Report statement for debug output in VHDL - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this VHDL report statement?

Consider the following VHDL process snippet. What will be printed on the simulator console when this process runs?

VHDL

process
begin
  report "Simulation started" severity note;
  wait;
end process;

A Simulation started

B No output

C Error: Simulation started

D

Simulation started
Severity: NOTE

💡 Hint

The report statement prints the message and the severity level in the simulator output.

The report statement outputs the message string followed by the severity level (note) in the simulation console.

❓ Predict Output

intermediate

What severity level is shown by this report statement?

Given this VHDL code snippet, what severity level will appear in the simulation output?

VHDL

process
begin
  report "Warning: Check signal" severity warning;
  wait;
end process;

A NOTE

B ERROR

C WARNING

D FAILURE

💡 Hint

The severity keyword defines the level shown in the output.

The severity specified is 'warning', so the output will show the severity level WARNING.

❓ Predict Output

advanced

What happens if you omit the severity in a report statement?

Examine this VHDL code. What is the default severity level if it is not specified?

VHDL

process
begin
  report "Default severity message";
  wait;
end process;

A Severity is ERROR by default

B Severity is NOTE by default

C Severity is WARNING by default

D No severity is shown

💡 Hint

Check the VHDL standard default severity for report statements.

If severity is omitted, the default severity is NOTE.

❓ Predict Output

advanced

What error occurs with this incorrect report statement?

What error will this VHDL code produce when compiled?

VHDL

process
begin
  report 123 severity warning;
  wait;
end process;

A Type error: report message must be a string

B Syntax error: missing semicolon

C No error, runs fine

D Runtime error: invalid severity

💡 Hint

The report message must be a string literal or string expression.

The report statement requires a string message. Using an integer causes a type error.

🧠 Conceptual

expert

How to use report statements for conditional debug output?

Which VHDL code snippet correctly uses a report statement to print a debug message only when signal 'error_flag' is true?

A if error_flag = '1' then report "Error detected" severity error; end if;

B report "Error detected" severity error when error_flag = '1';

C report "Error detected" severity error if error_flag = '1';

D when error_flag = '1' report "Error detected" severity error;

💡 Hint

Use an if statement to conditionally execute the report.

VHDL report statements do not support 'when' or 'if' clauses directly. You must use an if statement to conditionally execute the report.