Reputation: 462
I have 2 4-bit numbers (X0X1X2X3 and Y0Y1Y2Y3) and I want to combine them so that I would create an 8-bit number like that:
X0X1X2X3
Y0Y1Y2Y3 => X0Y0X1Y1X2Y2X3Y3
I know how to concatenate them in order to create the X0X1X1X3Y0Y1Y2Y3
but I am stuck on how to compose them in the pattern explained above.
Any hints?
Upvotes: 2
Views: 2718
Reputation: 1
You can also do this in one line.
uint8_t result = (varX << 4) | (varY);
Upvotes: 0
Reputation: 223787
Here's a fairly direct way of performing this transformation:
uint8_t result;
result |= (x & 8) << 4;
result |= (y & 8) << 3;
result |= (x & 4) << 3;
result |= (y & 4) << 2;
result |= (x & 2) << 2;
result |= (y & 2) << 1;
result |= (x & 1) << 1;
result |= (y & 1) << 0;
Upvotes: 3
Reputation: 726559
A very quick way of interleaving four-bit numbers is with a look-up table:
unsigned int spread[] = {
0x00, 0x01, 0x04, 0x05, 0x10, 0x11, 0x14, 0x15,
0x40, 0x41, 0x44, 0x45, 0x50, 0x51, 0x54, 0x55
};
You can use spread[]
array, which "spreads" bits of the original, to construct your output as follows:
unsigned int res = (spread[x] << 1) | spread[y];
The trick is in the construction of the look-up table. Its values are selected in such a way that the bits of the index are interleaved with zeros. For example, 0x07
, or 01112, becomes 0x15
, or 000101012.
Upvotes: 3