llvm / llvm-project

The LLVM Project is a collection of modular and reusable compiler and toolchain technologies.
http://llvm.org
Other
29.37k stars 12.14k forks source link

[clang] consteval constructor executes at runtime when invoked by escalated immediate function (or fails to link) #112677

Open rkjnsn opened 1 month ago

rkjnsn commented 1 month ago

Consider the following code:

#include <iostream>
#include <optional>

class ConstEval {
 public:
  consteval ConstEval(const char* value) {
    const char* current_char = value;
    while (*current_char) {
      if (*current_char == 'a') {
        throw "Disallowed character!";
      }
      ++current_char;
    }
    value_ = value;
  }
  constexpr const char* value() { return value_; }
 private:
  const char* value_;
};

void TakesOptional(std::optional<ConstEval> maybe_value) {
  if (maybe_value.has_value()) {
    std::cout << "Value: " << maybe_value->value() << std::endl;
  } else {
    std::cout << "No value" << std::endl;
  }
}

int main(int argc, const char*argv[]) {
  // 1
  TakesOptional("testing 123");
  // 2
  std::optional<ConstEval> foo2 = "testing 123";
  // 3
  constexpr std::optional<ConstEval> foo3 = "testing 123";
  // 4
  TakesOptional(ConstEval{"testing 123"});
}

// 5
std::optional<ConstEval> foo5 = "testing 123";
// 6
constinit std::optional<ConstEval> foo6 = "testing 123";

Assuming I'm reading things correctly, I would expect the std::optional constructor to be an immediate-escalating function (because it is "a function that results from the instantiation of a templated entity defined with the constexpr specifier"), and a contained call to ConstEval's constructor to be an immediate-escalating expression, causing the std::optional constructor to become an immediate function.

Thus, I would expect lines 1, 2, 3, 5, and 6 all to immediately invoke the std::optional constructor to create a compile-time constant std::optional (indirectly invoking the ConstEval constructor), while 4 would immediately invoke the ConstEval constructor to create a constant ConstEval, which would then passed to the (not immediate, in this case) std::optional constructor.

Given that, I would further expect that changing "testing" to "tasting" on any of the 6 lines would result in a compiler failure (specifically, one informing me that throw "Disallowed character!" is not valid in a constant expression).

However, with clang built from revision 3dbd929ea6af134650dd1d91baeb61a4fc1b0eb8, only lines 3, 4, and 6 fail to compile if "testing" is changed to "tasting". Lines 1, 2, and 5 unexpectedly continue to compile if "testing" in changed to "tasting", and instead cause an abort at runtime due to the uncaught exception. Thus, it appears that while clang properly considers the invocation of the explicitly-immediate ConstEval constructor to be a constant expression in 4, it erroneously does not consider the invocation of the escalated-to-immediate std::optional constructor to be a constant expression in 1, 2, and 5. Meanwhile, 3 and 6 invoke the constructor within an explicit constant expression, so those work as expected.

A further note: the observed behavior of ConstEval's constructor executing (and crashing if "testing" is changed to "tasting") at runtime for 1, 2, and 5 appears only to happen if it's small enough to inline into the generated std::optional specialization. If the constructor is more complicated and doesn't get inlined, lines 1, 2, and 5 will instead generate a linker error due to the symbol ConstEval::ConstEval(char const*) not being defined.

efriedma-quic commented 1 month ago

Needs reduced testcase that doesn't involve the full std::optional. (Looked briefly, but the issue seems to be deep inside the constructor for std::optional.)

rkjnsn commented 1 month ago

Unfortunately, the issue doesn't reproduce with a simple optional analog, so it's something within the complexity of std::optional's implementation that is triggering the bug.

efriedma-quic commented 1 month ago
class ConstEval {
 public:
  consteval ConstEval(const char* value) {
    throw "Disallowed character!";
  }
};
struct FakeOptionalBase {
    ConstEval val;
    template <class Anything = int> constexpr
    FakeOptionalBase(const char (&arg)[12]) : val(arg) {}
};
struct FakeOptional : FakeOptionalBase {
    using FakeOptionalBase::FakeOptionalBase;
};
void TakesOptional(FakeOptional maybe_value) {
}
int main(int argc, const char*argv[]) {
  TakesOptional(FakeOptional("tasting 123"));
}
efriedma-quic commented 1 month ago

The key here seems to be that the inherited constructor isn't marked as an immediate function in the AST.

efriedma-quic commented 1 month ago

(CC @cor3ntin)

llvmbot commented 1 month ago

@llvm/issue-subscribers-clang-frontend

Author: Erik Jensen (rkjnsn)

