Bit Field Packing

To pack a 3 bit number and a 5 bit number together into an 8 bit byte

X contains a 3 bit number and Y contains a 5 bit number

combine
	X =		X7	X6	X5	X4	X3	X2	X1	X0
	Y =		Y7	Y6	Y5	Y4	Y3	Y2	Y1	Y0
to produce
	R =		R7	R6	R5	R4	R3	R2	R1	R0
			X2	X1	X0	Y4	Y3	Y2	Y1	Y0

In XCSB this would be written as:

e.g.
	X = 0x05
	Y = 0x16
	R = (X << 5) | Y

Looking step by step at the computation we see

		X7	X6	X5	X4	X3	X2	X1	X0		Y7	Y6	Y5	Y4	Y3	Y2	Y1	Y0		R7	R6	R5	R4	R3	R2	R1	R0
X = 0x05		0	0	0	0	0	1	0	1
Y = 0x16											0	0	0	1	0	1	1	0
R = (X << 5)							x	x	x											1	0	1	0	0	0	0	0
R = R | Y														x	x	x	x	x		1	0	1	1	0	1	1	0

e.g.
	X = 0x05
	Y = 0x56
	R = (X << 5) | Y

Looking step by step at the computation we see

		X7	X6	X5	X4	X3	X2	X1	X0		Y7	Y6	Y5	Y4	Y3	Y2	Y1	Y0		R7	R6	R5	R4	R3	R2	R1	R0
X = 0x05		0	0	0	0	0	1	0	1
Y = 0x56											0	1	0	1	0	1	1	0
R = (X << 5)							x	x	x											1	0	1	0	0	0	0	0
R = R | Y												x		x	x	x	x	x		1	1	1	1	0	1	1	0

NOTE: here the result 11110110 is incorrect, it should be 10110110 but because Y did not actually fit into 6 bits the result was corrupted. The bit shown as 1 is in error, notice how this came from Y₆

Using a mask to fix the above problem

e.g.
	X = 0x05
	Y = 0x56
	R = (X << 5) | (Y & (0xff >> 3))

Looking step by step at the computation we see

		X7	X6	X5	X4	X3	X2	X1	X0		Y7	Y6	Y5	Y4	Y3	Y2	Y1	Y0		R7	R6	R5	R4	R3	R2	R1	R0		E7	E6	E5	E4	E3	E2	E1	E0
X = 0x05		0	0	0	0	0	1	0	1
Y = 0x56											0	1	0	1	0	1	1	0
R = (X << 5)							x	x	x											1	0	1	0	0	0	0	0
E = 0xff																													1	1	1	1	1	1	1	1
E = (E >> 3)																													0	0	0	1	1	1	1	1
E = Y & E														x	x	x	x	x											0	0	0	1	0	1	1	0
R = R | E																				1	0	1	1	0	1	1	0					x	x	x	x	x

To pack a 3 bit number, a 2 bit number and a 3 bit number together into an 8 bit byte

W contains a 3 bit number, X contains a 2 bit number, Y contains a 3 bit number

combine
	W =		W7	W6	W5	W4	W3	W2	W1	W0
	X =		X7	X6	X5	X4	X3	X2	X1	X0
	Y =		Y7	Y6	Y5	Y4	Y3	Y2	Y1	Y0
to produce
	R =		R7	R6	R5	R4	R3	R2	R1	R0
			W2	W1	W0	X1	X0	Y2	Y1	Y0

In XCSB this would be written as:

e.g.
	W = 0x05
	X = 0x02
	Y = 0x06
	R = (W << 5) | (X << 3) | Y

Looking step by step at the computation we see

		W7	W6	W5	W4	W3	W2	W1	W0		X7	X6	X5	X4	X3	X2	X1	X0		Y7	Y6	Y5	Y4	Y3	Y2	Y1	Y0		R7	R6	R5	R4	R3	R2	R1	R0		E7	E6	E5	E4	E3	E2	E1	E0
W = 0x05		0	0	0	0	0	1	0	1
X = 0x02											0	0	0	0	0	0	1	0
Y = 0x06																				0	0	0	0	0	1	1	0
R = (W << 5)							x	x	x																				1	0	1	0	0	0	0	0
E = (X << 3)																	x	x																				0	0	0	1	0	0	0	0
E = E | Y																									x	x	x											0	0	0	1	0	1	1	0
R = R | E																													1	0	1	1	0	1	1	0					x	x	x	x	x

e.g.
	W = 0x05
	X = 0x0A
	Y = 0x06
	R = (W << 5) | (X << 3) | Y

Looking step by step at the computation we see

		W7	W6	W5	W4	W3	W2	W1	W0		X7	X6	X5	X4	X3	X2	X1	X0		Y7	Y6	Y5	Y4	Y3	Y2	Y1	Y0		R7	R6	R5	R4	R3	R2	R1	R0		E7	E6	E5	E4	E3	E2	E1	E0
W = 0x05		0	0	0	0	0	1	0	1
X = 0x0A											0	0	0	0	1	0	1	0
Y = 0x06																				0	0	0	0	0	1	1	0
R = (W << 5)							x	x	x																				1	0	1	0	0	0	0	0
E = (X << 3)															x		x	x																				0	1	0	1	0	0	0	0
E = E | Y																									x	x	x											0	1	0	1	0	1	1	0
R = R | E																													1	1	1	1	0	1	1	0			x		x	x	x	x	x

NOTE: here the result 11110110 is incorrect, it should be 10110110 but because Y did not actually fit into 6 bits the result was corrupted. The bit shown as 1 is in error, notice how this came from X₃

Using a mask to fix the above problem

e.g.
	W = 0x05
	X = 0x0A
	Y = 0x06
	R = (W << 5) | ((X & (0xff >> 6)) << 3) | (Y & (0xff >> 3))

Looking step by step at the computation we see

		W7	W6	W5	W4	W3	W2	W1	W0		X7	X6	X5	X4	X3	X2	X1	X0		Y7	Y6	Y5	Y4	Y3	Y2	Y1	Y0		R7	R6	R5	R4	R3	R2	R1	R0		E7	E6	E5	E4	E3	E2	E1	E0
W = 0x05		0	0	0	0	0	1	0	1
X = 0x0A											0	0	0	0	1	0	1	0
Y = 0x06																				0	0	0	0	0	1	1	0
R = (W << 5)							x	x	x																				1	0	1	0	0	0	0	0
E = 0xff																																						1	1	1	1	1	1	1	1
E = (E >> 6)																																						0	0	0	0	0	0	1	1
E = E & X																	x	x																				0	0	0	0	0	0	1	0
E = (E << 3)																																						0	0	0	1	0	0	0	0
R = R | E																													1	0	1	1	0	0	0	0					x	x
E = 0xff																																						1	1	1	1	1	1	1	1
E = (E >> 5)																																						0	0	0	0	0	1	1	1
E = E & Y																									x	x	x											0	0	0	0	0	1	1	0
R = R | E																													1	0	1	1	0	1	1	0							x	x	x

NOTE: some of the steps shown here are combined by the XCSB compiler into single step. The real code generated by the compiler is shorter than shown, the full sequence is shown here for completeness