Only look for a variable's value in the constant expression evaluation activation frame, if the variable was indeed declared in that frame, otherwise it might be a constant expression and be usable within a nested local scope or emit an error.
void f(char c) {
struct X {
static constexpr char f() {
return c; // error gracefully here as opposed to crashing.
}
};
int I = X::f();
}
llvm-svn: 286748
Additionally, for pre-C++1z, instead of forbidding a lambda's closure type from being a literal type through circumlocutorily setting HasNonLiteralTypeFieldsOrBases falsely to true -- handle lambda's more directly in CXXRecordDecl::isLiteral().
One additional small step towards implementing constexpr-lambdas.
Thanks to Richard Smith for his review!
https://reviews.llvm.org/D22662
llvm-svn: 276514
Support the constexpr specifier on lambda expressions - and support its inference from the lambda call operator's body.
i.e.
auto L = [] () constexpr { return 5; };
static_assert(L() == 5); // OK
auto Implicit = [] (auto a) { return a; };
static_assert(Implicit(5) == 5);
We do not support evaluation of lambda's within constant expressions just yet.
Implementation Strategy:
- teach ParseLambdaExpressionAfterIntroducer to expect a constexpr specifier and mark the invented function call operator's declarator's decl-specifier with it; Have it emit fixits for multiple decl-specifiers (mutable or constexpr) in this location.
- for cases where constexpr is not explicitly specified, have buildLambdaExpr check whether the invented function call operator satisfies the requirements of a constexpr function, by calling CheckConstexprFunctionDecl/Body.
Much obliged to Richard Smith for his patience and his care, in ensuring the code is clang-worthy.
llvm-svn: 264513
Faisal Vali