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Let’s walk through the situation again with careful attention to the details, especially regarding pointer arithmetic and the memory addresses.

Week 1

Q6 Identify issues with code.

Recap of Code:

void main() {
    char array[10] = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50};
    char *p = array;

    p = p + (4);  // Line 4
    *p = *p + 5;  // Line 5

    p = p + (-5); // Line 6
    *p = *p + 5;  // Line 7

    p = p + (5);  // Line 8
    *p = *p + 5;  // Line 9
}

Key Assumption:

  • Memory starts at &array[0] = 256, or 0x100 in hexadecimal.
  • Each char takes up 1 byte in memory.

Step-by-Step Analysis with Memory Addresses:

  1. Initial Setup:

    • Array contents: {5, 10, 15, 20, 25, 30, 35, 40, 45, 50}.
    • char *p = array means p initially points to array[0], which is at address 0x100.
  2. Line 4 (p = p + (4);):
    • p is moved 4 positions forward, so it points to array[4], which is at address 0x104.

  3. Line 5 (*p = *p + 5;):

    • *p modifies the value at array[4]. Initially, array[4] = 25. After adding 5, array[4] = 30.
    • After line 5: p = 0x104 and *p = 30.

  4. Line 6 (p = p + (-5);):

    • p moves back 5 positions, so it now points to array[0] at address 0x100.

  5. Line 7 (*p = *p + 5;):

    • *p modifies the value at array[0]. Initially, array[0] = 5. After adding 5, array[0] = 10.
    • After line 7: p = 0x100 and *p = 10.

  6. Line 8 (p = p + (5);):

    • p moves forward 5 positions, so it now points to array[5] at address 0x105.
  7. Line 9 (*p = *p + 5;):
    • *p modifies the value at array[5]. Initially, array[5] = 30. After adding 5, array[5] = 35.
    • After line 9: p = 0x105 and *p = 35.

Evaluating Each Option:

  1. After line 5: p = 0x104, *p = 30, a[4] is modified:

    • This is true. After line 5, p is at address 0x104, *p = 30, and a[4] (or array[4]) is modified.

  2. After line 5: p = 0x114, *p = 35, a[5] is modified:

    • This is false. After line 5, p is at 0x104, not 0x114, and *p = 30, not 35.

  3. After line 9: p = 0x103, *p = 30, a[3] is modified:

    • This is false. After line 9, p is at 0x105, not 0x103, and a[3] is not modified.

  4. After line 7: p = 0x0fe, *p = Unknown, Index out of bounds:

    • This is false. After line 7, p is at 0x100, not 0x0fe, and *p = 10, so it’s not out of bounds.

  5. After line 7: p = 0x100, *p = 0, a[0] is modified:

    • This is false. After line 7, p = 0x100, but *p = 10, not 0. So this statement is incorrect.

  6. After line 9: p = 0x114, *p = 35, a[5] is modified:

    • This is false. After line 9, p = 0x105, not 0x114, though a[5] is modified.

  7. After line 7: p = 0x0ff, *p = Unknown, Index out of bounds:

    • This is false. After line 7, p is at 0x100, not 0x0ff, so this statement is incorrect.

Conclusion:

The only correct statement is:

  • After line 5: p = 0x104, *p = 30, a[4] is modified.

This matches the memory and pointer operations as detailed above.

Question 8: Find x = *((short*)array+13) where array = 0x1000, Little Endian

  1. Memory Layout: The memory contents are laid out byte by byte as shown in the question. Given that the system is little-endian, the least significant byte is stored first.

    Memory contents (in bytes):

    Address       0   1   2   3
0x1000       00  01  02  03
0x1004       04  05  06  07
0x1008       08  09  0a  0b
0x100c       0c  0d  0e  0f
0x1010       10  11  12  13
0x1014       14  15  16  17

  2. Pointer Arithmetic:

    • *((short*)array + 13) means we are treating the memory starting from array (address 0x1000) as an array of short values (2 bytes each).
    • Moving 13 positions forward (short* type moves by 2 bytes) means we are accessing the bytes at address 0x1000 + 13 * 2 = 0x101A.

  3. Little Endian Interpretation:

    • The value at address 0x101A is formed by the two bytes at 0x101A and 0x101B (little-endian means the least significant byte is stored first).
    • From the memory contents, at 0x101A, the value is 1a, and at 0x101B, the value is 1b.

  4. Forming the Result:

    • In little-endian, the least significant byte (1a) comes first, followed by the more significant byte (1b).
    • Therefore, the value is 0x1b1a (in hex).

