Converting a little endian number into long

[lucky6969b]

long x_size = buff2[3] << 24 | buff2[2] << 4 | buff2[1] >> 4 | buff2[0] >> 24;

The original value: 6630836
The value stored in file: b4 2d 65 00

I read the number as

char *buff2[5] = { 0 };
f.read(buff2, sizeof(char)*4);

The result is ffffffff (-1)

Are there any compact ways I can convert the number into the right form?
I guess I am not doing it right for the moment....
Thanks
Jack

[VictorN]

The size of the char is 8 bits. So how do you shift it 24 bits left (or right)?

/Edit: besides: why not just read this long value into the long variable?

[lucky6969b]

I thought I was doing this
00000000 1100101 00101101 10110100
-----------
|________________________>24bits???

Incorrect I think, what is your insight?

Going to tell my story though,
I have several x files to pack into one pck file to avoid multi-opening.
So I read the file size by ftell then write it to the pck file.
I must confess there is some inconsistency. I used fwrite (const void *)&filesize, sizeof(long),1, f_ptr);
then the data. Then reading it back into memory using the above format.
Any suggestions that I can do it better?
Thanks
Jack

[alanjhd08]

Also, you are reading into an array of 5 char pointers, not 5 chars. I think you meant char buff2[5].

The algorithm you are trying to use is also wrong. Try to get that right to start with, i.e. work out by how much you need to multiply each byte to get what you're looking for, then think about how to code that.

[VictorN]

Yes, you think right! You did it incorrect. See my previous post.

[lucky6969b]

Sorry Alan, Yes, i meant char buff2[5];

Victor, let me take a thought

x_size = (long) buff2;

The value seems to be off though...
Jack

[VictorN]

Sorry Alan, Yes, i meant char buff2[5];

So what is your actual code?
This one:

char *buff2[5] = { 0 };
f.read(buff2, sizeof(char)*4);

The result is ffffffff (-1)

Or this one:

char buff2[5] = { 0 };
f.read(buff2, sizeof(char)*4);

[VictorN]

Sorry Alan, Yes, i meant char buff2[5];

Victor, let me take a thought

x_size = (long) buff2;

The value seems to be off though...
Jack

No it is wrong!
Why not just _open a file and then _read to the long buffer exactly 4 bytes (or the sizeof(long))?

[lucky6969b]

This one:

char buff2[5] = { 0 };
f.read(buff2, sizeof(char)*4);

Thanks a lot

Update
Error 2 error C2664: 'std::basic_istream<_Elem,_Traits>::read' : cannot convert parameter 1 from 'long' to 'char *' l:\shader approach\partiii\chapter 14\xfiledemo\d3dutil.cpp 207 XFileDemo

long buff2;
f.read (buff2, sizeof(long));

Sorry, still couldn't make it

I am using this method because I need Unicode support.

Update:
I also tried this with no luck

x_size = (*buff2 >> 24) | ((*buff2 & 0x00ff0000) >> 8) | ((*buff2 & 0x0000ff00) << 8) | (*buff2 << 24);

Thanks

[lucky6969b]

I found a workaround this.. Thanks a lot

f.read(reinterpret_cast<char *>(&buff2), sizeof(long));