MPU region configuration in ARM Architecture - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to set the base address of an MPU region.

ARM Architecture

MPU->RNR = [1];  // Select region number

Drag options to blanks, or click blank then click option'

A5

B3

C8

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using an invalid region number outside the supported range.

The MPU Region Number Register (RNR) is set to select which region to configure. Here, setting it to 3 selects region 3.

2fill in blank

medium

Complete the code to enable an MPU region.

ARM Architecture

MPU->RASR = (1 << [1]);  // Enable region

Drag options to blanks, or click blank then click option'

A0

B1

C3

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Setting the wrong bit to enable the region.

The enable bit for the MPU region is bit 0 in the Region Attribute and Size Register (RASR). Setting bit 0 to 1 enables the region.

3fill in blank

hard

Fix the error in setting the region size to 256 bytes.

ARM Architecture

MPU->RASR = ([1] << 1) | (1 << 0);  // Set size and enable

Drag options to blanks, or click blank then click option'

A7

B5

C8

D6

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong size value causing incorrect region size.

The size field in RASR uses the formula: SIZE = log2(region size) - 1. For 256 bytes, log2(256) = 8, so SIZE = 7. Shifting 7 left by 1 positions it correctly in bits [5:1] of RASR, and OR with (1 << 0) enables the region.

4fill in blank

hard

Fill both blanks to configure an MPU region with size 512 bytes and full access permission.

ARM Architecture

MPU->RASR = ([1] << 1) | ([2] << 24) | (1 << 0);

Drag options to blanks, or click blank then click option'

A8

B3

C0

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing size value or access permission bits.

The size field is set by (log2(size) - 1). For 512 bytes, log2(512) = 9, so size = 8. The access permission field (AP) is bits 24-26; full access is represented by 3.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps region numbers to their sizes in bytes, filtering only regions larger than 256 bytes.

ARM Architecture

region_sizes = { [1]: 1 << ([2] + 1) for [3], [2] in regions.items() if (1 << ([2] + 1)) > 256 }

Drag options to blanks, or click blank then click option'

Aregion

Bsize_code

Cregion_sizes

Dregion_num

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing variable names or misunderstanding the size calculation.

This comprehension creates a dictionary where keys are region numbers and values are sizes calculated as 2^(size_code + 1). It filters regions with sizes greater than 256 bytes.

Practice

1. What is the primary purpose of configuring MPU regions in an ARM system?

easy

A. To increase CPU clock speed

B. To handle input/output device communication

C. To manage power consumption

D. To control memory access rights and protect memory

Solution

1. Step 1: Understand MPU function

   The Memory Protection Unit (MPU) is designed to control access to different memory areas.

2. Step 2: Identify the main goal of MPU regions

   MPU regions define access permissions to protect memory from unauthorized access.

3. Final Answer:

   To control memory access rights and protect memory -> Option D

4. Quick Check:

   MPU protects memory = B [OK]

Hint: MPU is about memory protection, not speed or power [OK]

Common Mistakes:

• Confusing MPU with CPU speed control
• Thinking MPU manages power or I/O devices
• Assuming MPU handles software tasks

2. Which of the following is the correct way to specify the size of an MPU region?

easy

A. Using a power-of-two size like 256, 512, 1024 bytes

B. Using the exact byte size like 1024 bytes

C. Using any arbitrary size value

D. Using the number of memory pages

Solution

1. Step 1: Recall MPU size rules

   MPU regions must have sizes that are powers of two (e.g., 256, 512, 1024 bytes).

2. Step 2: Eliminate incorrect options

   Exact byte sizes not power of two or arbitrary sizes are invalid; number of pages is unrelated.

