Processing byte arrays

[paul]

Hi all,
Could some kind soul peruse the following code and see if there is
anything wrong with it?
Its producing output, but its only occupying the first third of the
output array; to give an example, think of a TV screen where only the
top third shows anything, and what is does show is repeated 3 times.

The code should skip through an array of bytes, taking them in groups
of three, and then writing the results of a calculation on the three
bytes to a third array, so that

second array[0] = results of calculation on first array[ array elements
0,1,2]
second array[1] = results of calculation on first array[ array elements
3,4,5]

- in essence converting 24 bit colour to 8 bit greyscale.

There are elements in the code to allow jumping over redundant elements
in the first array;
for instance, you might only want to scan through the array elements
array[byte_offset ... byte_offset + m_i_total_bytes _to_read]
and then array[byte_offset + m_i_bytes_to_ju mp ... byte_offset +
m_i_total_bytes _to_read + m_i_bytes_to_ju mp]
etc.

TIA

Paul

NB: "int" is synonymous with 32 bit unsigned ints

int i_bytes_process ed=0;

byte * pixel;

pixel = image->image_data; // image data is defined as byte *
pixel+=byte_off set; // We might not have to start at
(0,0) in the image

// total bytes to read is 3*original
data width x height - 24 bit colour,
// ditto for "bytes_to_read_ per_row"
for( int bytes_read = 0; bytes_read < m_i_total_bytes _to_read;
bytes_read+=m_i _bytes_to_read_ per_row)
{

for(int i=0; i< m_i_bytes_to_re ad_per_row; i+=3)
{
byte red = pixel[2]; // last 8 bits is red data
byte green = pixel[1]; // middle 8 bits is green data
byte blue = pixel[0]; // first 8 bits is blue data

// Do some
processing on the values (omitted)

// Write out to the array -
this is a linked list
//
containing bytes elements

m_image_average _b_w.SetAt(i_by tes_processed, calc_result );

i_bytes_process ed++;

pixel+=3;
}

pixel+=m_i_byte s_to_jump; // We might not need to
// start processing
from the first byte of the next row.
}

[Frederick Gotham]

NB: "int" is synonymous with 32 bit unsigned ints

Not in C++, it ain't.
"int" is an abbreviation of "signed int", NOT "unsigned int".
If you want an abbreviation, use "unsigned".

unsigned i = 0;
byte * pixel;

Redudant statement -- does absolutely nothing. Perhaps you meant something
like:

byte *= pixel;
I'd need more context (and a fine-tooth comb) to tell you any more.
--

Frederick Gotham

[paul]

Frederick Gotham wrote:

NB: "int" is synonymous with 32 bit unsigned ints

Not in C++, it ain't.

I meant within the context of this example.


"int" is an abbreviation of "signed int", NOT "unsigned int".
If you want an abbreviation, use "unsigned".

unsigned i = 0;

byte * pixel;

Redudant statement -- does absolutely nothing. Perhaps you meant something
like:

byte *= pixel;

Don't you mean byte * pixel = image->image_data ?


I'd need more context (and a fine-tooth comb) to tell you any more.

What it does, simply put is
second_array[0] = f( first_array[0,1,2])
second_array[1]=f( first_array[3,4,5])
second_array[2]=f( first_array[6,7,8])
etc., where the function f(...) is just some junk I do on the 3 byte
array elements of the first array

[Frederick Gotham]

Paul posted:

Don't you mean byte * pixel = image->image_data ?

Think I might have found your error.

Here's a verbatim copy-paste from your original post:

byte * pixel;

pixel = image->image_data; // image data is defined as byte *
You've two problems here.
First of all, let's change it to what you intended:

byte * pixel = image->image_data;
Take out your favourite C++ operator precedence table (
http://www.difranco.net/cop2334/cpp_op_prec.htm ), and you'll see that
multiplication is performed before assignment -- so you're trying to
assign something to an R-value.

And even if you supply the necessary parentheses:

bytes * (pixel = image->image_data);

You're still left with a partly redundant statement, because the result
of the multiplication is discarded.

I'm still not quite sure what you're trying to do.
--

Frederick Gotham

[paul]

Frederick Gotham wrote:

Paul posted:

Don't you mean byte * pixel = image->image_data ?

Think I might have found your error.

Here's a verbatim copy-paste from your original post:

byte * pixel;

pixel = image->image_data; // image data is defined as byte *
You've two problems here.
First of all, let's change it to what you intended:

byte * pixel = image->image_data;
Take out your favourite C++ operator precedence table (
http://www.difranco.net/cop2334/cpp_op_prec.htm ), and you'll see that
multiplication is performed before assignment -- so you're trying to
assign something to an R-value.

And even if you supply the necessary parentheses:

bytes * (pixel = image->image_data);

Er, I was perhaps a bit glib when I use "byte" I mean, by "byte *
pixel", that pixel is
a pointer to a byte.

[Frederick Gotham]

Pul posted:

Er, I was perhaps a bit glib when I use "byte" I mean, by "byte *
pixel", that pixel is
a pointer to a byte.

Wups, it's me that's confused! I'm used to having a capital letter at the
start of a type's name, e.g.

Byte *pixel;
When I saw:

byte * pixel;
I (mistakenly) presumed that they were both objects, and that you were
performing a redundant multiplication.
--

Frederick Gotham