如何在C中将2个4位无符号数组合成1个8位数字

Question

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?

Answer 1

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 0111 ₂ , becomes 0x15 , or 00010101 ₂ .

Answer 2

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;

Answer 3

您也可以在一行中执行此操作。

 uint8_t result =  (varX << 4) | (varY);