.
I guess modern 32/64-bit processors work in registers of atleast 32-bits so they dont even care if the data is only 8-bits, or they are optimized more for a MAC operation.

 

[RandomCoder]

I also find it surprising that bit shifting is slower than using arithmatic. Very interesting!

 

[MaFu]

A real bit shift would be faster. But B4A hasn't it and the function call proceeds the overhead.