On 3 Aug 2010, at 18:54, Sasan Forghani wrote:
I have been writing a simple compiler for a CS class and have run
into a few
problems that I can't figure out how to solve. %left does not seem
to be
working as Bison is derivating immediately.
For example:
a := b + c + d
expr : expr ADDOP expr
expr: ref_name
ref_name: IDENTIFIER
ADDOP is specified with the %left
The directed translation in parse tree representation is
E
E + E
R E + E
b c d
Since ADDOP is specified with % left shouldn't the tree look like:
E
E + E
E + E R
R R d
b c
Bison uses LALR(1), so it build it bottom-up and from the right, as in
the first diagram.
Another problem I have run into is sending data back when a
recursive call
is finished. If the input is a := (b + c) + d, the code I have is as
follows:
stmt : ref_name BL_ASSIGNMENT expr {
fprintf(ptrToIrFile, "mov %s %s\n",
$<blIdentifier>3,
$<blIdentifier>1);
}
expr : expr BL_ADDOP expr {
if(strcmp($<blOperator>2, "+"))
strcpy($<blOperator>2, "add");
else
strcpy($<blOperator>2, "sub");
char bufferT[6];
changeToString(++amtOfTemps, "T", bufferT);
fprintf(ptrToIrFile, "%s %s %s %s\n",
$<blOperator>2, $<blIdentifier>1, $<blIdentifier>3,
bufferT);
strcpy($<blIdentifier>1, bufferT);
}
| ( expr)
as long as there isn't ( expr ) in the input input move up without
problem.
However once input does have an ( expr) part, $<blIdentifier>1 at
the ADDOP
level has the value of $<blIdentifier>1 at the ASSIGNMENT level.
When I add
the following action to the ( expr ) production-
{strcpy($<blIdentifier>1,
$<blIdentifier>2);}, $<blIdentifier>1 at the ADDOP level has the
correct
value but then the $<blIdentifier>1 at the ASSIGNMENT level value is
what I
changed $<blIdentifier>1 to be at the ( expr ) level. I'm very
confused.
Can't see the details, if you do strcpy() without properly allocating
first, there will be problems. Make sure its OK in the lexer, too.
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