Transmitting a byte over UART in Embedded C - Step-by-Step Execution

Concept Flow - Transmitting a byte over UART

Start Transmission

↓

Check UART TX Ready?

No→Wait

Yes↓

Load Byte into TX Register

↓

UART Sends Byte

↓

Transmission Complete

↓

End

The program waits until UART is ready, loads the byte to send, then UART transmits it and finishes.

Execution Sample

Embedded C

while(!UART_TX_READY());
UART_TX_REG = data_byte;
// UART hardware sends the byte automatically

Waits for UART ready, then writes the byte to the transmit register to send it.

Execution Table

Step	UART_TX_READY()	Condition	Action	UART_TX_REG	Output
1	false	No	Wait	unchanged	none
2	false	No	Wait	unchanged	none
3	true	Yes	Load byte	0x4A	start sending 0x4A
4	N/A	N/A	Byte sent by UART hardware	0x4A	byte transmitted
5	N/A	N/A	Transmission complete	0x4A	done

💡 UART_TX_READY() becomes true, byte loaded and sent, transmission completes

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
UART_TX_READY()	false	false	false	true	N/A	N/A
UART_TX_REG	0x00	0x00	0x00	0x4A	0x4A	0x4A
Output	none	none	none	start sending 0x4A	byte transmitted	done

Key Moments - 3 Insights

Why do we wait for UART_TX_READY() before loading the byte?

Because UART_TX_READY() tells us if the UART hardware is ready to accept a new byte. Loading before ready would overwrite data. See steps 1-3 in execution_table.

Does writing to UART_TX_REG immediately send the byte?

No, writing loads the byte into the UART hardware buffer. The hardware then sends it automatically, as shown in steps 3-4.

What happens if we skip the wait loop?

If we skip waiting, we might overwrite UART_TX_REG before the previous byte is sent, causing data loss. The wait ensures safe transmission (see steps 1-3).

Visual Quiz - 3 Questions

Look at the execution_table, what is UART_TX_READY() at step 3?

A N/A

B false

C true

D unknown

💡 Hint

Check the 'UART_TX_READY()' column at step 3 in execution_table.

At which step does the byte get loaded into UART_TX_REG?

A Step 1

B Step 3

C Step 4

D Step 5

💡 Hint

Look at the 'Action' and 'UART_TX_REG' columns in execution_table.

If UART_TX_READY() never becomes true, what happens?

A The program waits forever in the loop

B The byte is sent immediately

C UART_TX_REG is overwritten repeatedly

D Transmission completes without sending

💡 Hint

Refer to the wait loop in the code and steps 1-2 in execution_table.

Concept Snapshot

Transmitting a byte over UART:
1. Wait until UART_TX_READY() returns true.
2. Write the byte to UART_TX_REG.
3. UART hardware sends the byte automatically.
4. Do not write before ready to avoid data loss.
5. This ensures safe, sequential byte transmission.

Full Transcript

This visual execution shows how to send a byte over UART in embedded C. First, the program waits until the UART hardware signals it is ready to send a new byte by checking UART_TX_READY(). This prevents overwriting data. Once ready, the byte is loaded into the UART transmit register UART_TX_REG. The UART hardware then sends the byte automatically. The execution table traces each step, showing the wait, the loading of the byte, and the transmission completion. Key moments clarify why waiting is necessary and how the hardware handles sending. The quiz tests understanding of the readiness check, loading step, and consequences of skipping the wait. The snapshot summarizes the safe sequence to transmit a byte over UART.