Header chunk and the number of tracks

[kimyj1503]

Hello. I recently started studying MIDI messages and Fluidsynth.
After reading some document about the header chunk, I checked the related code on src/midi/fluid_midi.c.
I have some trouble understanding some codes and any help would be much appreciated.

 
The structure of the header chunk of the MIDI file.

	chunk ID	length	format type	number of tracks	time division
size in byte	4	4	2	2	2
value	"MThd"	6	0,1, or 2	1 to 65535	0 or 1 for MSB

The header chunk of the MIDI file is read through the function fluid_midi_file_read_mthd.
 

The below is how I understand the code about the time division, when the MSB equals 0.

mf->uses_smpte = 0;
mf->division = ((unsigned)mthd[12] << 8) | ((unsigned)mthd[13] & 0xff);

/*
The lower case alphabets here symbol the binary value of 0 or 1.
Lets say the binary values are as below
mthd[12] = 0abc defg
mthd[13] = hijk lmnp

Then
mf-> division will become 0abc defg hijk lmnp
*/

 
 

But I can't really understand the code for the number of tracks.
Why is mthd[10] shifted 16 bits instead of 8 bits?

mf->ntracks = (unsigned) mthd[11];
mf->ntracks += (unsigned int)(mthd[10]) << 16;

/*
The lower case alphabets here symbol the binary value of 0 or 1.
Lets say the binary values are as below
mthd[10] = abcd efgh
mthd[11] = ijkl mnpq

Then
mf-> ntracks will NOT become abcd efgh ijkl mnpq
*/

I understand that due to the below lines in the fluid_player_add_track function, ntracks value larger than 127 won't be used in other part of the code and ntracks value smaller than 128 will be converted to the unsigned integer as intended.

if(player->ntracks < MAX_NUMBER_OF_TRACKS)
    {
        player->track[player->ntracks++] = track;
        return FLUID_OK;
    }

Thank you in advance.

[derselbst]

Why is mthd[10] shifted 16 bits instead of 8 bits?

Because it's a bug in 20 year old legacy code. And it doesn't make a difference, because of the MAX_NUMBER_OF_TRACKS check you've correctly pointed out. Yet, you're welcome to draft a PR.