I have written a simple grammar:
operations :
/* empty */
| operations operation ';'
| operations operation_id ';'
;
operation :
NUM operator NUM
{
printf("%d\n%d\n",$1, $3);
}
;
operation_id :
WORD operator WORD
{
printf("%s\n%s\n%s\n",$1, $3, $<string>2);
}
;
operator :
'+' | '-' | '*' | '/'
{
$<string>$ = strdup(yytext);
}
;
As you can see, I have defined an operator that recognizes one of 4 symbols. Now, I want to print this symbol in operation_id. Problem is, that logic in operator works only for last symbol in alternative.
So if I write a/b; it prints ab/ and that's cool. But for other operations, eg. a+b; it prints aba. What am I doing wrong?
*I ommited new lines symbols in example output.
This non-terminal from your grammar is just plain wrong.
operator :
'+' | '-' | '*' | '/' { $<string>$ = strdup(yytext); }
;
First, in yacc/bison, each production has an action. That rule has four productions, of which only the last has an associated action. It would be clearer to write it like this:
operator : '+'
| '-'
| '*'
| '/' { $<string>$ = strdup(yytext); }
;
which makes it a bit more obvious that the action only applies to the reduction from the token '/'.
The action itself is incorrect as well. yytext should never be used outside of a lexer action, because its value isn't reliable; it will be the value at the time the most recent lexer action was taken, but since the parser usually (but not always) reads one token ahead, it will usually (but not always) be the string associated with the next token. That's why the usual advice is to make a copy of yytext, but the idea is to copy it in the lexer rule, assigning the copy to the appropriate member of yylval so that the parser can use the semantic value of the token.
You should avoid the use of $<type>$ =. A non-terminal can only have one type, and it should be declared in the prologue to the bison file:
%type <string> operator
Finally, you will find that it is very rarely useful to have a non-terminal which recognizes different operators, because the different operators are syntactically different. In a more complete expression grammar, you'd need to distinguish between a + b * c, which is the sum of a and the product of b and c, and a * b + c, which is the sum of c and the product of a and b. That can be done by using different non-terminals for the sum and product syntaxes, or by using different productions for an expression non-terminal and disambiguating with precedence rules, but in both cases you will not be able to use an operator non-terminal which produces + and * indiscriminately.
For what its worth, here is the explanation of why a+b results in the output of aba:
The production operator : '+' has no explicit action, so it ends up using the default action, which is $$ = $1.
However, the lexer rule which returns '+' (presumably -- I'm guessing here) never sets yylval. So yylval still has the value it was last assigned.
Presumably (another guess), the lexer rule which produces WORD correctly sets yylval.string = strdup(yytext);. So the semantic value of the '+' token is the semantic value of the previous WORD token, which is to say a pointer to the string "a".
So when the rule
operation_id :
WORD operator WORD
{
printf("%s\n%s\n%s\n",$1, $3, $<string>2);
}
;
executes, $1 and $2 both have the value "a" (two pointers to the same string), and $3 has the value "b".
Clearly, it is semantically incorrect for $2 to have the value "a", but there is another error waiting to occur. As written, your parser leaks memory because you never free() any of the strings created by strdup. That's not very satisfactory, and at some point you will want to fix the actions so that semantic values are freed when they are no longer required. At that point, you will discover that having two semantic values pointing at the same block of allocated memory makes it highly likely that free() will be called twice on the same memory block, which is Undefined Behaviour (and likely to produce very difficult-to-diagnose bugs).
Related
Trying to understand shift-reduce conflicts and fix them.
I have following YACC code, for which I was expecting shift-reduce conflict but Bison doesn't generate any such warnings
%%
lang_cons: /* empty */
| declaraion // SEMI_COLON
| func
;
declaraion : keyword ID
;
func : keyword ID SEMI_COLON
;
keyword : INT
| FLOAT
;
%%
But if I uncomment SEMI_COLON in 2nd rule (i.e, | declaraion SEMI_COLON ), I get shift-reduce conflict. I was expecting reduce-reduce conflict in this case. Please help me understand this mess!
PS: Consider input,
1) int varName
2) int func;
If you give bison the -v command line flag, it will produce an .output file containing the generated state machine, which will probably help you see what is going on.
Note that bison actually parses the augmented grammar, which consists of your grammar with the additional rule
start': start END
where END is a special token whose code is 0, indicating the end of input, and start is whatever your grammar uses as a start symbol. (That ensures that the bison parser will not silently ignore garbage at the end of an otherwise valid input.)
