Signed and unsigned integers in ARM Architecture - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Understanding Signed vs Unsigned Integers

In ARM architecture, what is the main difference between signed and unsigned integers?

A Unsigned integers are always stored in floating-point format.

B Unsigned integers can represent negative values, while signed integers can only represent positive values.

C Signed integers use more memory than unsigned integers of the same bit size.

D Signed integers can represent negative values, while unsigned integers can only represent zero and positive values.

💡 Hint

Think about how negative numbers are represented in binary.

Signed integers use a bit to represent the sign (positive or negative), allowing representation of negative values. Unsigned integers only represent zero and positive values.

📋 Factual

intermediate

Range of 8-bit Signed and Unsigned Integers

What is the range of values for an 8-bit signed integer and an 8-bit unsigned integer in ARM architecture?

A 0 to 255 for signed, -128 to 127 for unsigned

B -127 to 128 for signed, 0 to 256 for unsigned

C -128 to 127 for signed, 0 to 255 for unsigned

D -256 to 255 for signed, 0 to 127 for unsigned

💡 Hint

Recall that signed integers use one bit for sign and the rest for value.

An 8-bit signed integer uses one bit for sign, so it ranges from -128 to 127. An 8-bit unsigned integer uses all bits for value, so it ranges from 0 to 255.

🔍 Analysis

advanced

Effect of Overflow in Signed vs Unsigned Addition

In ARM assembly, what happens when you add two 8-bit unsigned integers and the result exceeds 255? How does this differ from adding two 8-bit signed integers where the result exceeds 127?

A Unsigned addition causes an error; signed addition clamps to 127.

B Unsigned addition wraps around modulo 256; signed addition causes overflow flag to set and result wraps around modulo 256.

C Unsigned addition clamps to 255; signed addition wraps around modulo 128.

D Both unsigned and signed addition cause the processor to halt.

💡 Hint

Consider how overflow flags and wrapping behavior work in binary arithmetic.

Unsigned addition wraps around modulo 256 because it uses all bits for value. Signed addition also wraps around modulo 256 but sets the overflow flag to indicate the result is invalid as a signed number.

❓ Comparison

advanced

Comparing Signed and Unsigned Integer Comparisons

In ARM assembly, how does the processor treat the comparison of two 16-bit integers differently when they are signed versus unsigned?

A Signed comparisons consider the sign bit and use signed flags; unsigned comparisons treat all bits as magnitude and use unsigned flags.

B Signed comparisons convert numbers to floating point before comparing.

C Both signed and unsigned comparisons treat numbers the same way.

D Signed comparisons ignore the sign bit; unsigned comparisons consider the sign bit.

💡 Hint

Think about how the sign bit affects ordering in comparisons.

Signed comparisons interpret the most significant bit as sign and set flags accordingly. Unsigned comparisons treat all bits as positive magnitude and set different flags for comparison results.

❓ Reasoning

expert

Interpreting a 32-bit Binary Pattern as Signed and Unsigned

You have a 32-bit binary value 0xFFFFFF9C. What are its decimal values when interpreted as a signed 32-bit integer and as an unsigned 32-bit integer in ARM architecture?

A -100 as signed, 4294967196 as unsigned

B 4294967196 as signed, -100 as unsigned

C -100 as signed, -100 as unsigned

D 100 as signed, 100 as unsigned

💡 Hint

Convert the hex to decimal considering two's complement for signed interpretation.

The hex 0xFFFFFF9C is negative in signed 32-bit two's complement and equals -100. As unsigned, it is a large positive number 4294967196.

Practice

1.

What is the main difference between signed and unsigned integers in ARM architecture?

easy

A. Signed integers use more memory than unsigned integers.

B. Signed integers can represent negative and positive numbers, unsigned only non-negative.

C. Unsigned integers can represent negative numbers, signed cannot.

D. Signed integers are faster to process than unsigned integers.

Solution

1. Step 1: Understand signed integers

   Signed integers use two's complement to represent both negative and positive numbers.

