Does the C++ standard provide any guarantee on the ordering of the size in bytes of char, wchar_t, char16_t, char32_t? (any extract from the standard is welcome)
For example do I have the guarantee that:
sizeof(char) <= sizeof(wchar_t) <= sizeof(char16_t) <= sizeof(char32_t)
It's 1 == sizeof(char) <= sizeof(wchar_t) and 1 == sizeof(char) <= sizeof(char16_t) <= sizeof(char32_t).
5.3.3/1 Sizeof [expr.sizeof]
... sizeof(char), sizeof(signed char) and sizeof(unsigned char) are 1. ... [ Note: in particular, sizeof(bool), sizeof(char16_t), sizeof(char32_t), and sizeof(wchar_t) are implementation-defined.75 — end note ].
3.9.1/5 Fundamental types [basic.fundamental]
... Type wchar_t shall have the same size, signedness, and alignment requirements (3.11) as one of the other integral types, called its underlying type. Types char16_t and char32_t denote distinct types with the same size, signedness, and alignment as uint_least16_t and uint_least32_t, respectively, in <cstdint>, called the underlying types.
Update: I haven't found it in the standard. cppreference says for uint_leastN_t:
smallest unsigned integer type with width of at least 8, 16, 32 and 64 bits respectively
Note that sizeof(char)==1 does not mean that a char has 8 bits. See also C++ FAQ. cppreference says about CHAR_BIT:
number of bits in byte
1.7/1 The C ++ memory model [intro.memory]
The fundamental storage unit in the C ++ memory model is the byte. A byte is at least large enough to contain any member of the basic execution character set (2.3) ...
The types char16_t and char32_t are defined to be the same size as uint_least16_t and uint_least32_t respectively.
No such constraints exist on wchar_t, except that it must be at least as large as char (which, of course, is true for all data types in C and C++). There exist actual implementations with 1-byte, 2-byte (MSVC++), and 4-byte (GCC) wide characters.
