Is it possible to parse an expression (without ambiguity) that can contains binary prefix, binary infix and binary postfix operators (let's assume that all the symbols are different) with precedence between them? For example:
a = 2 3 post+
b = pre+ 2 3*4
Then a would equal to 5 because = has lower precedence than the postfix post+ operator and b would be 14. I know that you can parse infix notated expressions with operator precedence parse or shunting yard but this problem seems far more complex for me.
Edit:
Parenthesis are allowed and pre/post variations of an operator have the same precedence as the infix one.
I would like to roll a hand-written algorithm.
Edit2:
By precedence I mean how much to consume. For example this:
a = 2 3 post+
Could result in these AST-s:
'=' has higher precedence than 'post+':
post+
/ \
= 3
/ \
a 2
'post+' has higher precedence than '=':
=
/ \
a post+
/ \
2 3
(The second one is what I need in this situation). I can't really use existing parser generators or fixed grammar for operands because the operators are loaded dynamically.
Related
I was having some trouble with Bison creating an operator as such:
<- = identity postfix operator with a low precedence to force evaluation of what's on the left first, e.g. 1+2<-*3 (equivalent (1+2)*3) as well as -> which is a prefix operator which does the same thing but to the right.
I was not able to get the syntax to work properly and tested with Python using - not False, which resulted in a syntax error (in Python, - has a greater precedence than not). However, this is not a problem in C or C++, where - and !/not have the same precedence.
Of course, the difference in precedence has nothing to do with the relationship between the 2 operators, only a relationship with other operators that result in the relative precedences between them.
Why is chaining prefix or postfix operators with different precedences a problem when parsing and how can implement the <- and -> operators while still having higher-precedence operators like !, ++, NOT, etc.?
Obligatory Bison (this pattern is repeated for all operators, where copy has greater precedence than post_unary):
post_unary:
copy
| post_unary "++"
| post_unary "--"
| post_unary '!'
;
Chaining operators in this category, e.g. x ! -- ! works fine syntactically.
Ok, let me suggest a possible erroneous grammar based on your sketch:
low_postfix:
mid_infix
| low_postfix "<-"
mid_infix:
high_postfix
| mid_infix '+' high_postfix
high_postfix:
term
| high_postfix "++"
term:
ID
'(' expr ')'
It should be clear just looking at those productions that var <- ++ is not part of the language. The only things that can be used as an operand to ++ are terms and other applications of ++. var <- is neither of these things.
On the other hand, var ++ <- is fine, because the operand to <- can be a mid_infix which can be a high_postfix which is an application of the ++ operator.
If the intention were to allow both of those postfix sequences, then that grammar is incorrect.
A version of that cascade is present in the Python grammar (albeit using prefix operators) which is why not - False is OK, but - not False is a syntax error. I'm reluctant to call that a bug because it may have been intentional. (Really, neither of those expressions makes much sense.) We could disagree about the value of such an intention but not on SO, which prefers to avoid opinionated discussions.
Note that what we might call "strict precedence" in this grammar and the Python grammar is by no means restricted to combinations of unary operators. Here's another one which you have likely never tried:
$ python3 -c 'print(41 + not False)'
File "<string>", line 1
print(41 + not False)
^
SyntaxError: invalid syntax
So, how can we fix that?
On some level, it would be nice to be able to just write an unambiguous grammar which conveyed our intention. And it is certainly possible to write an unambiguous grammar, which would convey the intention to bison. But it's at least an open question as to whether it would convey anything to a human reader, because the massive clutter of multiple rules necessary in order to keep track of what is and is not an acceptable grouping would be pretty daunting.
On the other hand, it's dead simple to do with bison/yacc precedence declarations. We just list the operators in order, and the parser generator resolves all the ambiguities accordingly. [See Note 1 below]
Here's a similar grammar to the above, with precedence declarations. (I left the actions in place in case you want to play with it, although it's by no means a Reproducible Example; the infrastructure it relies upon is much bigger than the grammar itself, and of little use to anyone other than me. So you'll have to define the three functions and fill in some of the bison type declarations. Or just delete the AST functions and use your own.)
