Why registers control hardware in Embedded C - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a hardware register at address 0x4000.

Embedded C

volatile unsigned int* reg = (volatile unsigned int*) [1];

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A0x6000

B0x4000

C0x5000

D0x3000

Attempts:

3 left

💡 Hint

Common Mistakes

Using an incorrect memory address for the register.

The hardware register is located at memory address 0x4000, so we cast that address to a pointer.

2fill in blank

medium

Complete the code to write the value 0xFF to the hardware register.

Embedded C

*reg = [1];

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A0xFF

B0xF0

C0x0F

D0x00

Attempts:

3 left

💡 Hint

Common Mistakes

Writing a wrong value that doesn't set all bits.

Writing 0xFF sets all bits in the register to 1, often enabling all features.

3fill in blank

hard

Fix the error in the code to read the value from the hardware register correctly.

Embedded C

unsigned int value = [1];

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Areg*

Breg

C&reg

D*reg

Attempts:

3 left

💡 Hint

Common Mistakes

Using the pointer variable instead of dereferencing it.

Dereferencing the pointer with *reg reads the value stored at the hardware register address.

4fill in blank

hard

Fill in the blank to set bit 3 of the register without changing other bits.

Embedded C

*reg = *reg [1] (1 << 3);

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A|

B&

C<<

D>>

Attempts:

3 left

💡 Hint

Common Mistakes

Using bitwise AND instead of OR to set bits.

To set a specific bit without affecting other bits, use the bitwise OR operator (|) with a mask where only that bit is set: (1 << 3).

5fill in blank

hard

Fill in the blanks to clear bit 2 and set bit 5 of the register.

Embedded C

*reg = (*reg [1] (1 << 2)) [2] (1 << 5);

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A& ~

B|

C&

D^

Attempts:

3 left

💡 Hint

Common Mistakes

Using XOR (^) instead of OR to set bits.

Not negating the mask when clearing bits.

First, clear bit 2 using AND with the negated mask (& ~ (1 << 2)), then set bit 5 using OR with (1 << 5).