2. Step 2: Understand unsigned integers

   Unsigned integers represent only zero and positive numbers, allowing a larger positive range.

3. Final Answer:

   Signed integers can represent negative and positive numbers, unsigned only non-negative. -> Option B

4. Quick Check:

   Signed = negative and positive, Unsigned = only positive [OK]

Hint: Signed = negatives allowed, unsigned = only zero or positive [OK]

Common Mistakes:

• Thinking unsigned can represent negative numbers
• Assuming signed uses more memory
• Believing signed is always faster

2.

Which of the following is the correct way to declare an unsigned 8-bit integer in ARM assembly?

easy

A. mov r0, #255 (assuming r0 is unsigned 8-bit)

B. movs r0, #255

C. mov r0, #-1

D. mov r0, #256

Solution

1. Step 1: Understand unsigned 8-bit range

   Unsigned 8-bit integers range from 0 to 255, so 255 is valid.

2. Step 2: Check instructions

   mov r0, #255 moves the immediate value 255 into register r0 correctly.

3. Final Answer:

   mov r0, #255 (assuming r0 is unsigned 8-bit) -> Option A

4. Quick Check:

   Unsigned 8-bit max = 255, mov loads immediate [OK]

Hint: Unsigned 8-bit max is 255, use mov with that value [OK]

Common Mistakes:

• Using movs which sets flags unnecessarily
• Trying to move negative values to unsigned
• Using values above 255 for 8-bit

3.

Consider a 4-bit signed integer using two's complement. What decimal value does the binary 1101 represent?

medium

A. -3

B. 13

C. -5

D. 5

Solution

1. Step 1: Identify two's complement sign

   In 4-bit two's complement, if the leftmost bit is 1, the number is negative.

2. Step 2: Calculate decimal value

   Binary 1101: invert bits -> 0010, add 1 -> 0011 (3 decimal), so value is -3. But careful: 1101 is -3 or -5? Let's check carefully:
   1101 binary = (1x-8) + (1x4) + (0x2) + (1x1) = -8 + 4 + 0 + 1 = -3

3. Final Answer:

   -3 -> Option A

4. Quick Check:

   Two's complement 1101 = -3 decimal [OK]

Hint: If MSB is 1, subtract 2^bits from value [OK]

Common Mistakes:

• Treating binary as unsigned
• Forgetting two's complement negative calculation
• Mixing bit positions

4.

In ARM assembly, a programmer tries to store the value -1 in an unsigned 8-bit register. What is the likely result?

medium

A. The program crashes at runtime.

B. Syntax error occurs during assembly.

C. The register stores 0.

D. The register stores 255 due to wrap-around.

Solution

1. Step 1: Understand unsigned storage

   Unsigned 8-bit registers store values 0 to 255. Negative values wrap around using modulo 256.

2. Step 2: Calculate wrap-around for -1

   -1 modulo 256 equals 255, so the register stores 255.

3. Final Answer:

   The register stores 255 due to wrap-around. -> Option D

4. Quick Check:

   Unsigned wrap-around: -1 = 255 [OK]

Hint: Negative to unsigned wraps modulo max+1 [OK]

Common Mistakes:

• Expecting syntax error for negative immediate
• Assuming register stores zero
• Thinking program crashes

5.

You have a 16-bit unsigned integer register holding the value 65535. You subtract 1 from it. What is the result stored in the register?

hard

A. 65536

B. -1

C. 65534

D. 0

Solution

1. Step 1: Understand 16-bit unsigned range

   16-bit unsigned integers range from 0 to 65535.

2. Step 2: Perform subtraction

   65535 - 1 = 65534, which fits in 16 bits without wrap-around.

3. Final Answer:

   65534 -> Option C

4. Quick Check:

   65535 - 1 = 65534 within unsigned 16-bit [OK]

Hint: Subtract within range stays positive, no wrap [OK]

Common Mistakes:

• Thinking result becomes negative
• Assuming overflow to zero
• Confusing unsigned with signed behavior