Consider this ridiculously icky template code:

template<class T> int f(int) { return 1; } // A template int f<int>(int); // B template<class T> int f(T) { return 2; } // C template int f(int); // D int main() { return f(0) + f<int>(0); }

According to GCC, Clang, ICC, and the paper standard, this program is valid C++, and it is supposed to return 3 . (According to MSVC, this program is awful and shouldn’t even link.)

First of all, I should mention that I gave a two-part talk at CppCon 2016 titled “Template Normal Programming.” Much of the stuff we’re about to dive into is covered in that talk. I highly recommend watching it… if I do say so myself.

Here we have an overload set consisting of two function templates, both named f . There’s template<class T> int f(int) (on line A) and there’s template<class T> int f(T) (on line C).

Each template has one explicit instantiation definition for T=int . The first instantiation definition (on line B) says, “I’m an instantiation of f with the first template parameter set to int .” There’s only one f visible at that point, so this is unambiguously an instantiation of template A.

The second instantiation definition (on line D) says “I’m an instantiation of f ,” but it doesn’t say which of our two f templates it’s an instantiation of, and also it doesn’t provide any template arguments. This is a little bit sneaky! The compiler has to do template argument deduction to figure out what the template arguments to f should be. First it looks at template A; but template A’s parameter T is not deducible, so the compiler discards that possibility. Then it considers template C. Template C’s parameter T is deducible from the first function parameter: we’re saying that the first parameter has type int , the primary template says that it has type T , therefore T=int and we’ve got ourselves a candidate! Since this is the only candidate (remember, template A was thrown out), the declaration is unambiguous: line D is an instantiation of template C.

So now we have explicit instantiations of two functions, both named f , both with a single template argument T=int , both taking a single function parameter of type int . Crazy!

In main , we call f(0) . The compiler does overload resolution and template argument deduction. This can’t be a call to template A, because template A’s parameter T is not deducible. So it must be a call to template C — that is, to the specialization instantiated on line D.

Then main calls f<int>(0) . Again the compiler does overload resolution. This might be a call to template A or to template C. But template A is more specialized than template C, and so it is a better match. So this is resolved into a call to template A — that is, to the specialization instantiated on line B.

So main returns 2 + 1, which is 3. Q.E.D.!

So why can’t MSVC compile this code? Well, it’s because of what I said three paragraphs ago: we have here two functions, both named f , both with a single template argument T=int , both taking a single function parameter of type int . So MSVC quite reasonably produces the same name-mangled symbol for both of these functions. Specifically, that mangling is

??$f@H@@YAHH@Z

(In MSVC mangling, H means int ; so f@H means f<int> . @@YAH means “function returning int . And the last H is the function parameter of type int . I don’t know who invented the MSVC mangling scheme, but I’m going to guess Jon Arbuckle.)

So we end up with two definitions for that symbol — one returning 1 and the other returning 2 — in the same assembly file! We get an error from the assembler, or perhaps from the linker.

MSVC’s logic seemed very reasonable. So why is it that GCC and Clang can compile this sneaky code? Well, they both use the Itanium mangling scheme, and Itanium says that we should mangle dependent function parameter types as they appear in the source code. In Itanium’s scheme, the specialization on line B mangles as

_Z1fIiEii

Here fIiE means f<int> , the next i means “returning int ,” and the last i is the function parameter of type int . But the specialization on line D mangles differently!

_Z1fIiEiT_

Here the function parameter type is not mangled as i ; it’s mangled as T_ , which means “the first template parameter type.” In this particular specialization, the first template parameter type happens to be int ; but we don’t mangle it as i , we mangle it as T_ . This rule is crafted precisely to permit valid-but-sneaky C++ code such as this example.

One unfortunate side effect of mangling dependent types is that we suddenly have to invent a mangling for all possible ways a type can depend on a template parameter. For example, Itanium will look at

template<class T> void f(int (*)[sizeof(T() * 1 + 2 ^ 3)]) {} template void f<int>(int (*)[4]);

and produce the mangling

_Z1fIiEvPAszeoplmlcvT__ELi1ELi2ELi3E_i

That is, the function parameter has type P ointer to A rray of size s i z eof e xclusive- o r of ( pl us of ( m u l tiply of ( c onstruct-from- v oid of T ) and the integer 1 ) and the integer 2 ) and the integer 3 ).

