将数字分为两部分

Question

I'm trying to split number (8bit) (0..255) into two numbers like this: 100 (binary 01100100) into 6(0110) and 4(0100).

In CI do it in such way:

short row = value >> 4; //value - uint8_t
short col = (row << 4) ^ value;

In Java I do it the other way (because C like code don't work properly):

short v1 = (byte)(value >> 4);
short v2 = (byte)((byte)(value << 4) >>> 4);

It works for numbers from 0 to 103. But for numbers greater than 103 it doesn't work.

ps Actually problem is here:

short v2 = (byte)((byte)(value) >>> 4);
=>104 (1101000). I expected to see something like this: 00001000, but result is 11111000 
=>103 (1100111). Result is 111;

Fix.

I fixed it in such way:

short value = (byte)255;
short v1 = 0;
short v2 = 0;

for(int i = 3; i >= 0; i--) {
    v2 = (short) (v2 << 1 | ((value >> i) & 1));
    v1 = (short) (v1 << 1 | (value >> (i + 4) & 1));
}
System.out.println(v1 + " " + v2);

But I prefer the C way, it's more simple.

Answer 1

short row = (value >> 4) & 0x0F;
short col = value & 0x0F;

This should also work in C.

& keeps only the bits which are 1 at the same position:

11111000   // 104
00001111   // 0x0F == 15
--------
00001000   // 104 & 0x0F == 8
           // because only the fourth bit from right is 1 in both numbers!

Answer 2

unsigned char f = 100;
unsigned char first = f >> 4;
unsigned char second = 0xF & f;
printf("%d %d", first, second);

00001111 if you bitwise and with number you will get last 4 bit

Answer 3

Try this.

short v1 = (byte)(value >> 4);
short v2 = (byte)((byte)(value << 4) >>> 4);

if(v2<0){
v2 = (short) -v2;
}