%left ','
%precedence "<-"
%precedence "->"
%left '+'
%left '*'
%precedence NEG
%right "++" '('
%%
expr: expr ',' expr { $$ = make_binop(OP_LIST, $1, $3); }
| "<-" expr { $$ = make_unop(OP_LARR, $2); }
| expr "->" { $$ = make_unop(OP_RARR, $1); }
| expr '+' expr { $$ = make_binop(OP_ADD, $1, $3); }
| expr '*' expr { $$ = make_binop(OP_MUL, $1, $3); }
| '-' expr %prec NEG { $$ = make_unop(OP_NEG, $2); }
| expr '(' expr ')' %prec '(' { $$ = make_binop(OP_CALL, $1, $3); }
| "++" expr { $$ = make_unop(OP_PREINC, $2); }
| expr "++" { $$ = make_unop(OP_POSTINC, $1); }
| VALUE { $$ = make_ident($1); }
| '(' expr ')' { $$ = $2; }
A couple of notes:
I used %prec NEG on the unary minus production in order to separate that production from the subtraction production. I also used a %prec declaration to modify the precedence of the call production (the default would be ')'), although in this particular case that's unnecessary. It is necessary to put '(' into the precedence list, though. ( is the lookahead symbol which is used in precedence comparisons.
For many unary operators, I used bison %precedence declaration in the precedence list, rather than %right or %left. Really, there is no such thing as associativity with unary operators, so I think that it's more self-documenting to use %precedence, which doesn't resolve conflicts involving reductions and shifts in the same precedence level. However, even though there is no such thing as associativity between unary operators, the nature of the precedence resolution algorithm is that you can put prefix operators and postfix operators in the same precedence level and choose whether the postfix or prefix operators have priority by using %right or %left, respectively. %right is almost always correct. I did that with ++, because I was a bit lazy by the time I got to that point.
This does "work" (I think). It certainly resolves all the conflicts; bison happily produces a parser without warnings. And the tests that I tried worked at least as I expected them to:
? a++->
=> [-> [++/post a]]
? a->++
=> [++/post [-> a]]
? 3*f(a)+2
=> [+ [* 3 [CALL f a]] 2]
? 3*f(a)->+2
=> [+ [-> [* 3 [CALL f a]]] 2]
? 2+<-f(a)*3
=> [+ 2 [<- [* [CALL f a] 3]]]
? 2+<-f(a)*3->
=> [+ 2 [<- [-> [* [CALL f a] 3]]]]
But there are some expressions where the operator precedence, while "correct", might not be easily explained to a novice user. For example, although the arrow operators look somewhat like parentheses, they don't group that way. Furthermore, the decision as to which of the two operators has higher precedence seems to me to be totally arbitrary (and indeed I might have done it differently from what you expected). Consider:
? <-2*f(a)->+3
=> [<- [+ [-> [* 2 [CALL f a]]] 3]]
? <-2+f(a)->*3
=> [<- [* [-> [+ 2 [CALL f a]]] 3]]
? 2+<-f(a)->*3
=> [+ 2 [<- [* [-> [CALL f a]] 3]]]
There's also something a bit odd about how the arrow operators override normal operator precedence, so that you can't just drop them into a formula without changing its meaning:
? 2+f(a)*3
=> [+ 2 [* [CALL f a] 3]]
? 2+f(a)->*3
=> [* [-> [+ 2 [CALL f a]]] 3]
If that's your intention, fine. It's your language.
Note that there are operator precedence problems which are not quite so easy to solve by just listing operators in precedence order. Sometimes it would be convenient for a binary operator to have different binding power on the left- and right-hand sides.
A classic (but perhaps controversial) case is the assignment operator, if it is an operator. Assignment must associate to the right (because parsing a = b = 0 as (a = b) = 0 would be ridiculous), and the usual expectation is that it greedily accepts as much to the right as possible. If assignment had consistent precedence, then it would also accept as much to the left as possible, which seems a bit strange, at least to me. If a = 2 + b = 7 is meaningful, my intuitions say that its meaning should be a = (2 + (b = 7)) [Note 2]. That would require differential precedence, which is a bit complicated but not unheard of. C solves this problem by restricting the left-hand side of the assignment operators to (syntactic) lvalues, which cannot be binary operator expressions. But in C++, it really does mean a = ((2 + b) = 7), which is semantically valid if 2 + b has been overloaded by a function which returns a reference.
Notes
Precedence declarations do not really add any power to the parser generator. The languages it can produce a parser for are exactly the same languages; it produces the same sort of parsing machine (a pushdown automaton); and it is at least theoretically possible to take that pushdown automaton and reverse engineer a grammar out of it. (In practice, the grammars produced by this process are usually monstrous. But they exist.)
All that the precedence declarations do is resolve parsing conflicts (typically in an ambiguous grammar) according to some user-supplied rules. So it's worth asking why it's so much simpler with precedence declarations than by writing an unambiguous grammar.
