llvmbot
commented
1 month ago @llvm/issue-subscribers-c-1
Author: Erik Schnetter (eschnett)
| | |
| --- | --- |
| Bugzilla Link | [22509](https://llvm.org/bz22509) |
| Version | 3.5 |
| OS | MacOS X |
## Extended Description
In the code below, I create a `shared_ptr<double>` that is ultimately supposed to hold the value `-1.0`, yet the actual value is `1.84467e+19`. This looks to me as if either wrong code is generated, or as if a `size_t` value is accidentally interpreted as double.
The code uses `auto` to declare arguments for a lambda expression. If I replace these `auto` manually by `size_t` -- which are they types that should be deduced -- the code works fine.
Strangely, the code builds without warnings. If a `size_t` is accidentally interpreted as double, I would have expected a warning.
I assume that the deduction of the auto parameters is not working correctly. I may be wrong about this; in this case, I would expect a warning from the compiler about mismatching types.
I am using:
```console
$ clang++ --version
clang version 3.5.1 (tags/RELEASE_351/final)
Target: x86_64-apple-darwin14.1.0
Thread model: posix
```
I build with:
```console
$ clang++ -std=c++14 -g -Wall -o lambda-auto lambda-auto.cc
```
The code is:
```cpp
#include <iostream>
#include <memory>
#include <utility>
template <typename> struct is_shared_ptr : std::false_type {};
template <typename T>
struct is_shared_ptr<std::shared_ptr<T>> : std::true_type {};
template <template <typename> class C, typename F, typename... Args,
typename R = std::result_of_t<F(std::size_t, Args...)>,
std::enable_if_t<is_shared_ptr<C<R>>::value> * = nullptr>
auto iota(F &&f, std::size_t s, Args &&... args) {
return std::make_shared<R>(
std::forward<F>(f)(std::size_t(0), std::forward<Args>(args)...));
}
int main(int argc, char **argv) {
auto r = iota<std::shared_ptr>([](auto x, auto y) { return double(x + y); },
std::size_t(1), std::size_t(-1));
static_assert(std::is_same<decltype(r), std::shared_ptr<double>>::value, "");
std::cout << "*r=" << *r << "\n";
return 0;
}
```
Endilll
EugeneZelenko
cor3ntin
Extended Description
In the code below, I create a
shared_ptr<double>that is ultimately supposed to hold the value-1.0, yet the actual value is1.84467e+19. This looks to me as if either wrong code is generated, or as if asize_tvalue is accidentally interpreted as double.The code uses
autoto declare arguments for a lambda expression. If I replace theseautomanually bysize_t-- which are they types that should be deduced -- the code works fine.Strangely, the code builds without warnings. If a
size_tis accidentally interpreted as double, I would have expected a warning.I assume that the deduction of the auto parameters is not working correctly. I may be wrong about this; in this case, I would expect a warning from the compiler about mismatching types.
I am using:
I build with:
The code is: