A compound literal does not yield an lvalue

[llvmbot]

Bugzilla Link	25424
Resolution	INVALID
Resolved on	Nov 05, 2015 18:16
Version	trunk
OS	Linux
Reporter	LLVM Bugzilla Contributor
CC	@zygoloid

Extended Description

Given

template <class A, class B> struct IsSame { static constexpr bool value = false; constexpr operator bool () const { return false; } };
template struct IsSame<A,A> { static constexpr bool value = true; constexpr operator bool () const { return true; } };
static_assert(IsSame<decltype(((const char []){ "string"})), const char (&) [7]>::value, "cxl1");

the test-case should pass, since C99/11 states in 6.5.2.5p4 that compound literals return lvalues. Clang preserves the prvalue-ness of the string literal.

Invocation:
clang++ -std=c++11 aa.cc

Clang version:
gentoo-x64 ~ # clang++ -v
clang version 3.8.0 (6215dbd9a8bf16e34b9b7a0cfda3d17e629a2d8d) (331a8c8)
Target: x86_64-pc-linux-gnu
Thread model: posix
InstalledDir: /usr/bin
Selected GCC installation: /usr/lib/gcc/x86_64-pc-linux-gnu/5.1.0

[llvmbot]

sorry, invocation is clang++ -std=c++11 -c
Alternately, adding int main(){} to the testcase should suffice.

Otherwise, there's the linker error about missing main. :)

[zygoloid]

Compound literals are not part of C++. Clang allows them as an extension, with the following semantics:

(T){init}

means the same thing as

T{init}

Thus the compound literal, in this case, a prvalue of type 'const char[7]'.