Currently sub-expressions like decltype(x){} break AST creation for
subsequent tokens in the whole expression. In some cases this triggers
validation checks in validateAst() and analysis on the file stops.
For example, code like this:
int x = decltype(0){} ? 0 : 1;
currently produces internalAstError.
To fix the issue iscpp11init_impl() was changed to recognize { preceded
by decltype(expr) as a start of C++11 brace initialization expression.
The previous fix for the issue (43b58dbc9e) didn't seem to actually fix
it because it added a check for noexcept without a condition, but when
AST is created noexcept always has a condition due to simplification
from "noexcept" to "noexcept(true)" in Tokenizer::simplifyKeyword().
The issue from the ticket couldn't be reproduced neither on 43b58dbc9e
nor on the previous commit, so it is hard to tell whether the fix was
effective or not.
The issue appeared again after a refactoring of AST code in ac67049661.
Test added with the original fix was unable to catch that because it
used testAst() helper function which skips most simplification steps.
To fix the issue we now check for noexcept with a condition and add a
proper regression test that:
1. Uses tokenizeAndStringify() to ensure that all simplifications are
performed before AST is created.
2. Parses the code snippet from the ticket, as having "if (cond)" is
crucial to reproducing the original issue (internalAstError).
Also fix AST creation for lambdas that have both constexpr and mutable
keywords.
This tries to decide a bit more properly when ':' can be part of a
ternary operator. More precisely, there are some times when we want to
delay the construction of the ast for ':', so that it is place
accordingly to the matching '?'.
Typically, this fixes an issue with
`return val < 0 ? throw 1 : val;`,
where the ast for ':' would be constructed during as part of the
`throw`, and the ast for `?` would be invalid.
This patch is a bit of a hardcode, stating that we don't expect ':'
inside a throw, unless there is a complete ternary operator in there
(there can't be a range based for loop, a case in a switch). When we
reach ':', we know we are and the end of the `throw`.
Daniel Marjamäki
Paul Fultz II
keinflue
dummyunit
Oliver Stöneberg
orbitcowboy
IOBYTE
Ken-Patrick Lehrmann
shaneasd
Daniel Friedrich