The simple hand-waving answer is that precedence rules only apply when there is a conflict. If the parser is in a state where only one action is possible, that's the action which remains, regardless of what the precedence rules might say. In a simple expression grammar, an infix operator followed by a prefix operator is not at all ambiguous: the prefix operator must be shifted, because there is no reduce action for a partial sequence ending with an infix operator.
But when we're writing a grammar, we have to specify explicitly what constructs are possible at each point in the grammar, which we usually do by defining a bunch of non-terminals, each corresponding to some parsing state. An unambiguous grammar for expressions already has split the expression non-terminal into a cascading series of non-terminals, one for each operator precedence value. But unary operators do not have the same binding power on both sides (since, as noted above, one side of the unary operator cannot take an operand). That means that a binary operator could well be able to accept a unary operator for one of its operands, and not be able to accept the same unary operator for its other operand. Which in turn means that we need to split all of our non-terminals again, corresponding to whether the non-terminal appears on the left or the right side of a binary operator.
That's a lot of work, and it's really easy to make a mistake. If you're lucky, the mistake will result in a parsing conflict; but equally it could result in the grammar not being able to recognise a particular construct which you would never think of trying, but which some irate language user feels is an absolute necessity. (Like 41 + not False)
It's possible that my intuitions have been permanently marked by learning APL at a very early age. In APL, all operators associate to the right, basically without any precedence differences.
So, the cppreference claims:
The expression in the middle of the conditional operator (between ? and :) is parsed as if parenthesized: its precedence relative to ?: is ignored.
However, it appears to me that the part of the expression after the ':' operator is also parsed as if it were between parentheses. I've tried to implement the ternary operator in my programming language (and you can see the results of parsing expressions here), and my parser pretends that the part of the expression after ':' is also parenthesized. For example, for the expression (1?1:0?2:0)-1, the interpreter for my programming language outputs 0, and this appears to be compatible with C. For instance, the C program:
#include <stdio.h>
int main() {
printf("%d\n",(1?1:0?2:0)-1);
}
Outputs 0.
Had I programmed the parser of my programming language that, when parsing the ternary operators, simply take the first already parsed node after ':' and take it as the third operand to '?:', it would output the same as ((1?1:0)?2:0)-1, that is 1.
My question is whether this would (pretending that the expression after the ':' is parenthesized) always be compatible with C?
"Pretends that it is parenthesised" is some kind of description of operator parenthesis. But of course that has to be interpreted relative to precedence relations (including associativity). So in a-b*c and a*b-c, the subtraction effectively acts as though its arguments are parenthesised, only the left-hand argument is treated that way in a-b-c and it is the comparison operator which causes grouping in a<b-c and a-b<c.
I'm sure you know all that since your parser seems to work for all these cases, but I say that because the ternary operator is right-associative and of lower precedence than any other operator [Note 1]. That means that the pseudo-parentheses imposed by operator precedence surround the right-hand argument (regardless of its dominating operator, since all operators have higher precedence), and also the left-hand argument unless its dominating operator is another conditional operator. But that wouldn't be the case in C, where the comma operator has lower precedence and would not be enclosed by the imaginary parentheses following the :.
It's important to understand what is meant by the precedence of a complex operator. In effect, to compute the precedence relations we first collapse the operator to a simple ?: which includes the enclosed (second) argument. This is not "as if the expression were parenthesized", because it is parenthesized. It is parenthesized between ? and :, which in this context are syntactically parenthetic.
In this sense, it is very similar to the usual analysis of the subscript operator as a postfix operator, although the brackets of the subscript operator enclose a second argument. The precedence of the subscript operator is logically what would result from considering it to be a single [], abstracting away the expression contained inside. This is also the same as the function call operator. That happens to be written with parentheses, but the precise symbols are not important: it is possible to imagine an alternative language in which function calls are written with different symbols, perhaps { and }. That wouldn't affect the grammar at all.
It might seem odd to think of ? and : to be "parenthetic", since they don't look parenthetic. But a parser doesn't see the shapes of the symbols. It is satisfied by being told that a ( is closed by a ) and, in this case, that a ? is closed by a :. [Note 2]
Having said all that, I tried your compiler on the conditional expression
d = 0 ? 0 : n / d
It parses this expression correctly, but the compiled code computes n / d before verifying whether d = 0 is true. That's not the way the conditional operator should work; in this case, it will lead to an unexpected divide by 0 exception. The conditional operator must first evaluate its left-hand argument, and then evaluate exactly one of the other two expressions.
