Quick Java Question

[clamum]

I have a question that I think isn't very difficult to do but I'm not sure how to do it: I want to take an int (it will be anywhere from 100 to 3000) and convert it to an array of bytes. After that, I would like to get the int back out of the byte array.

Thanks for any help

 

[Crusty]

bytearray[0] = (myInt & 0xff000000) >> 24;
bytearray[1] = (myInt & 0x00ff0000) >> 16;
bytearray[2] = (myInt & 0x0000ff00) >> 8;
bytearray[3] = (myInt & 0x000000ff);

^^ Didn't test it.... but should give you general idea.

& is the bit-wise AND operator. so it takes each bit of the operands and logically ANDs them together. >> is the right bit-shift operator... so after you & the integer with teh correct mask, you have to shift those bits over into the last 8 bits.

This would only work for a 32-bit integer.

 

[kamper]

Good call Crusty. As a side note, java guarantees its data type sizes so the above code is technically safe across all platforms. Kinda worries me, in case they ever decide to up it to 64, but it would have to be officially announced and everyone would know about it.

 

[Crusty]

Is it only Sun that guaruntees the data type sizes? I would think that whoever writes the JRE determines their data type sizes.

int mast = 0xff000000;

for (int i = 0; i <= 32 ; i+=8) {
bytearray[i/8] = (myInt & mask) >> (24 - i);
mask >> 8;
}

That loop should work too, just substitute your integer size for 32, byte size for 8, and (int size - byte size) for 24.... i think

 

[kamper]

No, it's part of the spec and if you write a jvm that doesn't conform, it's not Java.

 

[Crusty]

Originally posted by: kamper
No, it's part of the spec and if you write a jvm that doesn't conform, it's not Java.

Ah, that's nice to know 🙂.

 

[notfred]

or you can just do:

byte[0] = (byte)int;
byte[1] = (byte)(int >> 8);
byte[2] = (byte)(int >> 16);
byte[3] = (byte)(int >> 24);