Memory initialization with $readmemh in Verilog - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a memory array of 8-bit words with 256 entries.

Verilog

reg [7:0] memory [0:[1]];

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

A256

B128

C512

D255

Attempts:

3 left

💡 Hint

Common Mistakes

Using 256 as the highest index instead of 255.

The memory array indices go from 0 to 255 for 256 entries.

2fill in blank

medium

Complete the code to initialize the memory from a hex file named 'data.hex' inside an initial block.

Verilog

initial begin
  $readmemh([1], memory);
end

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

Adata.hex

B"data.hex"

C'data.hex'

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting quotes or using single quotes around the filename.

The filename must be a string literal enclosed in double quotes.

3fill in blank

hard

Fix the error in the code to correctly initialize the memory using $readmemh.

Verilog

initial begin
  $readmemh("data.hex", [1]);
end

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

Amem_array

Bmem

Cmemory

Dmemdata

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that does not match the declared memory array.

The memory array variable name must match the declared memory name.

4fill in blank

hard

Fill both blanks to create a memory initialization that reads from 'init.hex' and has 128 entries.

Verilog

reg [7:0] mem [0:[1]];
initial begin
  $readmemh([2], mem);
end

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

A127

B128

C"init.hex"

D'init.hex'

Attempts:

3 left

💡 Hint

Common Mistakes

Using 128 as highest index or single quotes for filename.

The memory size is 128 entries indexed 0 to 127, and the filename must be a double-quoted string.

5fill in blank

hard

Fill all three blanks to declare a 16-bit wide memory with 64 entries and initialize it from 'memdata.hex'.

Verilog

reg [[1]:0] mem_array [0:[2]];
initial begin
  $readmemh([3], mem_array);
end

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

A15

B63

C"memdata.hex"

D64

Attempts:

3 left

💡 Hint

Common Mistakes

Using 64 as highest index or missing quotes on filename.

For 16-bit width, bits go from 15 down to 0. For 64 entries, highest index is 63. Filename must be double-quoted.

Practice

1. What is the primary purpose of the $readmemh system task in Verilog?

easy

A. To write memory contents to a file in hexadecimal format.

B. To initialize registers with random values.

C. To convert binary data into hexadecimal format during simulation.

D. To load hexadecimal data from a file into a memory array before simulation.

Solution

1. Step 1: Understand the function of $readmemh

   $readmemh is used to read hexadecimal data from a file and load it into a memory array in Verilog.

2. Step 2: Compare options with the function

   To load hexadecimal data from a file into a memory array before simulation. correctly describes loading hex data into memory before simulation. Other options describe writing data, converting formats, or random initialization, which are incorrect.

3. Final Answer:

   To load hexadecimal data from a file into a memory array before simulation. -> Option D

4. Quick Check:

   $readmemh loads hex data [OK]

Hint: Remember: $readmemh reads hex data into memory [OK]

Common Mistakes:

• Confusing $readmemh with $writememh
• Thinking it initializes registers randomly
• Assuming it converts data formats

2. Which of the following is the correct syntax to initialize a memory array named mem using $readmemh with a file named data.hex?

easy

A. $readmemh(mem, "data.hex");

B. $readmemh("data.hex", mem);

C. readmemh("data.hex", mem);

D. $readmemh("mem", data.hex);

Solution

1. Step 1: Recall the correct order of arguments in $readmemh

   The syntax is $readmemh(filename, memory_array); where filename is a string and memory_array is the target memory.

2. Step 2: Match options to syntax

   $readmemh("data.hex", mem); matches the correct syntax. $readmemh(mem, "data.hex"); reverses arguments. readmemh("data.hex", mem); misses the $ sign. $readmemh("mem", data.hex); swaps arguments and uses incorrect quotes.

