Arduino programming

Greetings, I would appreciate help because I do not understand it means

** Wire.write(sintab2[lookup] >> 4); **
** Wire.write((sintab2[lookup] & 15) << 4);**

Attached the complete code

int lookup = 0;//varaible for navigating through the tables

int sintab2[512] =
{
2048, 2073, 2098, 2123, 2148, 2174, 2199, 2224,
2249, 2274, 2299, 2324, 2349, 2373, 2398, 2423,
2448, 2472, 2497, 2521, 2546, 2570, 2594, 2618,
2643, 2667, 2690, 2714, 2738, 2762, 2785, 2808,
2832, 2855, 2878, 2901, 2924, 2946, 2969, 2991,
3013, 3036, 3057, 3079, 3101, 3122, 3144, 3165,
3186, 3207, 3227, 3248, 3268, 3288, 3308, 3328,
3347, 3367, 3386, 3405, 3423, 3442, 3460, 3478,
3496, 3514, 3531, 3548, 3565, 3582, 3599, 3615,
3631, 3647, 3663, 3678, 3693, 3708, 3722, 3737,
3751, 3765, 3778, 3792, 3805, 3817, 3830, 3842,
3854, 3866, 3877, 3888, 3899, 3910, 3920, 3930,
3940, 3950, 3959, 3968, 3976, 3985, 3993, 4000,
4008, 4015, 4022, 4028, 4035, 4041, 4046, 4052,
4057, 4061, 4066, 4070, 4074, 4077, 4081, 4084,
4086, 4088, 4090, 4092, 4094, 4095, 4095, 4095,
4095, 4095, 4095, 4095, 4094, 4092, 4090, 4088,
4086, 4084, 4081, 4077, 4074, 4070, 4066, 4061,
4057, 4052, 4046, 4041, 4035, 4028, 4022, 4015,
4008, 4000, 3993, 3985, 3976, 3968, 3959, 3950,
3940, 3930, 3920, 3910, 3899, 3888, 3877, 3866,
3854, 3842, 3830, 3817, 3805, 3792, 3778, 3765,
3751, 3737, 3722, 3708, 3693, 3678, 3663, 3647,
3631, 3615, 3599, 3582, 3565, 3548, 3531, 3514,
3496, 3478, 3460, 3442, 3423, 3405, 3386, 3367,
3347, 3328, 3308, 3288, 3268, 3248, 3227, 3207,
3186, 3165, 3144, 3122, 3101, 3079, 3057, 3036,
3013, 2991, 2969, 2946, 2924, 2901, 2878, 2855,
2832, 2808, 2785, 2762, 2738, 2714, 2690, 2667,
2643, 2618, 2594, 2570, 2546, 2521, 2497, 2472,
2448, 2423, 2398, 2373, 2349, 2324, 2299, 2274,
2249, 2224, 2199, 2174, 2148, 2123, 2098, 2073,
2048, 2023, 1998, 1973, 1948, 1922, 1897, 1872,
1847, 1822, 1797, 1772, 1747, 1723, 1698, 1673,
1648, 1624, 1599, 1575, 1550, 1526, 1502, 1478,
1453, 1429, 1406, 1382, 1358, 1334, 1311, 1288,
1264, 1241, 1218, 1195, 1172, 1150, 1127, 1105,
1083, 1060, 1039, 1017, 995, 974, 952, 931,
910, 889, 869, 848, 828, 808, 788, 768,
749, 729, 710, 691, 673, 654, 636, 618,
600, 582, 565, 548, 531, 514, 497, 481,
465, 449, 433, 418, 403, 388, 374, 359,
345, 331, 318, 304, 291, 279, 266, 254,
242, 230, 219, 208, 197, 186, 176, 166,
156, 146, 137, 128, 120, 111, 103, 96,
88, 81, 74, 68, 61, 55, 50, 44,
39, 35, 30, 26, 22, 19, 15, 12,
10, 8, 6, 4, 2, 1, 1, 0,
0, 0, 1, 1, 2, 4, 6, 8,
10, 12, 15, 19, 22, 26, 30, 35,
39, 44, 50, 55, 61, 68, 74, 81,
88, 96, 103, 111, 120, 128, 137, 146,
156, 166, 176, 186, 197, 208, 219, 230,
242, 254, 266, 279, 291, 304, 318, 331,
345, 359, 374, 388, 403, 418, 433, 449,
465, 481, 497, 514, 531, 548, 565, 582,
600, 618, 636, 654, 673, 691, 710, 729,
749, 768, 788, 808, 828, 848, 869, 889,
910, 931, 952, 974, 995, 1017, 1039, 1060,
1083, 1105, 1127, 1150, 1172, 1195, 1218, 1241,
1264, 1288, 1311, 1334, 1358, 1382, 1406, 1429,
1453, 1478, 1502, 1526, 1550, 1575, 1599, 1624,
1648, 1673, 1698, 1723, 1747, 1772, 1797, 1822,
1847, 1872, 1897, 1922, 1948, 1973, 1998, 2023
};

void setup()
{
Wire.begin();

// Set A2 and A3 as Outputs to make them our GND and Vcc,
//which will power the MCP4725
pinMode(A2, OUTPUT);
pinMode(A3, OUTPUT);

digitalWrite(A2, LOW);//Set A2 as GND
digitalWrite(A3, HIGH);//Set A3 as Vcc
}
//---------------------------------------------------
void loop()
{
Wire.beginTransmission(MCP4725_ADDR);
Wire.write(64);
Wire.write(sintab2[lookup] >> 4);
Wire.write((sintab2[lookup] & 15) << 4);
Wire.endTransmission();
lookup = (lookup + 1) & 511;
}

sintab2 is an array.
lookup is an index. It selects one element of the array.

When lookup is 0, sintab2[llookup] is sintab2[0], which is 2048.

So, the statements in question boil down to:

  Wire.write(2048 >> 4);     
  Wire.write((2048 & 15) << 4);

is the right shift operator, so 2048 is shifted 4 places to the right. Since 2048 is 0x0800, shifting that 4 places to the looses the last 2 hex digits, result in the first write() method dealing with 0x08.

The second statement masks the low order bytes (which are 0), and shifts the remaining bits (still 0x0800) 4 places to the left, resulting in loosing the high order byte, leaving write() to deal with the 0x00 portion of the initial value.

Thank you so much for the clarification PaulS

Shifting 4 places to the right will loose the last hex digit, not the last two :wink:

The result is the same as divide by 16, by the way.

sterretje:
Shifting 4 places to the right will loose the last hex digit, not the last two :wink:

The result is the same as divide by 16, by the way.

I don't think that the code does what it is supposed to do. The code is breaking a byte into two nibbles, but it is operating on an int, so it is only dealing with 2 of the 4 nibbles, and throwing the other 2 away.