Thread: Sound & MIDI - what's a DWORD?

I just cludged this together today

Code:

#include <windows.h>
#include <cmath>
#include <iostream>
#include <iomanip>

using namespace std;

double Semitone = pow(2.0,(1.0/12.0));
double Pitch = 16.3516;


int main()
{
    cout<<left;
    cout<<"Chromatic Scale: Eight Octaves with frequencies";
    for (int n = 12; n<109; n++) // Corresponding to MIDI note numbers
    {if (!(n%6))
    {cout<<endl;
    if (!(n%12))
    {cout<<endl;}}
    cout<<setw(12)<<Pitch;
    Beep(Pitch, 75);
    Pitch = Pitch*Semitone;
    }
    cout<<"Done!";
    cin.get();
    return 0;
}

but I'd like to MIDI it, and then move on to truly useful stuff. I've got the simple "C-E-G" MIDI code working, but what the heck is a DWORD and how do I insert my desired note # into the 5th and 6th hex digits??

Thanks -

JM

(lots of books but no time or patience!)

Duh, that makes sense! Thanks - now how can I replace the 2nd (3rd?) byte with a variable?

Code:

DWORD NoteC = 0x007f3c90; 
DWORD NoteE = 0x007f4090; 
DWORD NoteG = 0x007f4390;

something like

Code:

DWORD NoteX = 0x007fX90

if X is two hex digits.

Sorry for the stupid n00b questions!

-JM

Code:

0x007f0090 | (myVariable << 8)

Where myVariable has values of 0 - 255. The | is the binary OR operator, and the << is a bitwise shift. Binary or takes each bit of the two operators, and if either one is 1, the corresponding bit in the result is 1, otherwise zero. Bitwise shifts just move bits. The binary number 10110 shifted two to the left is 1011000. (Bits will fall off of both ends. 1001 shifted right once is 100.) Google for more info.

Exactly what I needed - thanks!