Hello,
I am very new to spatch and I don't even know how to formulate my question,
let me try to explain it by example.
Suppose that I want to track the usage of "+" or "-" operators, this simple
script
@e@
expression l, r;
binary operator o = { -, + };
position p;
@@
l o@p r
@script:python@
l << e.l;
r << e.r;
p << e.p;
@@
print(p[0].line,p[0].column, ':', '(%s) %s (%s)' % (l, 'OP', r))
works as expected. For example, with the following C code
void func(void)
{
1 + 2 - 3;
4 - 5 + 6;
}
I get
3 3 : (1) OP (2)
3 7 : (1 + 2) OP (3)
4 3 : (4) OP (5)
4 7 : (4 - 5) OP (6)
Now. If I change the rule "e" above
@e@
expression l, r;
position p;
@@
(
l -@p r // PAT1
|
l +@p r // PAT2
)
I only get
3 7 : (1 + 2) OP (3)
4 3 : (4) OP (5)
I could probably understand why the output for line 3 doesn't report "(1) OP
(2)",
PAT1 matches and "eats" the "1 + 2" part, so PAT2 doesn't match. Not sure.
But. Could someone explain why I don't get
4 3 : (4) OP (5)
4 7 : (4 - 5) OP (6)
for "4 - 5 + 6" statement at line 4? IOW, why PAT2 can not match _after_ PAT1?
It looks as if only one pattern from disjunction can match in the same
statement.
IOW, (PAT1 | PAT2) actually means (PAT1* | PAT2*), not (PAT1 | PAT2)*. Say,
1 + 2 - 3 + 4 - 5 + 6 - 7;
results in
3 7 : (1 + 2) OP (3)
3 15 : (1 + 2 - 3 + 4) OP (5)
3 23 : (1 + 2 - 3 + 4 - 5 + 6) OP (7)
, only PAT1 matches. While in the case of
1 - 2 + 3 - 4 + 5 - 6 + 7;
PAT2 "wins" and blocks PAT2 till the end of the statement:
3 3 : (1) OP (2)
3 11 : (1 - 2 + 3) OP (4)
3 19 : (1 - 2 + 3 - 4 + 5) OP (6)
Is this all correct?
Thanks,
Oleg.
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