That makes your original grammar unambiguous; after varName is seen, the lookahead will be either END, in which case declaration is reduced, or ';', which will be shifted (followed by a reduction of func when the following END is seen).
In your second grammar, the conflict involves the choice between reducing declaration or shifting the semicolon. If the semicolon were part of declaration, then you would see a reduce/reduce conflict.
I have used flex and bison in order to make a lexical analyzer and a parser for an EBNF grammar. This work is done! I mean, when i put a file with a program I write, I can see if the program has mistakes. If it doesn't, I can see the whole program in my screen based on the grammar i have used. I have no problem in this.
Now, I want to use loop handling and loop unrolling. Which part should I change? The lexical analyzer? The parser? Or the main after the parser? And how?
Introduction
As we don't have sight of a piece of your code to see how you are handling a loop in the parser and outputting code, and an example of a specific loop that you might want unrolled it is difficult to give any more detailed advice than that already given. There are unlikely to be any more experienced compiler writers or teachers anywhere on the globe than those already reading your question! So we will need to explore other ways to explain how to solve a problem like this.
It often happens that people can't post examples of their code because they started with a significant code base provided as part of a class exercise or from an open source repository, and they do not fully understand how it works to be able to find appropriate code fragments to post. Let's imagine that you had the complete source of a working compiler for a real language and wanted to add some loop optimisations to that existing, working compiler, you might then say, as you did, "what source, how can I show some source?" (because in actuality it is many tens of thousands of lines of code).
An Example Compiler
In the absence of some code to reference the alternative is to create one, as an exemplar, to explain the problem and solution. This is often how it is done in compiler text books or compiler classes. I will use a similar simple example to demonstrate how such optimisations can be achieved using the tools flex and bison.
First, we need to define the language of the example. To keep within the reasonable size constraints of a SO answer the language must be very simple. I will use simple assignments of expressions as the only statement form in my language. The variables in this language will be single letters and the constants will be positive integers. The only expression operator is plus (+). An example program in my language might be:
i = j + k; j = 1 + 2
The output code generated by the compiler will be simple assembler for a single accumulator machine with four instructions, LDA, STO, ADD and STP. The code generated for the above statements would be:
LDA j
ADD k
STO i
LDA #1
ADD #2
STO j
STP
Where LDA loads a value or variable into the accumulator, ADD adds a variable or value to the accumulator, STO stores the accumulator back to a variable. STP is "stop" for the end-of-program.
The flex program
The language shown above will need the tokens for ID and NUMBER and should also skip whitespace. The following will suffice:
%{
#define yyterminate() return (END);
%}
digit [0-9]
id [a-z]
ws [\t\n\r ]
%%
{ws}+ /* Skip whitespace */
{digit}+ {yylval = (int)(0l - atol(yytext)); return(NUMBER); }
{id} {yylval = yytext[0]; return(ID); }
"+" {return('+'); }
"=" {return('='); }
Gory details
Just some notes on how this works. I've used atol to convert the integer to allow for deal with potential integer overflow that can occur in reading MAXINT. I'm negating the constants so they can be easily distinguished from the identifiers which will be positive in one byte. I'm storing single character identifiers to avoid having the burden of illustrating symbol table code and thus permit a very small lexer, parser and code generator.
The bison program
To parse the language and generate some code from the bison actions we can achieve this by the following bison program:
%{
#include <stdio.h>
%}
%token NUMBER ID END
%%
program : statements END { printf("STP\n"); return(0) ; }
;
statements : statement
| statements ';' statement
;
statement : ID '=' expression { printf("STO %c\n",$1); }
|
;
expression : operand {
/* Load operand into accumulator */
if ($1 <= 0)
printf("LDA #%d\n",(int)0l-$1);
else printf("LDA %c\n",$1);
}
| expression '+' operand {
/* Add operand to accumulator */
if ($3 <= 0)
printf("ADD #%d\n",(int)0l-$3);
else printf("ADD %c\n",$3);
}
;
operand : NUMBER
| ID
;
%%
#include "lex.yy.c"
Explanation of methodology
This paragraph is intended for those who know how to do this and might query the approach used in my examples. I've deliberately avoided building a tree and doing a tree walk, although this would be the orthodox technique for code generation and optimisation. I wanted to avoid adding all the necessary code overhead in the example to manage the tree and walk it. This way my example compiler can be really tiny. However, being restricted to only using bison action to perform the code generation limits me to the ordering of the bison rule matching. This meant that only pseudo-machine code could really be generated. A source-to-source example would be less tractable with this methodology. I've chosen an idealised machine that is a cross between MU0 and a register-less PDP/11, again with the bare minimum of features to demonstrate some optimisations of code.