Or again, Itanium distinguishes between these two templates whereas MSVC does not:

template<int> struct A {}; template<int I> A<I+0> f() { return {}; } template A<2> f<2>(); // _Z1fILi2EE1AIXplT_Li0EEEv template<int I> A<I> f() { return {}; } template A<2> f(); // _Z1fILi2EE1AIXT_EEv

Another unfortunate side effect is that when a function parameter’s type is itself the result of a template instantiation, we can’t just encode the final type; we have to encode how we got there. For example (Godbolt):

template<class> struct A { using type = int; }; template<class> using B = int; template<class T> T f(typename A<T>::type) { return 1; } template int f(int); template<class T> T f(B<T>) { return 2; } template int f(int);

Itanium mangles the first f as _Z1fIiET_N1AIS0_E4typeE and the second f as _Z1fIiET_i . Alias templates are “seen through” by the template machinery: B<T> is really just an alias for int in all respects, and therefore it is not considered to be template-dependent and can be mangled as simply i . On the other hand, typename A<T>::type depends on T and must be mangled as NAIS0_E4typeE . (MSVC mangles these two functions identically.)

This feature of the Itanium ABI is what sank Alisdair Meredith’s P0181 “Ordered By Default” (February 2016), a.k.a. std::default_order . He proposed essentially that the default template argument for std::set ’s comparator should be changed from std::less<T> to “ std::less<T> if it exists; otherwise std::default_strong_comparator<T> , which can be customized for types that want to be storeable-in-sets without providing operator< .” In modern terms:

template<class T> using void_if_lessable = decltype( (declval<T>() < declval<T>()), void() ); template<class T, class = void> struct default_strong_comparator { struct type { bool operator()(const T& a, const T& b) const { return std::strong_order(a, b) < 0; } }; }; template<class T> struct default_strong_comparator<T, void_if_lessable<T>> { using type = std::less<T>; }; template<class T, class C = typename default_set_comparator<T>::type, class A = std::allocator<T>> class set { // ... };

At first glance, this seems like a ABI-non-breaking, backward-compatible change. set<int> continues to mean set<int, std::less<int>, std::allocator<int>> just as it did before (because default_set_comparator<int>::type is a type alias for std::less<int> ). We check the mangling of

void f(set<int>);

and see that indeed it mangles to the same thing as before (Godbolt).

…But then the mangling of template-dependent function parameter types comes and bites us!

template<class T> int f(set<int, typename default_strong_comparator<T>::type>) { return 1; } // A template int f<int>(set<int>); // B template<class T> int f(set<int, std::less<T>>) { return 2; } // C template int f(set<int>); // D

Here are two distinct templates f . Template C is more specialized than template A. Instantiations B and D have different Itanium manglings:

_Z1fIiEi3setIT_N25default_strong_comparatorIS1_vE4typeESaIS1_EE // B _Z1fIiEi3setIT_St4lessIS1_ESaIS1_EE // D

Now consider this function template g (Godbolt):

template<class T> int g(set<T>) { return 3; } // E template int g(set<int>); // F

If set<T> means set<T, std::less<T>> , then the specialization on line F should be mangled as _Z1gIiEi3setIT_St4lessIS1_ESaIS1_EE . But if set<T> means set<T, default_strong_comparator<T>::type> , then the specialization on line F should be mangled as _Z1gIiEi3setIT_N25default_strong_comparatorIS1_vE4typeESaIS1_EE .

Thus, changing the default template argument of std::set can actually cause the name-mangling of a function to change! And g is a very plausible function template to write, too; this change wouldn’t go unnoticed in real codebases.

This subtle mangling issue sank P0181 default_order .

National Body Comment “FI 18” on the C++17 CD (N4664, March 2017) read in its entirety:

Comments: It was thought that using default_order as the default comparison for maps and sets was not abi-breaking but this is apparently not the case. Proposed change: Revert the change to the default comparison of maps and sets. Observations of the secretariat: Accepted

This comment was turned verbatim into LWG issue 2863 (opened February 2017, closed March 2017). I don’t think a code snippet specifically demonstrating the problem was ever published in any public venue.

So I wrote this blog post.