Thus, x equals 0x1b1a in hexadecimal.

If you are given a multidimensional array.

Q2: Let’s break down the expression char* arrayc = &array; *(short*)(array_c + 2);

Lets try to explain it in terms of pointer manipulation and how many bytes will be read Key idea:The +2 in the expression is pointer math based on the type of array_c a char* The number of bytes you read once the pointer is calculated is based on the cast type

Components of the Expression:

  1. array_c + 2:
    • array_c is a pointer to char
    • Adding 2 to array_c moves the pointer 2 bytes ahead because pointer arithmetic takes the size of the data type into account (in this case, char is 1 byte).
    • After this addition, the pointer now points to the third byte of the char array (array_c[2]).
  2. (short*):
    • This casts the resulting pointer from array_c + 2 (which is of type char*) to a short*.
  3. *(short*):
    • The * dereferences the short* pointer, meaning it reads the value stored at the memory location array_c + 2, but since the pointer has been cast to a short*, it reads 2 bytes of data starting from the location array_c + 2.

Q2. Explain array accesses array[i][j] for int a[16][16] in terms of pointer arithmetic.

In the expression array[i][j] for a 2D array int array[16][16], we can break down how the memory is accessed and calculate which byte is being accessed relative to the start of the array (array[0][0]).

Background:

  1. Memory Layout for 2D Arrays:
    • In C and C++, 2D arrays like array[16][16] are stored in row-major order. This means that elements of each row are stored contiguously in memory.
    • For an array of integers (int), the size of each element (in most systems) is 4 bytes since an int typically occupies 4 bytes of memory.
  2. Array Access:
    • The expression array[i][j] accesses the element at row i and column j of the 2D array.
    • The total number of columns in the array is 16, meaning each row contains 16 integers.
Memory layout for int array[16][16] (Row-Major Order):

     array[0][0]  array[0][1]  array[0][2]  ...  array[0][15]
     array[1][0]  array[1][1]  array[1][2]  ...  array[1][15]
     array[2][0]  array[2][1]  array[2][2]  ...  array[2][15]
     ...
     array[15][0] array[15][1] array[15][2] ...  array[15][15]
  1. Calculating the Memory Offset:
    • Each row has 16 integers, and each integer takes 4 bytes.
    • To access array[i][j], you first need to skip i rows and then skip j elements within the ith row.

Memory Address Calculation:

  • The total number of elements to skip before reaching array[i][j] is: $[\text{Offset in elements} = i \times 16 + j]$ This gives you the total number of elements before array[i][j].

  • To convert this to a byte offset, multiply by the size of each element (which is 4 bytes for int): $[\text{Byte offset} = (i \times 16 + j) \times 4]$ This is the number of bytes from the start of array[0][0] to the element array[i][j].

Example:

  • If you want to access array[2][3], the byte offset is: $[\text{Byte offset} = (2 \times 16 + 3) \times 4 = (32 + 3) \times 4 = 35 \times 4 = 140 \text{ bytes}]$ So, array[2][3] is 140 bytes from the start of array[0][0].

These are the detailed solutions to Questions 6 and 8. Let me know if you need further clarifications!

Week 2

Q6 How many bits are required to represent the number -1897? in 2’s complement and sign-magnitude? ?

Group of answer choices

12 (2’s complement), 12 (Sign-mag)
12 (2’s complement), 12 (Sign-mag)
11 (2’s complement), 13 (Sign-mag)
13 (2’s complement), 11 (Sign-mag)
14 (2’s complement), 12 (Sign-mag)

Let’s break down how many bits are required to represent the number -1897 in both 2’s complement and sign-magnitude.

1. 2’s Complement Representation:

In 2’s complement, to represent a negative number, we first find its binary form as a positive number, then invert all the bits and add 1.

Step-by-step process:

  • Convert 1897 (the absolute value) into binary:
    • 1897 in binary is 11101101001 (which is 11 bits long for the positive number).
  • In 2’s complement, to represent a negative number, we need an additional bit to denote the sign. Therefore, we need 12 bits to represent -1897 in 2’s complement.

Hence, 12 bits are needed for 2’s complement.

2. Sign-Magnitude Representation:

In sign-magnitude, one bit is used for the sign (0 for positive, 1 for negative), and the remaining bits represent the magnitude of the number (the absolute value).

  • Since the absolute value of 1897 is 11101101001 (11 bits long), we need an additional bit to represent the sign.
  • Therefore, 12 bits are required for sign-magnitude (1 bit for the sign and 11 bits for the magnitude).

