I mentioned that the microcontroller has several pins. For convenience, these pins are grouped in ports of 16 pins. Each port is named with a letter: Port A, Port B, etc. and the pins within each port are named with numbers from 0 to 15.

The first thing we have to find out is which pin is connected to which LED. This information is in the STM32F3DISCOVERY User Manual (You downloaded a copy, right?). In this particular section:

Section 6.4 LEDs - Page 18

The manual says:

• LD3, the North LED, is connected to the pin PE9. PE9 is the short form of: Pin 9 on Port E.
• LD7, the East LED, is connected to the pin PE11.

Up to this point, we know that we want to change the state of the pins PE9 and PE11 to turn the North/East LEDs on/off. These pins are part of Port E so we'll have to deal with the GPIOE peripheral.

Each peripheral has a register block associated to it. A register block is a collection of registers allocated in contiguous memory. The address at which the register block starts is known as its base address. We need to figure out what's the base address of the GPIOE peripheral. That information is in the following section of the microcontroller Reference Manual:

Section 3.2.2 Memory map and register boundary addresses - Page 51

The table says that base address of the GPIOE register block is 0x4800_1000.

Each peripheral also has its own section in the documentation. Each of these sections ends with a table of the registers that the peripheral's register block contains. For the GPIO family of peripheral, that table is in:

Section 11.4.12 GPIO register map - Page 243

We are interested in the register that's at an offset of 0x18 from the base address of the GPIOE peripheral. According to the table, that would be the register BSRR.

Now we need to jump to the documentation of that particular register. It's a few pages above in:

Section 11.4.7 GPIO port bit set/reset register (GPIOx_BSRR) - Page 240

Finally!

This is the register we were writing to. The documentation says some interesting things. First, this register is write only ... so let's try reading its value :-).

We'll use GDB's examine command: x.

(gdb) next
16              *(GPIOE_BSRR as *mut u32) = 1 << 9;

(gdb) x 0x48001018
0x48001018:     0x00000000

(gdb) # the next command will turn the North LED on
(gdb) next
19              *(GPIOE_BSRR as *mut u32) = 1 << 11;

(gdb) x 0x48001018
0x48001018:     0x00000000

Reading the register returns 0. That matches what the documentation says.

The other thing that the documentation says is that the bits 0 to 15 can be used to set the corresponding pin. That is bit 0 sets the pin 0. Here, set means outputting a high value on the pin.

The documentation also says that bits 16 to 31 can be used to reset the corresponding pin. In this case, the bit 16 resets the pin number 0. As you may guess, reset means outputting a low value on the pin.

Correlating that information with our program, all seems to be in agreement:

• Writing 1 << 9 (BS9 = 1) to BSRR sets PE9 high. That turns the North LED on.

• Writing 1 << 11 (BS11 = 1) to BSRR sets PE11 high. That turns the East LED on.

• Writing 1 << 25 (BR9 = 1) to BSRR sets PE9 low. That turns the North LED off.

• Finally, writing 1 << 27 (BR11 = 1) to BSRR sets PE11 low. That turns the East LED off.