Consider the following code: ```c++ #include <iostream> #include <optional> class ConstEval { public: consteval ConstEval(const char* value) { const char* current_char = value; while (*current_char) { if (*current_char == 'a') { throw "Disallowed character!"; } ++current_char; } value_ = value; } constexpr const char* value() { return value_; } private: const char* value_; }; void TakesOptional(std::optional<ConstEval> maybe_value) { if (maybe_value.has_value()) { std::cout << "Value: " << maybe_value->value() << std::endl; } else { std::cout << "No value" << std::endl; } } int main(int argc, const char*argv[]) { // 1 TakesOptional("testing 123"); // 2 std::optional<ConstEval> foo2 = "testing 123"; // 3 constexpr std::optional<ConstEval> foo3 = "testing 123"; // 4 TakesOptional(ConstEval{"testing 123"}); } // 5 std::optional<ConstEval> foo5 = "testing 123"; // 6 constinit std::optional<ConstEval> foo6 = "testing 123"; ``` Assuming I'm reading things correctly, I would expect the `std::optional` constructor to be an immediate-escalating function (because it is "a function that results from the instantiation of a templated entity defined with the constexpr specifier"), and a contained call to `ConstEval`'s constructor to be an immediate-escalating expression, causing the `std::optional` constructor to become an immediate function. Thus, I would expect lines 1, 2, 3, 5, and 6 all to immediately invoke the `std::optional` constructor to create a compile-time constant `std::optional` (indirectly invoking the `ConstEval` constructor), while 4 would immediately invoke the `ConstEval` constructor to create a constant `ConstEval`, which would then passed to the (not immediate, in this case) `std::optional` constructor. Given that, I would further expect that changing "testing" to "tasting" on any of the 6 lines would result in a compiler failure (specifically, one informing me that `throw "Disallowed character!"` is not valid in a constant expression). However, with clang built from revision 3dbd929ea6af134650dd1d91baeb61a4fc1b0eb8, only lines 3, 4, and 6 fail to compile if "testing" is changed to "tasting". Lines 1, 2, and 5 unexpectedly continue to compile if "testing" in changed to "tasting", and instead cause an abort at runtime due to the uncaught exception. Thus, it appears that while clang properly considers the invocation of the explicitly-immediate `ConstEval` constructor to be a constant expression in 4, it erroneously does _not_ consider the invocation of the escalated-to-immediate `std::optional` constructor to be a constant expression in 1, 2, and 5. Meanwhile, 3 and 6 invoke the constructor within an explicit constant expression, so those work as expected. A further note: the observed behavior of `ConstEval`'s constructor executing (and crashing if "testing" is changed to "tasting") at runtime for 1, 2, and 5 appears only to happen if it's small enough to inline into the generated `std::optional` specialization. If the constructor is more complicated and doesn't get inlined, lines 1, 2, and 5 will instead generate a linker error due to the symbol `ConstEval::ConstEval(char const*)` not being defined.
rkjnsn commented 3 weeks ago

To bring some relevant PR discussion to the bug,

In @cor3ntin's PR #112860, @efriedma-quic provides the following test case:

struct ConstEval {
  consteval ConstEval(int) {}
};
struct SimpleCtor { constexpr SimpleCtor(int) {}};
struct TemplateCtor {
  template <class Anything = int> constexpr
  TemplateCtor (int arg) {}
};
struct ConstEvalMember1 : SimpleCtor {
    int y = 10;
    ConstEval x = y;
    using SimpleCtor::SimpleCtor;
};
struct ConstEvalMember2 : TemplateCtor {
    int y = 10;
    ConstEval x = y;
    using TemplateCtor::TemplateCtor;
};
void f() {
  ConstEvalMember1 i1(0);
  ConstEvalMember2 i2(0);
}

If the inherited derived class constructor copies the immediate-escalating property of the base class, this leads to the counter-intuitive behavior that i1 produces an error while i2 doesn't.

Intuitively, given that a compiler-generated default constructor is immediately-escalating, and an inheriting constructor in a derived class effectively acts as a compiler-generated default constructor except that it initializes the relevant base class with the inherited constructor, I would expect the generated derived-class constructor always to be immediate-escalating, and for any immediate field or base class initialization (including, but not limited to, the base class from which the constructor is inherited) to cause the compiler-generated constructor to escalate to being immediate. Thus, I would expect both i1 and i2 to compile.

That said, I was not able to find any language in the standard confirming or denying that that's how immediate escalation with inherited constructors should work. However, I'm far from a standards expert, so I very well might be missing something.