Summary of Bits Required:

  • 2’s complement: 12 bits
  • Sign-magnitude: 12 bits

Correct Answer:

  • 12 (2’s complement), 12 (Sign-mag)

So, the correct option is:

12 (2’s complement), 12 (Sign-mag).

Sure! Let’s break this down step-by-step, showing exactly how the expression works in binary. The goal is to set every 10th bit in a 32-bit integer to 1, starting from the least significant bit (LSB).

Write 1-line expression for setting every 10th bit of int x to 1 (starting from the LSB) e.g., every 4th bit …0001001001 ? Legal ops: ! ~ & ^ | « »

Step 1: Understanding the Problem

We are dealing with a 32-bit integer. For each bit position, the index starts at 0 from the LSB (rightmost bit) to the MSB (leftmost bit).

  • The 10th bit is at position 9 (since we start counting from 0).
  • The 20th bit is at position 19.
  • The 30th bit is at position 29.

So we want to set these specific bits to 1.

Step 2: Creating the Bit Mask

We need to create a mask that has 1s in the 10th, 20th, and 30th bit positions (binary positions 9, 19, and 29, respectively).

This is done using left shifts (<<) and OR (|) operations.

  • 1 << 9: This shifts the number 1 (which is 00000001 in binary) to the left by 9 positions.
    • The result is 00000000 00000000 00000010 00000000 (in 32 bits).

  • 1 << 19: This shifts the number 1 to the left by 19 positions.

    • The result is 00000000 00000100 00000000 00000000 (in 32 bits).
  • 1 << 29: This shifts the number 1 to the left by 29 positions.
    • The result is 00100000 00000000 00000000 00000000 (in 32 bits).

Now, we combine these using the OR (|) operation.

  • OR operation in binary combines two bits as follows:
    • 0 | 0 = 0
    • 0 | 1 = 1
    • 1 | 0 = 1
    • 1 | 1 = 1

So, the result of the bitwise OR between these shifted values is:

  00000000 00000000 00000010 00000000  (1 << 9)
| 00000000 00000100 00000000 00000000  (1 << 19)
| 00100000 00000000 00000000 00000000  (1 << 29)
-----------------------------------------
  00100000 00000100 00000010 00000000  (final mask)

This gives us the binary mask with 1s at the 9th, 19th, and 29th positions, and 0s elsewhere.

Step 3: Applying the Mask to x

Once the mask is created, we need to apply it to the integer x using the bitwise OR (|) operation. This ensures that the bits in the positions we are interested in (9, 19, and 29) are set to 1, while the rest of the bits in x remain unchanged.

For example, if x is:

  x = 00000000 00000000 00000000 00000000  (initial value of x)

We apply the OR operation between x and the mask:

  00000000 00000000 00000000 00000000   (x)
| 00100000 00000100 00000010 00000000   (mask)
-------------------------------------------
  00100000 00000100 00000010 00000000   (result)

Now, the bits at positions 9, 19, and 29 are set to 1 in x.

Final Expression

The full expression to achieve this in a 32-bit integer x is:

x = x | (1 << 9 | 1 << 19 | 1 << 29);

Conclusion

  • 1 << 9 shifts 1 to the 9th position (10th bit).
  • 1 << 19 shifts 1 to the 19th position (20th bit).
  • 1 << 29 shifts 1 to the 29th position (30th bit).
  • We combine these with the bitwise OR to set all the necessary bits in x.

By using this approach, you can set every 10th bit (starting from the LSB) in a 32-bit integer x to 1.

Week 3

Q5: What is x7? ` lw x7, 380(x21)` . Assume that you have an int array[256] = 1,2…256; Assume that register x21 = &array[0]?

Let’s break this problem down step by step.

Assumptions:

  • You have an integer array array[256] = {1, 2, 3, ..., 256}; where each element in the array is an int (usually 4 bytes on most systems).
  • Register x21 contains the address of array[0], which means x21 = &array[0].
  • The instruction lw x7, 380(x21) is a load word instruction, which means “load the 32-bit value from memory at address x21 + 380 into register x7.”

Step 1: Understanding the Memory Layout

Since each element of array is an int, and an int typically occupies 4 bytes, we can calculate which element the memory address x21 + 380 corresponds to.

  • Each integer occupies 4 bytes.
  • The address of array[i] is given by &array[0] + i * 4.

To find out which element of the array the instruction is loading, we need to calculate the index corresponding to the offset of 380 bytes.

