Resolving a reduce/reduce conflict in a simple Yacc grammar - reduce

Question

I'd like to know how to resolve the reduce/reduce conflict in the grammar below. The problematic rule is: LDA HASH expression -when I remove it the conflict disappears. I have tried breaking the LDA xxx parsing rules up into separate rules, but I still get a conflict. What is the problem with this rule exactly and how do I fix it?
%%
program:
program statement NEWLINE { }
|
;
statement:
instruction {
printf("Opcode: %c Address Mode: %d Operand: %d\n", opcode[0], address_mode, operand);
numInstructions++;
}
| assignment {
}
;
instruction: ADC expression {
}
| CLC {
address_mode = IMPLIED_MODE;
opcode[0] = 0x18;
fwrite(opcode, 1, 1, output);
}
| CLD {
address_mode = IMPLIED_MODE;
opcode[0] = 0xd8;
fwrite(opcode, 1, 1, output);
}
| LDA expression {
opcode[0] = 0xad;
}
| LDA HASH expression {
opcode[0] = 0xa9;
address_mode = IMMEDIATE_MODE;
if( operand > 255 ) {
printf("syntax error. Immediate value too large. The value must lie in the range between 0..255. Line: %d\n", yylineno-1);
exit(1);
}
}
| LDA expression COMMA IDENTIFIER {
// LDA operand, x | LDA operand, y
opcode[0] = 0xbd;
}
| STA expression {
}
| RTS {
address_mode = IMPLIED_MODE;
opcode[0] = 0x60;
fwrite(opcode, 1, 1, output);
}
;
expression: number { $$ = $1;
operand = $1;
}
| HASH number {
operand = $2;
}
;
number:
NUMBER { $$ = $1; }
| HEXADECIMAL { $$ = $1; }
| BINARY { $$ = $1; }
;
assignment:
IDENTIFIER EQUALS expression { $$ = $3; }
;
%%

Answer 1

I have just spotted the error in the grammar. The problem is that the "expression: number" rule partially reduces to "HASH number" resulting in a reduce/reduce conflict with the "LDA HASH expression" rule. The way to resolve it is to delete the "HASH number" terms within the expression rule.