Here’s a fun little puzzle, courtesy of Richard Hodges on Slack.

template<class Vector> void test(Vector& vec) { using E = decltype(vec[0]); for (int i=0; i < 10; ++i) { vec.push_back(E(i)); } } int main() { std::vector<double> v; test(v); for (int i=0; i < 10; ++i) { printf("%f

", v[i]); } }

On any compiler you care to name, this code compiles with no warnings or errors — and produces utter garbage at runtime! Can you spot the bug? Spoiler below the break.

The trick is inside test() — that explicit conversion from int to E . If you saw this code verbatim in a real codebase, you’d probably smell something funky on the line vec.push_back(E(i)) . What is that explicit conversion doing there, when vec.push_back(i) would have done an implicit conversion anyway? Is it possible that this is our bug? …But double(i) should accomplish the same thing whether the conversion is explicit or implicit. Is it possible that we’re not converting i to double here?

Right! We’re converting i to E , where E is decltype(vec[0]) … and since vec[0] is a modifiable lvalue expression, decltype(vec[0]) is actually double& . (Remember that when we take decltype of the name of a variable or member, we get its declared type; but when we take decltype of an arbitrary non-name expression, we also get its value category. Stack Overflow has the details.)

So if E is a type alias for double& , then what is E(i) ? Well, this uses a little-known (and reviled) quirk of C++: the functional cast notation looks like you’re calling a nice safe constructor, but it’s actually a synonym for the bad old C-style cast, which means that in some cases it can actually be a reinterpret_cast ! (WG21 members can view this EWG reflector thread from June 2019 for the scary details.)

When we write E(i) , it means the same thing as (E)i , which is to say (double&)i , which is to say *(double*)&i . The result is that vec.push_back(E(i)) ends up treating the bit-pattern of i (and the four garbage bytes following it) as an eight-byte double value, and pushing a copy of that nonsense value onto our vector.

But you needn’t worry too much about hidden bugs of this type in your own codebase. The original sin here was that we were using functional-cast notation with a type E which was a reference type — and we got that reference type via a very suspect and unidiomatic application of decltype . If we had just written any of the following saner-looking lines instead, we’d have had no bug:

template<class Vector> void test(Vector& vec) { for (int i=0; i < 10; ++i) { vec.push_back(i); // OK, no explicit cast } } template<class Vector> void test(Vector& vec) { using E = typename Vector::value_type; // OK, not a reference type for (int i=0; i < 10; ++i) { vec.push_back(E(i)); } } template<class E> void test(std::vector<E>& vec) { // OK, not a reference type for (int i=0; i < 10; ++i) { vec.push_back(E(i)); } }

GCC has one extra wrinkle here. What do you suppose happens with this code?

template<class T> auto allocatorize(T&& t) { std::allocator<int> a; return T(std::forward<T>(t), a); } void test2() { std::vector<int> v; std::vector<int> w = allocatorize(v); }

GCC in -fpermissive mode complains:

warning: expression list treated as compound expression in functional cast [-fpermissive] 6 | return T(std::forward<T>(t), a); | ^~~~~~~~~~~~~~~~~~~~~~~~

This code has the same bug as our E(i) code: T is deduced as a reference type, and then T(...) is treated as a functional-style cast to a reference type. It definitely does not call the allocator-aware constructor of vector<int> ! Everyone but GCC (and GCC too, in non-permissive mode) rejects this attempt to initialize a reference type with the list of expressions (std::forward<T>(t), a) . But GCC in permissive mode has an extension that treats that expression-list as an application of the comma operator, so (std::forward<T>(t), a) ends up being treated as equivalent to a .

We don’t get a warning about “left operand of comma operator has no effect” because no compiler recognizes the function call std::forward<T>(t) as side-effect-free. I generally write std::forward<T>(t) as static_cast<T&&>(t) to save compilation time, but I guess this is another reason you might want to write static_cast<T&&>(t) — to improve the compiler’s ability to diagnose bad code!