Help with replacements on macros

Clang Frontend
#1

Hi All,

I have an AST matcher which has successfully matched all “if” statements in a large program:

StatementMatcher CondMatcher = ifStmt(hasCondition(expr().bind(“condition”)));

My goal is to use libtooling replacements to refactor all “if” conditions into my_func(). For instance,

if (true) { … }

should refactor to

if (my_func(true)) { … }

The only time I run into trouble is when I try to refactor complex macros. For instance, take the following code:

#define GUARD2 GUARD1(999)
#define GUARD1(X) if (1 | X) { return 11; }

int main() {
GUARD2
return 0;
}

My AST matcher successfully grabs the expanded if statement’s condition. To perform my replacement, I would like to obtain the following source locations:

#define GUARD2 GUARD1(999)
#define GUARD1(X) if (1 | X) { return 11; }

int main() {
GUARD2
return 0;
}

but if I do something like

SourceLocation LBegin = MySourceManager.getSpellingLoc(Condition->getBeginLoc());
SourceLocation LEnd = MySourceManager.getSpellingLoc(Condition->getEndLoc());

LEnd = Lexer::getLocForEndOfToken(LEnd, 0, MySourceManager, opts);

then the source locations are as follows:

#define GUARD2 GUARD1(999)
#define GUARD1(X) if (1 | X) { return 11; }

int main() {
GUARD2
return 0;
}

And thus the replacements turn out as

1 #define GUARD2 GUARD1(999))
2 #define GUARD1(X) if (my_func(1 | X) { return 11; }
3
4 int main() {
5 GUARD2
6 return 0;
7 }

which is incorrect. Note the extra parenthesis in line 1, and missing parenthesis in line 2. Any help at all would be very much appreciated.

Thanks in advance,

Scott Constable

#2

Hi Scott,

Hi All,

I have an AST matcher which has successfully matched all "if" statements in a large program:

StatementMatcher CondMatcher = ifStmt(hasCondition(expr().bind("condition")));

My goal is to use libtooling replacements to refactor all "if" conditions <cond> into my_func(<cond>). For instance,

if (true) { ... }

should refactor to

if (my_func(true)) { ... }

The only time I run into trouble is when I try to refactor complex macros. For instance, take the following code:

#define GUARD2 GUARD1(999)
#define GUARD1(X) if (1 | X) { return 11; }

int main() {
  GUARD2
  return 0;
}

My AST matcher successfully grabs the expanded if statement's condition. To perform my replacement, I would like to obtain the following source locations:

#define GUARD2 GUARD1(999)
#define GUARD1(X) if (<LBegin>1 | X<LEnd>) { return 11; }

int main() {
  GUARD2
  return 0;
}

but if I do something like

SourceLocation LBegin = MySourceManager.getSpellingLoc(Condition->getBeginLoc());
SourceLocation LEnd = MySourceManager.getSpellingLoc(Condition->getEndLoc());
LEnd = Lexer::getLocForEndOfToken(LEnd, 0, MySourceManager, opts);

There are some routines in lib/CodeGen/CoverageMappingGen.cpp which may help
with this.

You should be able to get the spelling location of the start of the conditional
by calling SM.getImmediateExpansionRange() in a loop. That's what the
"getStart(Stmt *)" method does.

There's a matching "getEnd(Stmt *)" routine which gets the precise end
location. It works by using a similar loop to find the location corresponding
to the last token in the statement, and then adding that token length to the
end location.

best,
vedant

#3

Hi All,

I have an AST matcher which has successfully matched all “if” statements in a large program:

StatementMatcher CondMatcher = ifStmt(hasCondition(expr().bind(“condition”)));

My goal is to use libtooling replacements to refactor all “if” conditions into my_func(). For instance,

if (true) { … }

should refactor to

if (my_func(true)) { … }

The only time I run into trouble is when I try to refactor complex macros. For instance, take the following code:

#define GUARD2 GUARD1(999)
#define GUARD1(X) if (1 | X) { return 11; }

int main() {
GUARD2
return 0;
}

My AST matcher successfully grabs the expanded if statement’s condition. To perform my replacement, I would like to obtain the following source locations:

#define GUARD2 GUARD1(999)
#define GUARD1(X) if (1 | X) { return 11; }

int main() {
GUARD2
return 0;
}

but if I do something like

SourceLocation LBegin = MySourceManager.getSpellingLoc(Condition->getBeginLoc());
SourceLocation LEnd = MySourceManager.getSpellingLoc(Condition->getEndLoc());

LEnd = Lexer::getLocForEndOfToken(LEnd, 0, MySourceManager, opts);

then the source locations are as follows:

#define GUARD2 GUARD1(999)
#define GUARD1(X) if (1 | X) { return 11; }

int main() {
GUARD2
return 0;
}

And thus the replacements turn out as

1 #define GUARD2 GUARD1(999))
2 #define GUARD1(X) if (my_func(1 | X) { return 11; }
3
4 int main() {
5 GUARD2
6 return 0;
7 }

which is incorrect. Note the extra parenthesis in line 1, and missing parenthesis in line 2. Any help at all would be very much appreciated.

That’s what Lexer::makeFileCharRange solves - for a given range, it will give you the range on the same macro expansion level, if such a range exists.