convert conditional expression to string

Clang Frontend
1

Hi all,

How can I get string representations of conditional expressions of IfStmt?
The problem is as follows:
For example;

2

Hi Kihong,
The problem in your particular case is that the begin location of integer literal is reported as the end location (should we fix it in AST)?
As a workaround, you can try to use Lexer::getLocForEndOfToken() to get the token end location for the location reported by getLocEnd().

02.12.2018 8:36, Kihong Heo via cfe-dev пишет:

3

Hi Kihong,
The problem in your particular case is that the begin location of
integer literal is reported as the end location (should we fix it in AST)?

No, that’s just how clang represents source ranges: source ranges are inclusive and source locations are the locations of tokens. So the start and end location of an integer literal should be the same, and the user needs to find the end of the last token themselves.

4

Thanks.

It seems Expr::getSourceRange() returns the range of the first token.
Then, in this particular case, what is the most common way to get the last token of the expression? How can I get “x < 10”?

Best,
Kihong

5

Hello!

There is a cool tutorial how to work with Clang Tidy:
https://youtu.be/1S2A0VWGOws (by Daniel Jasper and Manuel Klimek)

I have already implemented this based on the video.

Clang Tidy version:

static StringRef exprToStr(const Expr *E,
const MatchFinder::MatchResult &Result) {
return Lexer::getSourceText(
CharSourceRange::getTokenRange(E->getSourceRange()),
*Result.SourceManager, Result.Context->getLangOpts(), 0);
}

Static Analyzer version:

static StringRef exprToStr(const Expr *E, BugReporterContext &BRC) {
return Lexer::getSourceText(
CharSourceRange::getTokenRange(E->getSourceRange()),
BRC.getSourceManager(), BRC.getASTContext().getLangOpts(), 0);
}