Move changes inline

Author: Blargian

README.md

@@ -257,7 +257,11 @@ GPIOA is at address 0x40020000, GPIOB is at 0x40020400, and so on:
 
 We can create pin numbering that includes the bank and the pin number.
 To do that, we use 2-byte `uint16_t` value, where upper byte indicates
+<<<<<<< HEAD
 GPIO bank, and lower byte indicates pin number (see the [appendix](#pin-function) for further explanation of the functions below):
+=======
+GPIO bank, and lower byte indicates pin number (see the [appendix](#Appendix) for further explanation ):
+>>>>>>> 43d1ea3 (Move changes inline)
 
 ```c
 #define PIN(bank, num) ((((bank) - 'A') << 8) | (num))
@@ -272,6 +276,18 @@ This way, we can specify pins for any GPIO bank:
   uint16_t pin2 = PIN('G', 11);   // G11  - GPIOG pin 11
 ``` 
 
+Let's look first at what happens for `PIN('A', 3)`:
+
+- `(bank) - 'A'` results in `'A' - 'A'` which will evaluate to `0`. As a 16 bit binary value this would be `0b00000000,00000000`.
+- Next we bit shift this value left by 8 bits because we want to store `bank` in the upper byte of this 16 bit, or 2 byte value. In this case the result remains the same: `0b00000000,00000000`.
+- Finally we bitwise OR the value above with `num`, in our case `3` which has a 16 bit binary representation of `0b00000000,00000011`. The result in binary is `0b00000000,00000011`.
+
+Let's take a look at what happens for `PIN('G',11)`:
+
+- `(bank) - 'G'` results in `'G' - 'A'` which will evaluate to `6`. As a 16 bit binary value this would be `0b00000000,00000110`.
+- Next we bit shift this value left by 8 bits because we want to store `bank` in the upper byte of this 16 bit, or 2 byte value. This results in a binary value of: `0b00000110,00000000`.
+- Finally we bitwise OR the value above with `num`, in our case `11` which has a 16 bit binary representation of `0b00000000,00001011`. The result of the bitwise OR in binary is `0b00000110,00001011` which is a combination of `bank` in the upper byte and `pin` in the lower byte.
+
 Let's rewrite the `gpio_set_mode()` function to take our pin specification:
 
 ```c
@@ -1755,37 +1771,6 @@ print the result to the UART, and check for the expected output in the test.
 
 Happy testing!
 
-## Appendix
-
-In this section you will find some further explanations for select points in this guide. 
-
-### PIN function
-
-We defined `PIN` as below:
-
-```c
-#define PIN(bank, num) ((((bank) - 'A') << 8) | (num))
-```
-
-This function is perhaps most easily understood through a worked example. Let's take pins `A3` and `G11` and see what is happening in the function above, step by step.
-
-```c
-uint16_t pin1 = PIN('A', 3);    // A3   - GPIOA pin 3
-uint16_t pin2 = PIN('G', 11);   // G11  - GPIOG pin 11
-```
-
-Let's look first at what happens for `PIN('A', 3)`:
-
-- `(bank) - 'A'` results in `'A' - 'A'` which will evaluate to `0`. As a 16 bit binary value this would be `0b00000000,00000000`.
-- Next we bit shift this value left by 8 bits because we want to store `bank` in the upper byte of this 16 bit, or 2 byte value. In this case the result remains the same: `0b00000000,00000000`.
-- Finally we bitwise OR the value above with `num`, in our case `3` which has a 16 bit binary representation of `0b00000000,00000011`. The result in binary is `0b00000000,00000011`  
-
-Let's take a look at what happens for `PIN('G',11)`
-
-- `(bank) - 'G'` results in `'G' - 'A'` which will evaluate to `6`. As a 16 bit binary value this would be `0b00000000,00000110`.
-- Next we bit shift this value left by 8 bits because we want to store `bank` in the upper byte of this 16 bit, or 2 byte value. This results in a binary value of: `0b00000110,00000000`.
-- Finally we bitwise OR the value above with `num`, in our case `11` which has a 16 bit binary representation of `0b00000000,00001011`. The result of the bitwise OR in binary is `0b00000110,00001011` which is a combination of `bank` in the upper byte and `pin` in the lower byte.  
-
 ## About me
 
 I am Sergey Lyubka, an engineer and entrepreneur. I hold a MSc in Physics from