Optimisation
Now we have a working compiler for a language in a few lines of code we can start to demonstrate how the process of adding code optimisation might work.
As has already been said by the esteemed #Chris Dodd:
If you want to do program transformations after parsing, you should do them after parsing. You can do them incrementally (calling transform routines from your bison code after parsing part of your input), or after parsing is complete, but either way, they happen after parsing the part of the program you are transforming.
This compiler works by emitting code incrementally after parsing part of the input. As each statement is recognised the bison action (within the {...} clause) is invoked to generate code. If this is to be transformed into more optimal code it is this code that has to be changed to generate the desired optimisation. To be able to achieve effective optimisation we need a clear understanding of what language features are to be optimised and what the optimal transformation should be.
Constant Folding
A common optimisation (or code transformation) that can be done in a compiler is constant folding. In constant folding the compiler replaces expressions made entirely of numbers by the result. For example consider the following:
i = 1 + 2
An optimisation would be to treat this as:
i = 3
Thus the addition of 1 + 2 was made by the compiler and not put into the generated code to occur at run time. We would expect the following output to result:
LDA #3
STO i
Improved Code Generator
We can implement the improved code by looking for the explicit case where we have a NUMBER on both sides of expression '+' operand. To do this we have to delay taking any action on expression : operand to permit the value to be propagated onwards. As the value for an expression might not have been evaluated we have to potentially do that on assignment and addition, which makes for a slight explosion of if statements. We only need to change the actions for the rules statement and expression however, which are as shown below:
statement : ID '=' expression {
/* Check for constant expression */
if ($3 <= 0) printf("LDA #%d\n",(int)0l-$3);
else
/* Check if expression in accumulator */
if ($3 != 'A') printf("LDA %c\n",$3);
/* Now store accumulator */
printf("STO %c\n",$1);
}
| /* empty statement */
;
expression : operand { $$ = $1 ; }
| expression '+' operand {
/* First check for constant expression */
if ( ($1 <= 0) && ($3 <= 0)) $$ = $1 + $3 ;
else { /* No constant folding */
/* See if $1 already in accumulator */
if ($1 != 'A')
/* Load operand $1 into accumulator */
if ($1 <= 0)
printf("LDA #%d\n",(int)0l-$1);
else printf("LDA %c\n",$1);
/* Add operand $3 to accumulator */
if ($3 <= 0)
printf("ADD #%d\n",(int)0l-$3);
else printf("ADD %c\n",$3);
$$ = 'A'; /* Note accumulator result */
}
}
;
If you build the resultant compiler, you will see that it does indeed generate better code and perform the constant folding transformation.
Loop Unrolling
The transformation that you specifically asked about in your question was that of loop unrolling. In loop unrolling the compiler will look for some specific integer expression values in the loop start and end conditions to determine if the unrolled code transformation should be performed. The compiler can will then generate two possible code alternative sequences for loops, the unrolled and standard looping code. We can demonstrate this concept in this example mini-compiler by using integer increments.
If we imagine that the machine code has an INC instruction which increments the accumulator by one and is faster that performing an ADD #1 instruction, we can further improve the compiler by looking for that specific case. This involves evaluating integer constant expressions and comparing to a specific value to decide if an alternative code sequence should be used - just as in loop unrolling. For example:
i = j + 1
should result in:
LDA j
INC
STO i
Final Code Generator
To change the code generated for n + 1 we only need to recode part of the expression semantics and just test that when not folding constants wether the constant to be used would be 1 (which is negated in this example). The resultant code becomes:
expression : operand { $$ = $1 ; }
| expression '+' operand {
/* First check for constant expression */
if ( ($1 <= 0) && ($3 <= 0)) $$ = $1 + $3 ;
else { /* No constant folding */
/* Check for special case of constant 1 on LHS */
if ($1 == -1) {
/* Swap LHS/RHS to permit INC usage */
$1 = $3;
$3 = -1;
}
/* See if $1 already in accumulator */
if ($1 != 'A')
/* Load operand $1 into accumulator */
if ($1 <= 0)
printf("LDA #%d\n",(int)0l-$1);
else printf("LDA %c\n",$1);
/* Add operand $3 to accumulator */
if ($3 <= 0)
/* test if ADD or INC */
if ($3 == -1) printf("INC\n");
else printf("ADD #%d\n",(int)0l-$3);
else printf("ADD %c\n",$3);
$$ = 'A'; /* Note accumulator result */
}
}
;
Summary
In this mini-tutorial we have defined a whole language, a complete machine code, written a lexer, a compiler, a code generator and an optimiser. It has briefly demonstrated the process of code generation and indicated (albeit generally) how code transformation and optimisation could be performed. It should enable similar improvements to be made in other (as yet unseen) compilers, and has addressed the issue of identifying loop unrolling conditions and generating specific improvements for that case.
