Go Binary Protocols Checkpoint

1. An unsigned-varint length prefix is 96 01 in hexadecimal. What is the payload length in decimal? Enter one integer.

2. The buffer is [03, 'c', 'a', 't', 02, 'o', 'k'] and maxLen=8. How many bytes does one successful DecodeFrame call consume? Enter one integer.

3. The entire current input consists of ten 0x80 bytes. maxLen=1024. What does this frame contract require now?

Select one answer.

ErrNeedMore, because the terminating byte has not arrived ErrFrameTooLarge, because every long varint is too large ErrMalformedVarint, because no legal tenth byte has terminated the prefix Success with a zero-length body, because every data group is zero

4. With maxLen=6, the buffer contains only [08] in hexadecimal. Which result is correct?

Select one answer.

ErrNeedMore with consumed=1 ErrNeedMore with consumed=0 Success with an empty payload and consumed=1 ErrFrameTooLarge with consumed=0

5. A complete prefix declares MaxInt bytes, maxLen=MaxInt, and the input has no body. Which availability check avoids overflow after the safe uint64-to-int conversion?

Select one answer.

length > len(buf)-prefixBytes prefixBytes+length > len(buf) int(uint64(prefixBytes)+uint64(length)) > len(buf) length+prefixBytes < 0 means a valid empty frame

6. A caller keeps the zero-copy payload and overwrites the same receive array for its next read. What can preserve the saved payload independently?

Select one answer.

Reslice the original receive slice to length zero Copy the payload into separately owned storage before overwriting Keep the consumed count alongside the alias Decode the first frame a second time after overwriting

7. The buffer starts with [80,00] in hexadecimal and maxLen=0. What does this exercise accept?

Select one answer.

Malformed varint, because all encodings must be shortest form Need more, because no payload byte follows A zero-length frame consuming 2 bytes A 128-byte frame exceeding the limit

8. Two correct parsers each report zero allocations on complete two-byte payloads. What conclusion is justified?

Select one answer.

They necessarily have identical latency on every input A custom parser is faster because it uses no standard-library helper Neither needs tests for partial input once allocations are zero Allocation counts alone do not establish speed; benchmark representative sizes and failure paths