41Implementation

Split a frame by result width

Problem

FrameSplitter.fields must divide a 12-bit frame into a 4-bit kind (the high bits) and an 8-bit payload (the low bits), returned as a tuple in that order. The starter returns tuple2(0, 0), so the public run prints kind = 0 payload = 00 for the sample 12'ha5c.

Implement the method body with split, as the starter's TODO asks. Keep the interface, the method's result type, the module structure, and mkTop unchanged. Do not spell out bit indices.

Read and predict

1. Read the method signature. The input is Bit#(12) and the result is Tuple2#(Bit#(4), Bit#(8)). Write down the two result widths and check that they add up to the input width.
2. Read mkTop. A match assignment receives the tuple and prints kind first, then payload.
3. Before pressing Run, predict the starter's line, then predict the line after a correct repair for 12'ha5c: which hex digit belongs to kind and which two belong to payload?

Run the starter once to confirm that both fields are zero.

Make the repair

split takes one vector and returns a pair whose widths come from the type the caller expects, not from arguments. Because the method already declares that type, the result relation is fixed before you write any expression. Replace only the returned value, then run again and compare with your prediction. Explain why the first field receives the most significant bits.

Completion contract

• For every 12-bit frame, the first field equals its high four bits and the second equals its low eight bits.
• No input bit is lost or duplicated between the two fields.
• The width relation comes from the declared result type, not from hand-written bit ranges.
• Returning the fields for 12'ha5c as constants matches only the sample.

Progressive hints

Transfer experiment

On paper, change the result type to Tuple2#(Bit#(5), Bit#(7)). Predict where the boundary moves for 12'ha5c, and which type-level sum must still hold.