Pass a string value in a recursive bison rule - recursion

Question

i'm having some issues on bison (again).
I'm trying to pass a string value between a "recursive rule" in my grammar file using the $$,
but when I print the value I have passed, the output looks like a wrong reference ( AU�� ) instead the value I wrote in my input file.
line: tok1 tok2
| tok1 tok2 tok3
{
int len=0;
len = strlen($1) + strlen($3) + 3;
char out[len];
strcpy(out,$1);
strcat(out," = ");
strcat(out,$3);
printf("out -> %s;\n",out);
$$ = out;
}
| line tok4
{
printf("line -> %s\n",$1);
}
Here I've reported a simplified part of the code.
Giving in input the token tok1 tok2 tok3 it should assign to $$ the out variable (with the printf I can see that in the first part of the rule the out variable has the correct value).
Matching the tok4 sequentially I'm in the recursive part of the rule. But when I print the $1 value (who should be equal to out since I have passed it trough $$), I don't have the right output.

Answer 1

You cannot set:
$$ = out;
because the string that out refers to is just about to vanish into thin air, as soon as the block in which it was declared ends.
In order to get away with this, you need to malloc the storage for the new string.
Also, you need strlen($1) + strlen($3) + 4; because you need to leave room for the NUL terminator.
It's important to understand that C does not really have strings. It has pointers to char (char*), but those are really pointers. It has arrays (char []), but you cannot use an array as an aggregate. For example, in your code, out = $1 would be illegal, because you cannot assign to an array. (Also because $1 is a pointer, not an array, but that doesn't matter because any reference to an array, except in sizeof, is effectively reduced to a pointer.)
So when you say $$ = out, you are making $$ point to the storage represented by out, and that storage is just about to vanish. So that doesn't work. You can say $$ = $1, because $1 is also a pointer to char; that makes $$ and $1 point to the same character. (That's legal but it makes memory management more complicated. Also, you need to be careful with modifications.) Finally, you can say strcpy($$, out), but that relies on $$ already pointing to a string which is long enough to hold out, something which is highly unlikely, because what it means is to copy the storage pointed to by out into the location pointed to by $$.
Also, as I noted above, when you are using "string" functions in C, they all insist that the sequence of characters pointed to by their "string" arguments (i.e. the pointer-to-character arguments) must be terminated with a 0 character (that is, the character whose code is 0, not the character 0).
If you're used to programming in languages which actually have a string datatype, all this might seem a bit weird. Practice makes perfect.
The bottom line is that what you need to do is to create a new region of storage large enough to contain your string, like this (I removed out because it's not necessary):
$$ = malloc(len + 1); // room for NUL
strcpy($$, $1);
strcat($$, " = ");
strcat($$, $3);
// You could replace the strcpy/strcat/strcat with:
// sprintf($$, "%s = %s", $1, $3)
Note that storing mallocd data (including the result of strdup and asprintf) on the parser stack (that is, as $$) also implies the necessity to free it when you're done with it; otherwise, you have a memory leak.

Answer 2

I've solved it changin the $$ = out; line into strcpy($$,out); and now it works properly.