It should also have made it clear, how difficult it is to answer questions without specific examples of some program code to refer to.
...
IF LP assignment-expression RP marker statement {
backpatch($3.tlist,$5.instr);
$$.nextList = mergeList($3.flist,$6.nextList);
}
|IF LP assignment-expression RP marker statement ELSE Next statement {
backpatch($3.tlist,$5.instr);
backpatch($3.flist,$8.instr);
YYSTYPE::BackpatchList *temp = mergeList($6.nextList,$8.nextList);
$$.nextList = mergeList(temp,$9.nextList);
}
...
Assignment-expression is any assignment expression that is possible using the C operators =, +=, -=, *=, /=.
LP = (
RP = )
marker and Next are both EMPTY rule
The problem with above grammar rule and implementation is that it can't generate correct code when expression is as
bool a;
if(a){
printf("hi");
}
else{
prinf("die");
}
It expects that assignment-expression must contain relop or OR or AND to generate correct code .
Since in this case we do comparison for relop same case apply to OR and AND.
But as in above code doesn't contain any thing out of this , So it's unable to generate correct code.
The correct code can be generated by using following rule but this leads to two reduce-reduce conflict .
...
IF LP assignment-expression {
if($3.flist == NULL && $3.tlist == NULL)
...
} RP marker statement {
...
}
|IF LP assignment-expression{
if($3.flist == NULL && $3.tlist == NULL)
...
} RP marker statement ELSE Next statement {
...
}
...
What are the modification I should do in the grammar rule so that it will work as expected ?
I tried IF ELSE grammar rule
from here as well as from dragon book but unable to solve this .
Whole grammar can be found here Github
In order to insert the mid-rule action, you need to left-factor; otherwise, the bison-generated parser cannot decide which of the two MRAs to reduce. (Even though they are presumably identical, bison doesn't know that.)
if_prefix: "if" '(' expression ')' { $$ = $3; /* Normalize the flist */ }
if: if_prefix marker statement { ... }
| if_prefix marker statement "else" Next statement { ... }
(You could left factor differently; that's just one suggestion.)
It looks like your grammar has an incorrect definition of expression.
An assignment expression is only one of many non-terminals that should be able to reduce to an expression. For an if/then/else construction you generally need to allow any expression to occur between the parens. Your first example, as you point out, is perfectly valid C but doesn't contain an assignment.
In your grammar, you have this line:
/*Expression list**/
expression:
assignment-expression{}
|expression COMMA assignment-expression{}
;
However, an expression should be able to have more than assignment-expressions. Not being terribly familiar with yacc/bison, I would guess you need to change this to something like the following:
/*Expression **/
expression:
assignment-expression{}
|logical-OR-expression{}
|logical-AND-expression{}
|inclusive-OR-expression{}
|exclusive-OR-expression{}
|inclusive-AND-expression{}
|equality-expression{}
|relational-expression{}
|additive-expression{}
|multiplicative-expression{}
|exponentiation-expression{}
|unary-expression{}
|postfix-expression{}
|primary-expression{}
|expression COMMA expression{}
;
I can't really validate that this is going to work for you, and it may be imperfect, but hopefully you get the idea. Each different type of expression needs to be able to reduce to an expression. You have something very similar for statement earlier in your grammar, so this should hopefully make sense.
It might be helpful to do some reading or watch some tutorials on how LR grammars work.
I have to write code for checking if an arithmetic expression is valid or not , in lex. I am aware that I could do this very easily using yacc but doing only in lex is not so easy.
I have written the code below, which for some reason doesn't work.
