Antlr4 parser not parsing reassignment statement correctly - antlr

Question

I've been creating a grammar parser using Antlr4 and wanted to add variable reassignment (without having to declare a new variable)
I've tried changing the reassignment statement to be an expression, but that didn't change anything
Here's a shortened version of my grammar:
grammar MyLanguage;
program: statement* EOF;
statement
: expression EOC
| variable EOC
| IDENTIFIER ASSIGNMENT expression EOC
;
variable: type IDENTIFIER (ASSIGNMENT expression)?;
expression
: STRING
| INTEGER
| IDENTIFIER
| expression MATH expression
| ('+' | '-') expression
;
MATH: '+' | '-' | '*' | '/' | '%' | '//' | '**';
ASSIGNMENT: MATH? '=';
EOC: ';';
WHITESPACE: [ \t\r\n]+ -> skip;
STRING: '"' (~[\u0000-\u0008\u0010-\u001F"] | [\t])* '"' | '\'' (~[\u0000-\u0008\u0010-\u001F'] | [\t])* '\'';
INTEGER: '0' | ('+' | '-')? [1-9][0-9]*;
IDENTIFIER: [a-zA-Z_][a-zA-Z0-9_]*;
type: 'str';
if anything else might be of relevance, please ask
so I tried to parse
str test = "empty";
test = "not empty";
which worked, but when I tried (part of the fibbionaci function)
temp = n1;
n1 = n1 + n2;
n2 = temp;
it got an error and parsed it as
temp = n1; //statement
n1 = n1 //statement - <missing ';'>
+n2; //statement
n2 = temp; //statement

Answer 1

Your problem has nothing to do with assignment statements. Additions simply don't work at all - whether they're part of an assignment or not. So the simplest input to get the error would be x+y;. If you print the token stream for that input (using grun with the -tokens option for example), you'll get the following output:
[#0,0:0='x',<IDENTIFIER>,1:0]
[#1,1:1='+',<'+'>,1:1]
[#2,2:2='y',<IDENTIFIER>,1:2]
[#3,3:3=';',<';'>,1:3]
[#4,4:3='<EOF>',<EOF>,1:4]
line 1:1 no viable alternative at input 'x+'
Now compare this to x*y;, which works fine:
[#0,0:0='x',<IDENTIFIER>,1:0]
[#1,1:1='*',<MATH>,1:1]
[#2,2:2='y',<IDENTIFIER>,1:2]
[#3,3:3=';',<';'>,1:3]
[#4,4:3='<EOF>',<EOF>,1:4]
The important difference here is that * is recognized as a MATH token, but + isn't. It's recognized as a '+' token instead.
This happens because you introduced a separate '+' (and '-') token type in the alternative | ('+' | '-') expression. So whenever the lexer sees a + it produces a '+' token, not a MATH token, because string literals in parser rules take precedence over named lexer rules.
If you turn MATH into a parser rule math (or maybe mathOperator) instead, all of the operators will be literals and the problem will go away. That said, you probably don't want a single rule for all math operators because that doesn't give you the precedence you want, but that's a different issue.
PS: Something like x+1 still won't work because it will see +1 as a single INTEGER token. You can fix that by removing the leading + and - from the INTEGER rule (that way x = -2 would be parsed as a unary minus applied to the integer 2 instead of just the integer -2, but that's not a problem).