没有#include wchar的GCC C中的前缀L“…”代表什么？

Question

That is, why does unsigned short var= L'ÿ' work, but unsigned short var[]= L"ÿ"; does not?

Answer 1

L'ÿ' is of type wchar_t , which can be implicitly converted into an unsigned short . L"ÿ" is of type wchar_t[2] , which cannot be implicitly converted into unsigned short[2] .

Answer 2

L is the prefix for wide character literals and wide-character string literals. This is part of the language and not a header. It's also not GCC-specific. They would be used like so:

wchar_t some_wchar = L'ÿ';
wchar_t *some_wstring = L"ÿ"; // or wchar_t some_wstring[] = L"ÿ";

You can do unsigned short something = L'ÿ'; because a conversion is defined from wchar_t to short. There is not such conversion defined between wchar_t* and short.

Answer 3

wchar_t is just a typedef to one of the standard integer types. The compiler implementor choses such a type that is large enough to hold all wide characters. If you don't include the header, this is still true and L'ß' is well defined, only that you as a programmer don't know what type it has.

Your initialization to an integer type works because there are rules to convert one into another. Assigning a wide character string (ie the address of the first address of a wide character array) to an integer pointer is only possible if you guess the integer type to which wchar_t corresponds correctly. There is no automatic conversion of pointers of different types, unless one of them is void* .

Answer 4

Chris has already given the correct answer, but I'd like to offer some thoughts on why you may have made the mistake to begin with. On Windows, wchar_t was defined as 16-bit way back in the early days of Unicode where it was intended to be a 16-bit character set. Unfortunately this turned out to be a bad decision (it makes it impossible for the C compiler to support non-BMP Unicode characters in a way that conforms to the C standard), but they were stuck with it.

Unix systems from the beginning have used 32-bit wchar_t , which of course means short * and wchar_t * are incompatible pointer types.

Answer 5

For what I remember of C

• 'Y' or whatever is a char and you can cast it into an int and therefore convert it into a L,
• "y" is a string constant and you can't translate it into a integer value