3. Final Answer:

   Using a power-of-two size like 256, 512, 1024 bytes -> Option A

4. Quick Check:

   MPU size = power of two [OK]

Hint: MPU sizes must be powers of two, not arbitrary [OK]

Common Mistakes:

• Using sizes not in powers of two
• Confusing size with memory pages
• Specifying exact byte sizes without power-of-two check

3. Given an MPU region configured with base address 0x20000000 and size 512 bytes, which of the following addresses is NOT inside this region?

medium

A. 0x20000200

B. 0x200001FF

C. 0x20000000

D. 0x20000100

Solution

1. Step 1: Calculate region address range

   Base address is 0x20000000, size is 512 bytes, so region covers 0x20000000 to 0x200001FF (512 bytes).

2. Step 2: Check each address

   Addresses 0x20000100, 0x200001FF, and 0x20000000 fall inside the range; 0x20000200 is outside.

3. Final Answer:

   0x20000200 -> Option A

4. Quick Check:

   Address outside 0x20000000-0x200001FF = D [OK]

Hint: Region ends at base + size - 1 address [OK]

Common Mistakes:

• Off-by-one errors in address range
• Misunderstanding base address as inclusive/exclusive
• Confusing hex address increments

4. A developer configures an MPU region with base address 0x10000000 and size 300 bytes. What is the problem with this configuration?

medium

A. Base address must be zero

B. Size must be a power of two, 300 is invalid

C. Base address must be aligned to size, 0x10000000 is not aligned

D. No problem, configuration is valid

Solution

1. Step 1: Check size validity

   MPU region sizes must be powers of two; 300 is not a power of two.

2. Step 2: Verify base address alignment

   Base address must be aligned to the size, but since size is invalid, alignment check is secondary.

3. Final Answer:

   Size must be a power of two, 300 is invalid -> Option B

4. Quick Check:

   MPU size power-of-two rule violated = A [OK]

Hint: Always check size is power of two before alignment [OK]

Common Mistakes:

• Ignoring size power-of-two requirement
• Assuming any base address is valid
• Confusing alignment rules with size validity

5. You want to configure two MPU regions: one for a 1 KB read-only memory area starting at 0x08000000, and another for a 2 KB read-write memory area starting at 0x08000400. Which of the following configurations correctly sets the base addresses and sizes without overlap?

hard

A. Region1: base=0x08000000, size=1024 bytes; Region2: base=0x08000400, size=2048 bytes

B. Region1: base=0x08000000, size=2048 bytes; Region2: base=0x08000400, size=1024 bytes

C. Region1: base=0x08000000, size=1024 bytes; Region2: base=0x08000800, size=2048 bytes

D. Region1: base=0x08000000, size=512 bytes; Region2: base=0x08000200, size=2048 bytes

Solution

1. Step 1: Calculate address ranges for each region

   Region1: base 0x08000000, size 1024 bytes covers 0x08000000 to 0x080003FF.
   Region2: base 0x08000400, size 2048 bytes covers 0x08000400 to 0x08000BFF.

2. Step 2: Check for overlap and alignment

   Region1: base=0x08000000, size=1024 bytes; Region2: base=0x08000400, size=2048 bytes -- invalid MPU configuration because Region2 base address 0x08000400 is not aligned to its size 2048 bytes (0x08000400 mod 2048 = 1024 ≠ 0).
   Region1: base=0x08000000, size=1024 bytes; Region2: base=0x08000800, size=2048 bytes sets Region2 base at 0x08000800, which is after Region1 ends, no overlap.
   Region1: base=0x08000000, size=2048 bytes; Region2: base=0x08000400, size=1024 bytes swaps sizes causing overlap.
   Region1: base=0x08000000, size=512 bytes; Region2: base=0x08000200, size=2048 bytes sizes and bases cause overlap and incorrect sizes.

3. Final Answer:

   Region1: base=0x08000000, size=1024 bytes; Region2: base=0x08000800, size=2048 bytes -> Option C

4. Quick Check:

   No overlap and correct alignment = C [OK]

Hint: Ensure regions do not overlap by checking address ranges [OK]

Common Mistakes:

• Overlapping regions due to incorrect base or size
• Ignoring alignment to power-of-two sizes
• Swapping sizes and bases incorrectly