It's disallowed by the language:
[C++11: 7.1.6.4]:
1 The auto
type-specifier signifies that the type of a variable being declared shall be deduced from its initializer or that a function declarator shall include a trailing-return-type.
2 The auto
type-specifier may appear with a function declarator with a trailing-return-type (8.3.5) in any context where such a declarator is valid.
3 Otherwise, the type of the variable is deduced from its initializer. The name of the variable being declared shall not appear in the initializer expression. This use of auto
is allowed when declaring variables in a block (6.3), in namespace scope (3.3.6), and in a for-init-statement (6.5.3). auto
shall appear as one of the decl-specifiers in the decl-specifier-seq and the decl-specifier-seq shall be followed by one or more init-declarators, each of which shall have a non-empty initializer.
4 The auto
type-specifier can also be used in declaring a variable in the condition of a selection statement (6.4) or an iteration statement (6.5), in the type-specifier-seq in the new-type-id or type-id of a new-expression (5.3.4), in a for-range-declaration, and in declaring a static data member with a brace-or-equal-initializer that appears within the member-specification of a class definition (9.4.2).
5 A program that uses auto
in a context not explicitly allowed in this section is ill-formed.
It's hard to prove a negative, but there's simply no explicit rule in the standard to allow auto
in your case.
However, the same rules mean that the following is valid:
struct Foo {
static constexpr auto constant_string = "foo";
};
int main() {}
(Note that the type of Foo::constant_string
is char const* const
rather than, say, char const[3]
; this is an effect of using auto
.)
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