3. Final Answer:

   $readmemh("data.hex", mem); -> Option B

4. Quick Check:

   Syntax is $readmemh(filename, memory) [OK]

Hint: Filename first, memory second in $readmemh [OK]

Common Mistakes:

• Swapping filename and memory arguments
• Omitting the $ sign
• Using incorrect quotes or missing parentheses

3. Given the following Verilog code snippet and a file memfile.hex containing:

0A
1B
2C
3D

What will be the value of memory[2] after initial block executes?

reg [7:0] memory [0:3];
initial begin
  $readmemh("memfile.hex", memory);
end

medium

A. 8'h2C

B. 8'h1B

C. 8'h0A

D. 8'h3D

Solution

1. Step 1: Understand memory indexing and file data

   The file has 4 hex values: 0A at index 0, 1B at index 1, 2C at index 2, 3D at index 3.

2. Step 2: Identify value at memory[2]

   memory[2] will hold the third value, which is 2C in hex.

3. Final Answer:

   8'h2C -> Option A

4. Quick Check:

   memory[2] = 2C hex [OK]

Hint: Index matches line number starting at 0 [OK]

Common Mistakes:

• Confusing index with line number (off by one)
• Misreading hex values
• Assuming decimal instead of hex

4. Consider this Verilog code snippet:

reg [7:0] mem [0:3];
initial begin
  $readmemh("data.hex", mem, 1, 3);
end

If data.hex contains:

AA
BB
CC
DD
EE

What values will mem[0] and mem[1] hold after initialization?

medium

A. mem[0] = 8'h00, mem[1] = 8'hAA

B. mem[0] = 8'hAA, mem[1] = 8'hBB

C. mem[0] = 8'hBB, mem[1] = 8'hCC

D. mem[0] = 8'h00, mem[1] = 8'hCC

Solution

1. Step 1: Understand $readmemh with start and end addresses

   The call $readmemh("data.hex", mem, 1, 3); loads file lines starting at mem[1] up to mem[3].

2. Step 2: Map file lines to memory

   File lines: 0=AA, 1=BB, 2=CC, 3=DD, 4=EE.
   mem[1] = AA, mem[2] = BB, mem[3] = CC.
   mem[0] remains uninitialized (default 8'h00).

3. Final Answer:

   mem[0] = 8'h00, mem[1] = 8'hAA -> Option A

4. Quick Check:

   Start=1 loads from line 0 to mem[1] [OK]

Hint: Start address shifts file lines to memory indices [OK]

Common Mistakes:

• Assuming mem[0] is loaded when start address is 1
• Confusing file line numbers with memory indices
• Ignoring default zero initialization

5. You want to initialize a 16-word memory array mem with 8-bit width using $readmemh. The hex file init.hex contains 20 lines of data. Which approach correctly loads only the first 16 lines into mem without causing simulation errors?

hard

A. Use $readmemh("init.hex", mem, 0, 19); and truncate memory size to 16.

B. Use $readmemh("init.hex", mem, 0, 19); and ignore extra lines automatically.

C. Use $readmemh("init.hex", mem, 0, 15); to load lines 0 to 15.

D. Manually edit the file to 16 lines before loading.

Solution

1. Step 1: Understand memory size and file length

   The memory has 16 words (indices 0 to 15). The file has 20 lines, more than memory size.

2. Step 2: Use $readmemh with start and end addresses

   Using $readmemh("init.hex", mem, 0, 15); loads only the first 16 lines safely into memory without overflow.

3. Step 3: Evaluate other options

   Use $readmemh("init.hex", mem, 0, 19); and ignore extra lines automatically. tries to load 20 lines into 16 words, causing out-of-bounds error.
   Use $readmemh("init.hex", mem, 0, 19); and truncate memory size to 16. tries to load 20 lines into 16 words, causing overflow.
   Manually edit the file to 16 lines before loading. is manual and not always practical.

4. Final Answer:

   Use $readmemh("init.hex", mem, 0, 15); -> Option C

5. Quick Check:

   Limit loading to memory size with start/end [OK]

Hint: Limit $readmemh range to memory size to avoid errors [OK]

Common Mistakes:

• Loading more lines than memory size causing errors
• Assuming $readmemh ignores extra lines automatically
• Editing files manually instead of using range parameters