Question: I have a processor trace output that has the following format: Time Cycle PC Instr Decoded instruction Register and memory contents 905ns 86 00000e36 00a005b3 c.add x11, x0, x10 x11=00000e5c x10:00000e5c 915ns 87 00000e38 00000693 c.addi x13, x0, 0 x13=00000000 925ns 88 00000e3a 00000613 c.addi x12, x0, 0 x12=00000000 935ns 89 00000e3c 00000513 c.addi x10, x0, 0 x10=00000000 945ns 90 00000e3e 2b40006f c.jal x0, 692 975ns 93 000010f2 0d01a703 lw x14, 208(x3) x14=00002b20 x3:00003288 PA:00003358 985ns 94 000010f6 00a00333 c.add x6, x0, x10 x6=00000000 x10:00000000 995ns 95 000010f8 14872783 lw x15, 328(x14) x15=00000000 x14:00002b20 PA:00002c68 1015ns 97 000010fc 00079563 c.bne x15, x0, 10 x15:00000000  Allegedly, this is \t separated, however this is not the case, as inline spaces are found here and there. I want to transform this into a .csv format with a header row and the entries following. For example: Time,Cycle,PC,Instr,Decoded instruction,Register and memory contents 905ns,86,00000e36,00a005b3,"c.add x11, x0, x10", x11=00000e5c x10:00000e5c 915ns,87,00000e38,00000693,"c.addi x13, x0, 0", x13=00000000 ...  To do that, I am using Lark in python3 (>=3.10). And I came up with the following grammar for the source format: Lark Grammar start: header NEWLINE entries+ # Header is expected to be # Time\tCycle\tPC\tInstr\tDecoded instruction\tRegister and memory contents header: HEADER_FIELD+ # Entries are expected to be e.g., # 85ns 4 00000180 00003197 auipc x3, 0x3000 x3=00003180 entries: TIME \ CYCLE \ PC \ INSTR \ DECODED_INSTRUCTION \ reg_and_mem? NEWLINE reg_and_mem: REG_AND_MEM+ /////////////// // TERMINALS // /////////////// HEADER_FIELD: / [a-z ]+ # Characters that are optionally separated by a single space /xi TIME: / [\d\.]+ # One or more digits [smunp]s # Time unit /x CYCLE: INT PC: HEXDIGIT+ INSTR: HEXDIGIT+ DECODED_INSTRUCTION: / [a-z\.]+ # Instruction mnemonic ([-a-z0-9, ()]+)? # Optional operand part (rd,rs1,rs2, etc.) (?= # Stop when x[0-9]{1,2}[=:] # Either you hit an xN= or xN: |PA: # or you meet PA: |\s+$ # or there is no REG_AND_MEM and you meet a \n ) /xi REG_AND_MEM: / (?:[x[0-9]+|PA) [=|:] [0-9a-f]+ /xi /////////////// // IMPORTS // /////////////// %import common.HEXDIGIT %import common.NUMBER %import common.INT %import common.UCASE_LETTER %import common.CNAME %import common.NUMBER %import common.WS_INLINE %import common.WS %import common.NEWLINE /////////////// // IGNORE // /////////////// %ignore WS_INLINE  Here is my simple driver code: import lark class TraceTransformer(lark.Transformer): def start(self, args): return lark.Discard def header(self, fields): return [str(field) for field in fields] def entries(self, args): print(args) ... # the grammar provided above # stored in the same directory # as this file parser = lark.Lark(grammar=open("grammar.lark").read(), start="start", parser="lalr", transformer=TraceTransformer()) # This is parsed by the grammar without problems # Note that I omit from the c.addi the operand # part and its still parsed. This is ok as some # mnemonics do not have operands (e.g., fence). dummy_text_ok1 = r"""Time Cycle PC Instr Decoded instruction Register and memory contents 905ns 86 00000e36 00a005b3 c.add x11, x0, x10 x11=00000e5c x10:00000e5c 915ns 87 00000e38 00000693 c.addi x13, x0, 0 x13=00000000 925ns 88 00000e3a 00000613 c.addi x12=00000000 935ns 89 00000e3c 00000513 c.addi x10, x0, 0 x10=00000000""" # Now here starts trouble. Note that here we don't # have a REG_AND_MEM part on the jump instruction. # However this is still parsed with no errors. dummy_text_ok2 = r"""Time Cycle PC Instr Decoded instruction Register and memory 945ns 90 00000e3e 2b40006f c.jal x0, 692 """ # But here, when the parser meets the line of cjal # where there is no REG_AND_MEM part and a follow # up entry exists we have an issue. dummy_text_problematic = r"""Time Cycle PC Instr Decoded instruction Register and memory contents 905ns 86 00000e36 00a005b3 c.add x11, x0, x10 x11=00000e5c x10:00000e5c 915ns 87 00000e38 00000693 c.addi x13, x0, 0 x13=00000000 925ns 88 00000e3a 00000613 c.addi x12, x0, 0 x12=00000000 935ns 89 00000e3c 00000513 c.addi x10, x0, 0 x10=00000000 945ns 90 00000e3e 2b40006f c.jal x0, 692 975ns 93 000010f2 0d01a703 lw x14, 208(x3) x14=00002b20 x3:00003288 PA:00003358 985ns 94 000010f6 00a00333 c.add x6, x0, x10 x6=00000000 x10:00000000 995ns 95 000010f8 14872783 lw x15, 328(x14) x15=00000000 x14:00002b20 PA:00002c68 1015ns 97 000010fc 00079563 c.bne x15, x0, 10 x15:00000000 """ parser.parse(dummy_text_ok1) parser.parse(dummy_text_ok2) parser.parse(dummy_text_problematic)  The Runtime Error No terminal matches 'c' in the current parser context, at line 6 col 45 945ns 90 00000e3e 2b40006f c.jal x0, 692 ^ Expected one of: * DECODED_INSTRUCTION  So this indicates that the DECODED_INSTRUCTION rule is not behaving as expected. The Rule DECODED_INSTRUCTION: / [a-z\.]+ # Instruction mnemonic ([-a-z0-9, ()]+)? # Optional operand part (rd,rs1,rs2, etc.) (?= # Stop when x[0-9]{1,2}[=:] # Either you hit an xN= or xN: |PA: # or you meet PA: |\s+$ # or there is no REG_AND_MEM and you meet a \n ) /xi  This rule is really heavy, it has to match the whole ISA of the processor, which is in RISC-V btw. So here step-by-step I have  The instruction mnemonic regex as a sequence of a-z characters and optional dots (.) The optional operand part (there exist instructions in the ISA with no operands).  Now, this was tricky. Instead of accounting from every possible instruction variation in my rules above, I thought to leverage the fact that there exist characters in the following column (Register and memory contents) which do not exist in any instruction variation of the ISA. This is where the look-ahead part of the regex comes in place. I stop when  Either I have reached the xN= part or the xN: part of the field Either I have reached the PA: part of the field OR I have reached the end of the line ($) as the field does not exist.  However, the last case does not seem to work as intended, as shown in the above example. The way I see it, this seems OK to either stop when you meet one of the two criteria, OR you have encountered a new line (implying that the following part is omitted for the current entry). Did I blunder something in the regex part?

Answer: For $ to mean end-of-line, you need to add the m, i.e. MULTILINE flag DECODED_INSTRUCTION: / ... /xim