Audio and video files are already binary code. So I think you should be a bit more explicit.

I know - sorry.

I just want to make the codes visible. I've done it using the 'terminal' on a Mac computer, but that is only files that are already stored.

I want to make an audio/visual installation, were the video and audio are projected as binary numbers instead of picture and music. The video and audio are produced by the audience in real time, so the convert process has to be in real time.

Is there any way that 'Processing' (or other programs - perhaps 'MAX/MSP') that could communicate with the 'terminal'?

Best regards.

You can get the binary data and convert it to string using the hex() function, for example.
Then you can use text() to display it on the sketch area.

Okay, thank you! - That worked well, but I tried something else.

Now I got the code:

byte b[] = loadBytes("testimage.jpg");
 for (int i = 0; i < b.length; i++) {
   String a = binary(b[i] & 0xff);
   print(a);

}


And I'm getting a lot of beautiful numbers. But how to I 'stream' a file into processing, so I can get the binary string in real time?

Best regards

Aah, you wanted binary digits! 0s and 1s...

I don't know how you get the real time data, so it is hard to answer.

Exactly! ;-)

Okay, I'm diggin' the world wide web for answers then - but.

Can you tell me how to get the numbers into a text box? I can't figure that out..

Should it be:

size(100, 100, P3D);
PFont font;
font = loadFont("FFScala-32.vlw");
textFont(font);
text(print(b));
fill(0, 102, 153);
text(print(b));

I'm pretty new to all this - sorry for my ignorance.

More like text(str(binary(b[i])), posX, posY);

Thank you!

Hm - now I got:


byte b[] = loadBytes("EP.JPG");
 for (int i = 0; i < b.length; i++) {
   String a = binary(b[i] & 0xff);
   print(a);
}

PFont font;
font = loadFont("Meta-Bold.vlw");
textFont(font);
text(str(binary(b[i])), posX, posY);


And it says error: The method str(boolean) in the type PApplet is not applicable for the argument (String)

whats the point, it would be indistinguishable from random(0,1); unless you were broadcasting this to androids

The binary() function returns a string so it should be text(binary(b[i]), posX, posY); without the str() conversion function.