语音压缩器配置:8位Network.NodeID转4段冒号格式的转换规则咨询
Breaking Down the NodeID to 4-Part Value Conversion Rule
Got it, let's figure out this conversion step by step. First, let's lay out all your known NodeID ↔ converted value pairs clearly:
Network.NodeID 33716246→2:2:120:22Network.NodeID 13716256→0:209:75:32Network.NodeID 33716156→2:2:119:188Network.NodeID 33716296→2:2:120:72
The Pattern Uncovered
Let's convert each decimal NodeID to its 32-bit hexadecimal representation, split into 4 byte-sized chunks, then convert back to decimal—here's how it lines up:
For
33716246(decimal):- Hexadecimal:
0x02027816 - Split into big-endian bytes:
02,02,78,16 - Decimal conversion:
2,2,120,22→ matches the given result exactly.
- Hexadecimal:
For
13716256(decimal):- Hexadecimal:
0x00D14B20 - Split into bytes:
00,D1,4B,20 - Decimal conversion:
0,209,75,32→ perfect match.
- Hexadecimal:
For
33716156(decimal):- Hexadecimal:
0x020277BC - Split into bytes:
02,02,77,BC - Decimal conversion:
2,2,119,188→ spot on.
- Hexadecimal:
For
33716296(decimal):- Hexadecimal:
0x02027848 - Split into bytes:
02,02,78,48 - Decimal conversion:
2,2,120,72→ 100% correct.
- Hexadecimal:
Formal Conversion Rule
Forward Conversion (NodeID → w:x:y:z):
- Treat the decimal
Network.NodeIDas a 32-bit unsigned integer. - Convert it to an 8-character hexadecimal string (pad with leading zeros if necessary to hit 8 characters).
- Split the hex string into 4 consecutive 2-character segments (big-endian order: first segment = most significant byte).
- Convert each 2-character hex segment to its decimal equivalent.
- Join the four decimal values with colons (
:) to get the final result.
- Treat the decimal
Reverse Conversion (w:x:y:z → NodeID):
- Convert each of the four decimal values (
w,x,y,z) to a 2-character hex string (pad with leading zeros if needed). - Concatenate the four hex strings in order (
w_hex+x_hex+y_hex+z_hex). - Convert the resulting 8-character hex string back to a decimal integer—this is your target
Network.NodeID.
- Convert each of the four decimal values (
内容的提问来源于stack exchange,提问作者phifer2088
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