Step 2: Calculate the Index

Since each integer is 4 bytes, the offset of 380 bytes corresponds to: $\frac{380}{4}$ = 95

So, 380(x21) corresponds to array[95].

Step 3: Value of array[95]

Given that array[i] = i + 1, the value of array[95] is: array[95] = 95 + 1 = 96

Step 4: Result in x7

After the instruction lw x7, 380(x21) is executed, the value loaded into x7 is the value of array[95], which is 96.

Final Answer:

The value of register x7 after the instruction is executed will be 96.

Q6 What is the final int value of register x20 ?

Assume that you have an int array[16][16] = 1,2...256; Assume that register x21 = &array[0]?

1. addi x22, x0, 1        // x22 = 1
2. slli x22, x22, 4       // x22 = x22 << 4 = 16
3. addi x22, x22, 15      // x22 = x22 + 15 = 31
4. slli x22, x22, 2       // x22 = x22 << 2 = 124
5. add  x22, x21, x22     // x22 = x21 + 124
6. lw   x20, 0(x22)       // x20 = memory[x22]

Let’s analyze the assembly code backwards, from the final instruction up to the first one. This way, we can build an understanding of how the state of the registers evolves throughout the code.

Here’s the assembly code:

1. addi x22, x0, 1        // x22 = 1
2. slli x22, x22, 4       // x22 = x22 << 4 = 16
3. addi x22, x22, 15      // x22 = x22 + 15 = 31
4. slli x22, x22, 2       // x22 = x22 << 2 = 124
5. add  x22, x21, x22     // x22 = x21 + 124
6. lw   x20, 0(x22)       // x20 = memory[x22]

Backward Breakdown:

Line 6: lw x20, 0(x22)

  • Effect: This instruction loads the 32-bit word from the memory address stored in x22 into register x20.
  • Interpretation: At this point, x22 contains the address of a specific element in the array[16][16] (which is 256 integers). The instruction retrieves the value from that memory location into x20.
  • Question: How did x22 get this address?

Line 5: add x22, x21, x22

  • Effect: This adds the value of x22 (which holds an offset in bytes) to x21 (which holds the base address of the array).
  • Interpretation: x21 is the base address of the array (i.e., &array[0]). The current value in x22 represents the byte offset of the desired element relative to the base address. This instruction computes the final memory address (&array[31] in this case) by adding the offset to the base address.
  • Question: How did x22 get the byte offset?

Line 4: slli x22, x22, 2

  • Effect: This shifts x22 left by 2 bits (which is equivalent to multiplying x22 by 4).

  • Interpretation: The reason for this shift is that each integer (element) in the array occupies 4 bytes. So, shifting by 2 bits adjusts the element index into a byte offset. At this point, x22 holds the index 31 (from line 3), and after this shift, it becomes 31 * 4 = 124. This is the byte offset needed to reach the 31st element in the array.

  • Question: How did x22 get the element index of 31? (now work forwards)

To access the element at row i and column j, the memory address is computed as:

$\text{Address of } array[i][j] = \text{Base Address} + (i \times \text{Number of Columns} + j) \times \text{Size of an Element}$

Line 1: addi x22, x0, 1

  • Effect: This initializes x22 with the value 1 by adding 1 to x0 (which is always 0).
  • Interpretation: This sets x22 = 1, which can be interpreted as selecting row 1 of the array in a row-major 2D array layout. The value 1 means you are working with the second row of the array (since array[0] is the first row and array[1] is the second row).

Line 2: slli x22, x22, 4

$i \times \text{Number of Columns}$

  • Effect: This shifts the value of x22 left by 4 bits (which is equivalent to multiplying it by 16).
  • Interpretation: Before this, x22 = 1 (from line 1). After this shift, x22 = 1 << 4 = 16. The shift converts a row index of 1 into the start of the second row, assuming each row has 16 elements.
  • Question: How did x22 get the value 1?

Line 3: addi x22, x22, 15

$(i \times \text{Number of Columns} + j)$

  • Effect: This adds 15 to the value already in x22.
  • Interpretation: Before this instruction, x22 holds the value 16 (from line 2). Adding 15 gives x22 = 16 + 15 = 31. This is the index of the element you are accessing in the array (array[31]).
  • Question: How did x22 get the value 16?

In C and many assembly languages, a 2D array is stored in row-major order. This means that:

  • All elements of row 0 are stored first, followed by all elements of row 1, and so on.
  • Even though array is conceptually 2D, in memory, it is stored as a 1D linear block of 256 integers (each taking 4 bytes). You can think of a 2D array as a continuous line of memory, with an index for each element. The address of an element array[i][j] can be computed based on its position in this linear block.

