When to use blocking (combinational) in Verilog - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to assign the output as the AND of inputs using blocking assignment.

Verilog

always @(*) begin
  out = [1] & b;
end

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Aa

Bout

Cc

Dd

Attempts:

3 left

💡 Hint

Common Mistakes

Using the output variable on the right side causes incorrect logic.

Using non-input variables in the expression.

Use blocking assignment for combinational logic. Here, a is the input to be ANDed with b.

2fill in blank

medium

Complete the code to use blocking assignment for combinational logic that sums two inputs.

Verilog

always @(*) begin
  sum = [1] + b;
end

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Aa

Bc

Cd

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Using the output variable on the right side.

Using a variable not declared as input.

Use blocking assignment with inputs on the right side. Here, a and b are inputs summed to produce sum.

3fill in blank

hard

Fix the error in the combinational block by choosing the correct blocking assignment variable.

Verilog

always @(*) begin
  [1] = a & b;
  c = [1] | d;
end

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Ad

Bc

Ca

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Using output variable c on the left side before it is assigned.

Using input variables incorrectly.

Use a temporary variable temp with blocking assignment to hold intermediate combinational results.

4fill in blank

hard

Fill both blanks to create a combinational block that calculates sum and carry using blocking assignments.

Verilog

always @(*) begin
  [1] = a ^ b;
  [2] = a & b;
end

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

Asum

Bcarry

Ctemp

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up sum and carry variables.

Using non-blocking assignments in combinational logic.

Use blocking assignments to calculate sum as XOR and carry as AND of inputs a and b.

5fill in blank

hard

Fill all three blanks to write a combinational block that computes max, min, and mid values using blocking assignments.

Verilog

always @(*) begin
  [1] = (a > b) ? a : b;
  [2] = (a < b) ? a : b;
  [3] = (a != b) ? ([1] + [2]) >> 1 : a;
end

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

Amax_val

Bmin_val

Cmid_val

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Using non-blocking assignments in combinational blocks.

Mixing variable names or forgetting to assign all outputs.

Use blocking assignments to assign max_val, min_val, and mid_val in combinational logic.