Writing and Reading data from RTC module EEPROM_Arduino

Question

i have took below code from https://lastminuteengineers.com/ds1307-rtc-arduino-tutorial/. i am not understand what these codes are doing. please help me to understand.

Wire.write((int)(eeaddress >> 8)); // MSB
Wire.write((int)(eeaddress & 0xFF)); // LSB

Thanks

• jsotola: how many bits long is eeaddress?

Answer 1

These two lines write a 16 bit (2 8-bit) value, which is signed -32768..32767 or unsigned 0..65535.

Assume a value in bits: 10101010 11001100

Statement 1:

Wire.write((int)(eeaddress >> 8)); // MSB

The first statement shifts the value 8 bits (positions) to the right; this is called the MSB (Most Significant Byte, the left 8 bits). Bits on the right are clipped / fall off. Thus

10101010 11001100

will become

00000000 10101010

When casted to an 8 bit value you get:

10101010

Statement 2:

Wire.write((int)(eeaddress & 0xFF)); // LSB

The second statement will get the LSB (Least Significant Byte). This used the AND operator, meaning a bit is only 1 if both values are 1. 0xFF is the 'mask'

Original value: 10101010 11001100
Mask: 00000000 11111111
 ----------------- AND (&) operator
Result: 00000000 11001100

As you can see only the right 8 bits are kept. When casting this to an 8 bit value:

11001100

• Gerben: Do you also know why there is an int cast ((int)...)?
• Michel Keijzers: @Gerben; actually I would either expect a 8 bit type cast (uint8_t or int8_t cast as the result is always 8 bits, or no cast at all, as the caller probably expects an 8 bit variable. Even if the caller expects 16 bit (which would be unnecessary much), the 8 bit argument will automatically be casted to 16 bits, without a need for a cast in the two lines provided in this code.
• Gerben: Thank you. Seems like the Wire library will cast it (back) into an uint8_t anyways.
• Michel Keijzers: @Gerben, true, thus casting it within the calling write method is not needed.
• Suman Ponmathan: Thanks @Michel Keijzers