Where:

  • Base Address is the address of array[0][0] (which is held in x21).
  • i is the row index.
  • j is the column index.
  • Number of Columns is the total number of columns (in this case, 16).
  • Size of an Element is 4 bytes for an int.

Putting It All Together (In Reverse):

  • Line 1 initializes x22 to 1, meaning we are working with the second row of the array.
  • Line 2 shifts x22 left by 4, multiplying x22 by 16. This moves the index from “row 1” to “the start of row 1,” because each row contains 16 elements.
  • Line 3 adds 15 to x22, meaning we are accessing the 15th element of the second row. After this, x22 = 31 (this represents the 31st element in the array, since row-major ordering linearizes the 2D array).
  • Line 4 shifts x22 left by 2, multiplying by 4 to convert the element index (31) into a byte offset (31 * 4 = 124 bytes).
  • Line 5 adds x21 (the base address of the array) to this byte offset, calculating the final memory address of array[31] (the 31st element).
  • Line 6 loads the value from memory at this calculated address (&array[31]) into x20. Since array[31] = 32, x20 = 32.

Intuition for 2D Array Access:

  • Row-Major Layout: The array is stored linearly in memory, with all elements of the first row, then all elements of the second row, and so on.
  • Indexing: To access an element, you need to compute its linear index by considering both the row and column. For example, array[1][15] is actually the 31st element in a linear view of the array.
  • Shifting and Multiplying: The left shifts in this code adjust the index to account for the number of elements per row (16 elements) and the size of each element (4 bytes).

By understanding this backwards flow, you can see how the code efficiently calculates the memory address of array[1][15] (the 31st element) and loads its value into a register.

Week 4

What is the rd of the instruction add s11, s11, s3 after the code below is run ?

•text
lw t0, 16(zero) 
addi t1,zero, 128
add t0,t0,t1
sw t0,16(zero)
add s11,s11,s3 
exit:

Overview

The self-modifying code is designed to change the destination register (rd) in the add s11, s11, s3 instruction. This is achieved by loading the instruction from memory, modifying it by adding 128, and writing it back to memory. The key point is that adding 128 flips bit 7 of the instruction word, which corresponds to the least significant bit of the rd field. This changes the rd register from s11 (x27) to s10 (x26).

1. Understanding the Load and Store Instructions

Instructions:

  1. lw t0, 16(zero)

    • Operation: Load Word
    • Action: Loads the 32-bit instruction at memory address 16 into register t0.

  2. sw t0, 16(zero)

    • Operation: Store Word
    • Action: Writes the modified 32-bit instruction from t0 back into memory address 16.
  • Loading an Instruction as Data: The lw instruction treats the instruction at address 16 as data, allowing us to manipulate it.
  • Writing Back the Modified Instruction: The sw instruction overwrites the original instruction with the modified version, effectively altering the program’s code.

What is at memory address 16?. If the first instruction is located at address 0x0, the instruction at address 16 is the fifth instruction in the program. Note that each instruction is 32 bits (4 bytes) long in RISC-V.

**2. How is the instruction modified

Instructions Involved:

  1. addi t1, zero, 128

    • Operation: Add Immediate
    • Action: Sets t1 to 128 (0b10000000 in binary). 7th bit is a 1.

  2. add t0, t0, t1

    • Operation: Add
    • Action: Adds 128 to the instruction word in t0.
Effect on the Instruction:

3. Understanding the add s11, s11, s3 Instruction

Instruction Encoding:
Bits 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
Bits 0 0 0 0 0 0 0 1 0 0 1 1 1 1 0 1 1 0 0 0 1 1 0 1 1 0 0 1 1 0 0 1
  • Opcode (bits [6:0]): 0110011 (for R-type instructions)
  • rd (bits [11:7]): x27 (s11). 11011
  • Funct3 (bits [14:12]): 000 (for add)
  • rs1 (bits [19:15]): x27 (s11). 11011
  • rs2 (bits [24:20]): x19 (s3). 10011
  • Funct7 (bits [31:25]): 0000000
Binary Representation of rd:
Bits 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
Bits 0 0 0 0 0 0 0 1 0 0 1 1 1 1 0 1 1 0 0 0 1 1 0 1 1 0 0 1 1 0 0 1

+

Bits 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0    
Bits 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0

4. Effect of Adding 128 to the Instruction Word

Flipping Bit 7:
  • Bit Positions:
    • Bit 7: Least significant bit of the rd field.
  • Original Bit 7:
    • Value: 1 (from x27 which is 11011)
  • After Adding 1Modified rd:
    • x28
  • Change in Destination Register:
  • From: x27 (s11)
  • To: x28 (t3)

Week 6

Represent decimal value -0.5 in a IEEE-like floating-point number with a 5-bit exponent and a 4-bit mantissa that represents the ?

