The following piece of code compiles under clang++ 3.7.0 but is denied by g++ 5.3.1. Both have -std=c++14
option. Which compiler is correct? Anyone knows where in the standard talks about this? Thanks.
#include <stdexcept>
using namespace std;
constexpr int f(int n) {
if (n <= 0) throw runtime_error("");
return 1;
}
int main() {
char k[f(1)];
}
Output
[hidden] g++ -std=c++14 c.cpp
c.cpp: In function ‘constexpr int f(int)’:
c.cpp:7:1: error: expression ‘<throw-expression>’ is not a constant-expression
}
^
[hidden] clang++ -std=c++14 c.cpp
[hidden]
[hidden] g++ -v
Using built-in specs.
COLLECT_GCC=/usr/bin/g++
COLLECT_LTO_WRAPPER=/usr/libexec/gcc/x86_64-redhat-linux/5.3.1/lto-wrapper
Target: x86_64-redhat-linux
Configured with: ../configure --enable-bootstrap --enable-languages=c,c++,objc,obj-c++,fortran,ada,go,lto --prefix=/usr --mandir=/usr/share/man --infodir=/usr/share/info --with-bugurl=http://bugzilla.redhat.com/bugzilla --enable-shared --enable-threads=posix --enable-checking=release --enable-multilib --with-system-zlib --enable-__cxa_atexit --disable-libunwind-exceptions --enable-gnu-unique-object --enable-linker-build-id --with-linker-hash-style=gnu --enable-plugin --enable-initfini-array --disable-libgcj --with-isl --enable-libmpx --enable-gnu-indirect-function --with-tune=generic --with-arch_32=i686 --build=x86_64-redhat-linux
Thread model: posix
gcc version 5.3.1 20151207 (Red Hat 5.3.1-2) (GCC)
[hidden]
[hidden] clang++ -v
clang version 3.7.0 (http://llvm.org/git/clang.git 2ddd3734f32e39e793550b282d44fd71736f8d21)
Target: x86_64-unknown-linux-gnu
Thread model: posix
Found candidate GCC installation: /usr/lib/gcc/x86_64-redhat-linux/3.4.6
Found candidate GCC installation: /usr/lib/gcc/x86_64-redhat-linux/5.3.1
Selected GCC installation: /usr/lib/gcc/x86_64-redhat-linux/5.3.1
Candidate multilib: .;@m64
Candidate multilib: 32;@m32
Selected multilib: .;@m64
clang is correct, note the HEAD revision of gcc accepts also accepts this code. This is a well-formed constexpr function, as long as there is value for the argument(s) that allows the function to be evaluated as a core constant expression. In your case 1
is such a value.
This is covered in the draft C++14 standard section 7.1.5
The constexpr specifier [dcl.constexpr] which tells us what is allowed in a constexpr function:
The definition of a constexpr function shall satisfy the following constraints:
it shall not be virtual (10.3);
its return type shall be a literal type;
each of its parameter types shall be a literal type;
its function-body shall be = delete, = default, or a compound-statement that does not contain
no restriction on throw
and it also says (emphasis mine):
For a non-template, non-defaulted constexpr function or a non-template, non-defaulted, non-inheriting
constexpr constructor, if no argument values exist such that an invocation of the function or constructor
could be an evaluated subexpression of a core constant expression (5.19), the program is ill-formed; no
diagnostic required.
and below this paragraph we have the following example, similar to yours:
constexpr int f(bool b)
{ return b ? throw 0 : 0; } // OK
constexpr int f() { return f(true); } // ill-formed, no diagnostic required
throw
is not allowed in a core constant expression, which is covered in section 5.19
[expr.const] paragraph 2
which says:
A conditional-expression e is a core constant expression unless the evaluation of e, following the rules of the
abstract machine (1.9), would evaluate one of the following expressions
and includes the following bullet:
- a throw-expression (15.1).
and so f
would not be usable in a core constant expression when n <= 0
.
Update
As TemplateRex points out, there are two gcc bugs reports for this:
- Never executed "throw" in constexpr function fails to compile
- C++14] throw-expression is not a valid constant-expression
TemplateRex also notes the fixes are not applied to to 5.3.0
and are only in trunk. No, work arounds are provided.
As shown by Shafik Yaghmour it's a gcc bug, which will hopefully be fixed in v6.
Until then, you can revert to the c++11
constexpr style:
constexpr auto foo(int n) -> int
{
return n <= 0 ? throw runtime_error("") : 1;
}
However there is a better workaround, still retaining all of the c++14
constexpr extensions:
// or maybe name it
// throw_if_zero_or_less
constexpr auto foo_check_throw(int n) -> void
{
n <= 0 ? throw std::runtime_error("") : 0;
}
constexpr auto foo(int n) -> int
{
foo_check_throw(n);
// C++14 extensions for constexpr work:
if (n % 2)
return 1;
return 2;
}