Notes:
In C, this is not quite correct. The comma operator has lower precedence, and there is a more complex interaction with assignment operators, which logically have the same precedence and are also right-associative.
In C-like languages those symbols are not used for any other purpose, so it's OK to just regard them as strange-looking parentheses and leave it at that. But as the case of the function-call operator shows (or, for that matter, the unary - operator), it is sometimes possible to reuse operator symbols for more than one purpose.
As a curiosity, it is not strictly necessary that open and close parentheses be different symbols, as long as they are not used for any other purpose. So, for example, if | is not used as an operator symbol (as it is in C), then you could use | a | to mean the absolute value of a without creating any ambiguities.
A precise analysis of the circumstances in which symbol reuse leads to actual ambiguities is beyond the scope of this answer.
Why i get error constantly on that?
var rotation:Float= Double(arc4random_uniform(50))/ Double(100-0.2)
Actually i try this one too:
var rotation:Double= Double(arc4random_uniform(50))/ Double(100-0.2)
thank you
Swift has strict rules about the whitespace around operators. Divide '/' is a binary operator.
The important rules are:
If an operator has whitespace around both sides or around neither
side, it is treated as a binary operator. As an example, the +
operator in a+b and a + b is treated as a binary operator.
If an operator has whitespace on the left side only, it is treated as a
prefix unary operator. As an example, the ++ operator in a ++b is
treated as a prefix unary operator.
If an operator has whitespace on
the right side only, it is treated as a postfix unary operator. As an
example, the ++ operator in a++ b is treated as a postfix unary
operator.
That means that you need to add a space before the / or remove the space after it to indicate that it is a binary operator:
var rotation = Double(arc4random_uniform(50)) / (100.0 - 0.2)
If you want rotation to be a Float, you should use that instead of Double:
var rotation = Float(arc4random_uniform(50)) / (100.0 - 0.2)
There is no need to specify the type explicitly since it will be inferred from the value you are assigning to. Also, you do not need to explicitly construct your literals as a specific type as those will conform to the type you are using them with.
Why i get error constantly on that?
var rotation:Float= Double(arc4random_uniform(50))/ Double(100-0.2)
Actually i try this one too:
var rotation:Double= Double(arc4random_uniform(50))/ Double(100-0.2)
thank you
Swift has strict rules about the whitespace around operators. Divide '/' is a binary operator.
The important rules are:
If an operator has whitespace around both sides or around neither
side, it is treated as a binary operator. As an example, the +
operator in a+b and a + b is treated as a binary operator.
If an operator has whitespace on the left side only, it is treated as a
prefix unary operator. As an example, the ++ operator in a ++b is
treated as a prefix unary operator.
If an operator has whitespace on
the right side only, it is treated as a postfix unary operator. As an
example, the ++ operator in a++ b is treated as a postfix unary
operator.
That means that you need to add a space before the / or remove the space after it to indicate that it is a binary operator:
var rotation = Double(arc4random_uniform(50)) / (100.0 - 0.2)
If you want rotation to be a Float, you should use that instead of Double:
var rotation = Float(arc4random_uniform(50)) / (100.0 - 0.2)
There is no need to specify the type explicitly since it will be inferred from the value you are assigning to. Also, you do not need to explicitly construct your literals as a specific type as those will conform to the type you are using them with.
let tolerance = 0.00000001
let (~=) x1 x2 = abs(x1 - x2) < tolerance
This throws the error:
"invalid operator definition. Prefix operator definitions must use a valid prefix operator name"
This isn't even a prefix operator, I don't get why it thinks so.
However the following is fine:
let (=~) x1 x2 = abs(x1 - x2) < tolerance
I only switched the order, so "=" comes before "~".
Is there any document online stating some rules regarding this?
I'm using Visual Studio 2013 with "F# 2013". The interactive console says "F# Interactive version 12.0.21005.1"
You can't define an infix operator starting with ~ in F#.
The F# 3.0 specification, section Categorization of Symbolic Operators explains the reason quite clearly:
The operators +, -, +., -., %, %%, &, && can be used as both prefix and infix operators. When these operators are used as prefix operators, the tilde character is prepended internally to generate the operator name so that the parser can distinguish such usage from an infix use of the operator. For example, -x is parsed as an application of the operator ~- to the identifier x. This generated name is also used in definitions for these prefix operators. Consequently, the definitions of the following prefix operators include the ~ character.
In F#, the ~ character in first position denotes a prefix operator. For example, (~-) is the prefix "opposite" operator: (~-) 3 is equivalent to - 3.