Watching register values in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of register value after bitwise operation

What is the output of the following code that manipulates a hardware register value?

Embedded C

volatile unsigned char REG = 0x3C;  
REG |= 0x05;  
printf("0x%X", REG);

A 0x39

B 0x05

C 0x38

D 0x3D

💡 Hint

Remember that |= sets bits that are 1 in the mask without clearing others.

The original value 0x3C (00111100) OR 0x05 (00000101) results in 00111101 which is 0x3D.

❓ Predict Output

intermediate

Value of register after clearing bits

What will be the value of REG after clearing bits 0 and 1 using bitwise AND?

Embedded C

volatile unsigned char REG = 0xFF;  
REG &= ~0x03;  
printf("0x%X", REG);

A 0xFF

B 0x03

C 0xFC

D 0x00

💡 Hint

Use bitwise NOT on 0x03 to clear bits 0 and 1.

0x03 is 00000011, ~0x03 is 11111100. AND with 0xFF clears bits 0 and 1, resulting in 0xFC.

🔧 Debug

advanced

Identify the error in register assignment

What error will this code cause when trying to assign a value to a hardware register?

Embedded C

volatile unsigned char *REG = 0x40021000;  
*REG = 0xFF;

A Runtime error: segmentation fault

B Compilation error: incompatible types in assignment

C No error, value 0xFF assigned correctly

D Compilation error: missing semicolon

💡 Hint

Check the pointer assignment syntax for hardware registers.

Assigning an integer directly to a pointer variable without casting causes a compilation error.

❓ Predict Output

advanced

Result of shifting register bits

What is the output of this code that shifts bits in a register?

Embedded C

volatile unsigned char REG = 0x0F;  
REG = REG << 2;  
printf("0x%X", REG);

A 0x3C

B 0x3F

C 0xF0

D 0x0F

💡 Hint

Left shift moves bits to higher positions, filling with zeros on the right.

0x0F is 00001111. Shifting left by 2 bits gives 00111100 which is 0x3C.

🧠 Conceptual

expert

Why use volatile keyword with hardware registers?

Why is the volatile keyword important when declaring variables that map to hardware registers?

A It prevents the compiler from optimizing out reads and writes to the register

B It makes the variable read-only

C It initializes the register to zero automatically

D It allows the variable to be used in interrupt service routines only

💡 Hint

Think about how the compiler treats variables that can change outside program control.

Volatile tells the compiler the value can change anytime, so it must not optimize away accesses.