To represent the decimal value $-0.5$ in an IEEE-like floating-point format with a 5-bit exponent and a 4-bit mantissa, follow these steps:

1. Define the Bias for the Exponent:

For a 5-bit exponent, the bias is calculated as $2^{(5-1)} - 1 = 15$.

2. Convert $-0.5$ to Binary Scientific Notation:**

$-0.5$ in binary is $-0.1$. Normalizing this, we get:

$-0.1_2 = -1.0_2 \times 2^{-1}$

So, the exponent is $-1$, and the mantissa (fractional part after the leading 1) is $0.0$.

3. Determine the Sign Bit:**

Since the number is negative, the sign bit is $1$.

4. Calculate the Exponent Field:**

Add the bias to the actual exponent: $\text{Exponent Field} = -1 + 15 = 14$ In binary (5 bits), $14$ is $01110$.

5. Determine the Mantissa Field:

With a 4-bit mantissa and a fractional part of $0.0$, the mantissa field is $0000$.

6. Assemble the Bits:

Combine the sign bit, exponent field, and mantissa field: $\text{Bits} = 1 \text{ (sign bit)} | 01110 \text{ (exponent)} | 0000 \text{ (mantissa)}$

7. Convert to Hexadecimal:

Pad the 10 bits to 12 bits to make full nibbles: $\text{Bits} = 10 1110 0000$ So, the hexadecimal representation is $2E0$.

Answer: 2E0**

What is the decimal value represented by the floating-point number FP: 0xF8 in an IEEE-like format with a 5-bit exponent and a 4-bit mantissa?

To determine the decimal value represented by the floating-point number FP: 0xF8 in an IEEE-like format with a 5-bit exponent and a 4-bit mantissa, let’s carefully analyze the given information and proceed step by step.

1. Understand the Floating-Point Format:

  • Total Bits: The format consists of 1 sign bit, 5 exponent bits, and 4 mantissa bits, totaling 10 bits.

  • Bias Calculation: The bias for the exponent is calculated as:

    $\text{Bias} = 2^{(5-1)} - 1 = 2^4 - 1 = 15$

2. Convert the Hexadecimal to Binary:

The hexadecimal number 0xF8 is:

  • Binary Representation (8 bits): 1111 1000

However, since our format requires 10 bits, we need to pad the binary number with two leading zeros:

  • Padded Binary (10 bits): 00 1111 1000

3. Split the Binary into Fields:

Assign bits to the sign, exponent, and mantissa fields:

  • Sign Bit (Bit 9): The first bit.
  • Exponent Bits (Bits 8-4): The next 5 bits.
  • Mantissa Bits (Bits 3-0): The last 4 bits.

So, we have:

  • Sign Bit: 0
  • Exponent Bits: 01111
  • Mantissa Bits: 1000

4. Interpret Each Field:

Sign Bit:

  • 0 indicates the number is positive.

Exponent Field:

  • Binary: 01111
  • Decimal Exponent Field: $ (0 \times 2^4) + (1 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) = 15 $
  • Actual Exponent: Exponent Field minus Bias $\text{Exponent} = 15 - 15 = 0$

Mantissa Field:

  • Binary: 1000
  • Fractional Value: Since the mantissa represents the fractional part after the leading 1 (implied in normalized numbers), we calculate the mantissa value as: $\text{Mantissa Fraction} = \frac{1 \times 2^{-1} + 0 \times 2^{-2} + 0 \times 2^{-3} + 0 \times 2^{-4}}{1} = 0.5$
  • Normalized Mantissa: $1 + \text{Mantissa Fraction} = 1 + 0.5 = 1.5$

5. Calculate the Decimal Value:

Using the IEEE floating-point formula: $\text{Value} = (-1)^{\text{Sign Bit}} \times \text{Normalized Mantissa} \times 2^{\text{Exponent}}$

Plugging in the values: $\text{Value} = (-1)^0 \times 1.5 \times 2^{0} = 1 \times 1.5 \times 1 = 1.5$

Answer: 1.5