What happens with adjacent string literal concatenation when there is a modifier(L, u8, etc.)
It is valid in C and C++ to break a string literal because the preprocessor or the compiler will concatenate adjacent string literals.
const char *zStr = "a" "b"; // valid
What happens when string literals are prefixed with L
(wide characters), u
(UTF-16), U
(UTF-32), u8
(UTF-8), and raw string l开发者_运维知识库iterals (R"foo(this is a "raw string literal" with double quotes)foo"
)?
For example, is the following allowed:
const wchar_t *zStr = L"a" "b"; // valid?
In C++0x your example is valid according to [lex.string]/p13:
... If one string literal has no encoding-prefix, it is treated as a string literal of the same encoding-prefix as the other operand. ...
In C++03 this same section said that this code had undefined behavior:
... If a narrow string literal token is adjacent to a wide string literal token, the behavior is undefined. ...
Yes, that particular example is allowed by C++0x. Any combination of prefixless and L-prefixed literals will be treated as though all are L-prefixed.
EDIT: Citation -- N3242 (current C++0x working draft) §2.14.5/13:
In translation phase 6 (2.2), adjacent string literals are concatenated. If both string literals have the same encoding-prefix, the resulting concatenated string literal has that encoding-prefix. If one string literal has no encoding-prefix, it is treated as a string literal of the same encoding-prefix as the other operand.
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