Besides this, i also don't get how to handle binary operators .
My wrong code:
%{
#include <stdio.h>
/* Will be using stack to check the validity of arithetic expressions */
char stack[100];
int top = 0;
int validity =0;S
%}
operand [a-zA-Z0-9_]+
%%
/* Will consider unary operators (++,--), binary operators(+,-,*,/,^), braces((,)) and assignment operators (=,+=,-=,*=,^=) */
"(" { stack[top++]='(';}
")" { if(stack[top]!=')') yerror(); else top--;}
[+|"-"|*|/|^|%] { if(stack[top]!='$') yerror(); else stack[top]=='&';}
"++" { if(stack[top]!='$') yerror(); else top--;}
[+"-"*^%]?= { if(top) yerror();}
operand { if(stack[top]=='&') top--; else stack[top++]='$';}
%%
int yerror()
{
printf("Invalid Arithmetic Expression\n");
}
First, learn how to write regular expressions in Flex. (Patterns, Flex manual).
Inside a character class ([…]), neither quotes nor stars nor vertical bars are special. To include a - or a ], you can escape them with a \ or put them at the beginning of the list, or in the case of - at the end.
So in:
[+|"-"|*|/|^|%]
The | is just another character in the list, and including it five times doesn't change anything. "-" is a character range consisting only of the character ", although I suppose the intention was to include a -. Probably you wanted [-+*/^%] or [+\-*/^%].
There is no way that the flex scanner can guess that a + (for example) is a unary operator instead of a binary operator, and putting it twice in the list of rules won't do anything; the first rule will always take effect.
Finally, if you use definitions (like operand) in your patterns, you have to enclose them in braces: {operand}; otherwise, flex will interpret it as a simple keyword.
And a hint for the assignment itself: A valid unparenthesized arithmetic expression can be simplified into the regular expression:
term {prefix-operator}*{operand}{postfix-operator}*
expr {term}({infix-operator}{term})*
But you can't use that directly because (a) it doesn't deal with parentheses, (b) you probably need to allow whitespace, and (c) it doesn't correctly reject a+++++b because C insists on the "maximal munch" rule for lexical scans, so that is not the same as the correct expression a++ + ++b.
You can, however, translate the above regular expression into a very simple two-state state machine.
This question already has answers here:
how to resolve 2+2 and 2++2 conflict
(2 answers)
Closed 8 years ago.
I have some difficulty figuring out how to fix this.
Basically (and this could be valable for any operator, yet I'm using the '+' as an example), say we had this rule in the lexer source :
[+-]?[0-9]+ { yylval = atoi(yytext); return INTEGER; }
And, in the paser, we'd have
exp: INTEGER
| exp '+' exp { $$ = $1 + $3; }
| // etc etc
Then, in the resulting calculator, if I do
2 + 2
It would work as expected and give me the number 4.
But if I do
2+2
i.e. without spaces between 2, + and the other 2, I have a syntax error. The reason is that "+2" itself is a token, so bison reads "exp exp" and doesn't find anything since it's not part of the parser rules.
But, the line
2++2
is fine, since bison does "2" + "+2".
My question is... how could we fix that behavior so that "2+2" works the same way as "2 + 2"?
EDIT: It seems this question, as is, was a duplicate of another one, as pointed out in a comment below. Well, I have partically found the answer, but still.
If we make it the parser's job, and define a custom precedence level for the unary rules like this:
exp:
| // bla bla bla
| '+' exp %prec UPLUS { $$ = +$2; }
| '-' exp %prec UMINUS { $$ = -$2; }
I still see a problem. Indeed, we can technically do this, in the calculator:
2+++++2
4
2+++++++++++2
4
2++++3
5
Is there a way to avoid such an ugly syntax and trigger an error or at least a warning, so that only 2+2 is allowed, and, at worse, only 2+2 and 2++2, which are the only two choices that make sense there!
Thanks!
Unary operators are best handled in the grammar, not the scanner. There's no reason to do it the hard way. Just allow unary operators '+" and '-' in the productions for 'primary'; ignore unary '+'; and output code to negate the operand if the number of unary '-' operators is odd.
And get rid of [-+]? in the lex specification. At present you seem to be trying to handle it in both places.
There's also no reason to prohibit spaces between unary operators and their operands, or to only allow one unary operator, which is what handling it in the lexer condemns you to doing. Do it in the grammar. Only.