So, I'm working with Java, and testing out various bitwise functions, shifts, whatnot. When I do whatever bitwise operations, I want to be able to see things on a bit level - that is, after doing:

Code:
```byte a = (byte) 0b10011001; a = ~a; System.out.print( a );```

Instead of having it display 01100110 like I would want it to, it displays 102. Is there a routine somewhere that displays a decimal number as a binary one?

EDIT: Welp, this is embarrassing. At the time of writing I had written a routine that does what I want, but it had some errors; so I posted about it, and fixed the errors less than five seconds after making the post. Anyways, here's the routine.

Code:
```public static String Binarify( byte ByteToCheck ) {         String binaryCode = "";         byte[] reference = new byte[]{ (byte) 0x80, 0x40, 0x20, 0x10, 0x08, 0x04, 0x02, 0x01 };                 for ( byte z = 0; z < 8; z++ ) {             //if bit z of byte a is set, append a 1 to binaryCode. Otherwise, append a 0 to binaryCode             if ( ( reference[z] & ByteToCheck ) != 0 ) {                 binaryCode += "1";             }             else {                 binaryCode += "0";             }         }                 return binaryCode;     }```

The routine accepts a byte as a parameter, and returns a string version of the binary digits. If someone has a much better/more optimized way of doing this, please let me know!
Too bad you're not using Python to turn integer x into a string:

Code:
`binstring = "".join(map(lambda n: "1" if (x&(1<<n)) else "0",range(ceil(log(x,2)))))`

The optimal Java form would be pretty close, I think. You should make the mask be your iteration variable, in my opinion, rather than the bit position.
KermMartian wrote:

Code:
`binstring = "".join(map(lambda n: "1" if (x&(1<<n)) else "0",range(ceil(log(x,2)))))`
Ugh.
Code:
```binstring = bin(0x2f) # '0b101111' binstring = bin(0x2f)[2:].rjust(8,'0') # '00101111'```

Would be better to have a single variable as your mask, initialize to 0x80 and simply shift right (watch for sign extension). Also somewhat faster to use a StringBuilder preallocated to the right size rather than storing a raw String (which involves creating a new object every time you append to it).
Yeah, benryves will tell you and I'll be the first to confirm that I love re-inventing the wheel. I was pretty confident that I was solving a solved problem as far as Python, but I charged ahead anyway. Also, Lincoln, are you sure you're always dealing precisely with 8-bit numbers?
Kerm, in Java, isn't a byte guaranteed to be 8 bits? And, for the original question, a quick google search found http://stackoverflow.com/questions/5263187/how-can-i-print-a-integer-in-binary-format-in-java

So, basically:

Code:
```public static String binarify(byte b) {   return Integer.toBinaryString((int)b); }```
Or, just use ints and Integer.toBinaryString()
This doesn't seem to be working for me. This is exactly how I thought I should do it, but it is still printing in base 10. Here is the code. Any help would be appreciated. It's not a big deal because it's just for debugging, but it's really annoying me.

byte[] sys_mem = m_sysinfo.getSystemMemory();
int index = frame_num * bytes_per_page;
byte b1 = sys_mem[index];
byte b2 = sys_mem[index + 1];
Debug.user(" 0:" + Integer.toBinaryString((int)b1));
Debug.user(" 1:" + Integer.toBinaryString((int)b2));

This prints 0:0 and 1:0

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