Shift operators in Verilog - Mini Project: Build & Apply

Shift Operators in Verilog

📖 Scenario: You are working on a simple digital circuit that needs to shift bits in a binary number to the left or right. This is common in hardware design for tasks like multiplying or dividing by powers of two.

🎯 Goal: Build a Verilog module that uses shift operators to shift an 8-bit input number left and right by a fixed number of bits, then display the results.

📋 What You'll Learn

Create an 8-bit input register with a specific binary value

Create a 3-bit shift amount register

Use the left shift operator (<<) to shift the input left by the shift amount

Use the right shift operator (>>) to shift the input right by the shift amount

Display the original input, left shifted, and right shifted values

💡 Why This Matters

🌍 Real World

Shift operators are used in digital circuits for fast multiplication, division, and bit manipulation.

💼 Career

Understanding shift operators is essential for hardware engineers and FPGA/ASIC designers working with Verilog.

1

Create the input data

Create an 8-bit register called data_in and set it to the binary value 8'b10110011.

Verilog

// Your code here

Hint

Use reg [7:0] data_in = 8'b10110011; to create the 8-bit input.

2

Create the shift amount

Create a 3-bit register called shift_amt and set it to 3'b010 (which is decimal 2).

Verilog

reg [7:0] data_in = 8'b10110011;
// Your code here

Hint

Use reg [2:0] shift_amt = 3'b010; to create the shift amount.

3

Apply shift operators

Create two 8-bit wires called shift_left and shift_right. Assign shift_left to data_in shifted left by shift_amt using the << operator. Assign shift_right to data_in shifted right by shift_amt using the >> operator.

Verilog

reg [7:0] data_in = 8'b10110011;
reg [2:0] shift_amt = 3'b010;
// Your code here

Hint

Use wire [7:0] shift_left = data_in << shift_amt; and wire [7:0] shift_right = data_in >> shift_amt;.

4

Display the results

Write an initial block that uses $display to print the values of data_in, shift_left, and shift_right in binary format.

Verilog

reg [7:0] data_in = 8'b10110011;
reg [2:0] shift_amt = 3'b010;
wire [7:0] shift_left = data_in > shift_amt;

// Your code here

Hint

Use initial block and $display with %b to print binary values.

Practice

1. What does the left shift operator << do to a binary number in Verilog?

easy

A. Moves bits to the left, multiplying the number by powers of 2

B. Moves bits to the right, dividing the number by powers of 2

C. Reverses the bits in the number

D. Sets all bits to zero

Solution

1. Step 1: Understand the left shift operator

   The left shift operator << moves all bits in a binary number to the left by a specified number of positions.

2. Step 2: Effect on the number's value

   Moving bits left multiplies the number by 2 for each position shifted.

3. Final Answer:

   Moves bits to the left, multiplying the number by powers of 2 -> Option A

4. Quick Check:

   Left shift = multiply by 2 [OK]

Hint: Left shift multiplies by 2 powers quickly [OK]

Common Mistakes:

• Confusing left shift with right shift
• Thinking it reverses bits
• Assuming it clears bits

2. Which of the following is the correct syntax to shift a 4-bit register reg [3:0] a; two bits to the right in Verilog?

easy

A. a >>> 2;

B. a >> 2;

C. a << 2;

D. a >> 2 =;

Solution

1. Step 1: Identify the right shift operator syntax

   In Verilog, the right shift operator is >> and is used as variable >> number_of_bits;.

2. Step 2: Check the options for correct syntax

   a >> 2; uses a >> 2; which correctly shifts a two bits to the right. a << 2; is left shift, a >>> 2; is arithmetic right shift (signed), and a >> 2 =; has invalid syntax.

3. Final Answer:

   a >> 2; -> Option B

4. Quick Check:

   Right shift syntax = a >> number [OK]

Hint: Use >> for right shift, << for left shift [OK]

Common Mistakes:

• Using left shift operator instead of right shift
• Confusing logical and arithmetic shifts
• Writing invalid syntax like assignment inside shift

3. Given reg [7:0] x = 8'b00011010; what is the value of x << 3 in binary?

medium

A. 8'b11010000

B. 8'b11010000000

C. 8'b00011010

D. 8'b00000000

Solution

1. Step 1: Understand the initial value and shift

   The initial value is 00011010 (decimal 26). Left shifting by 3 moves bits 3 places left, adding zeros on the right.

2. Step 2: Perform the left shift

   Shifting left by 3: 00011010 becomes 11010000 (the leftmost bits shifted out are lost, and zeros fill from right).

3. Final Answer:

   8'b11010000 -> Option A

4. Quick Check:

   Left shift 3 bits = 11010000 [OK]

Hint: Left shift drops left bits, adds zeros right [OK]

Common Mistakes:

• Expecting longer bit width after shift
• Not losing bits shifted out
• Confusing left and right shift effects

4. What is wrong with this Verilog code snippet?

reg [3:0] a = 4'b1010;
reg [3:0] b;
b = a >> 5;

medium

A. Right shift operator should be << instead of >>

B. Syntax error due to missing semicolon

C. Shifting by more bits than width causes all zeros

D. Cannot assign shifted value to a register

Solution

1. Step 1: Analyze the shift amount vs register width

   The register a is 4 bits wide. Shifting right by 5 bits exceeds its width.

2. Step 2: Effect of shifting more than width

   Shifting more bits than the width results in all bits shifted out, so the result is zero.

3. Final Answer:

   Shifting by more bits than width causes all zeros -> Option C

4. Quick Check:

   Shift > width = zero result [OK]

Hint: Shift count > width clears all bits [OK]

Common Mistakes:

• Thinking syntax error occurs
• Confusing left and right shift operators
• Believing assignment to register is invalid

5. You want to multiply a 6-bit unsigned number num by 8 using shift operators in Verilog. Which expression correctly does this without overflow?

hard

A. num >>> 3

B. num >> 3

C. num << 2

D. num << 3

Solution

1. Step 1: Understand multiplication by powers of 2 using shifts

   Multiplying by 8 equals multiplying by 2^3, so shifting left by 3 bits achieves this.

2. Step 2: Check for overflow considerations

   Since num is 6 bits, shifting left by 3 bits results in a 9-bit value. To avoid overflow, the result should be stored in a wider register. The expression num << 3 correctly multiplies by 8.

3. Final Answer:

   num << 3 -> Option D

4. Quick Check:

   Multiply by 8 = shift left 3 [OK]

Hint: Multiply by 8 = shift left 3 bits [OK]

Common Mistakes:

• Using right shift instead of left shift
• Using arithmetic right shift (>>>)
• Shifting by wrong number of bits