Are there machines, where sizeof(char) != 1, or at least CHAR_BIT > 8

cc99charsizeofstandards

Are there machines (or compilers), where sizeof(char) != 1?

Does C99 standard says that sizeof(char) on standard compliance implementation MUST be exactly 1? If it does, please, give me section number and citation.

Update:
If I have a machine (CPU), which can't address bytes (minimal read is 4 bytes, aligned), but only 4-s of bytes (uint32_t), can compiler for this machine define sizeof(char) to 4? sizeof(char) will be 1, but char will have 32 bits (CHAR_BIT macros)

Update2:
But sizeof result is NOT a BYTES ! it is the size of CHAR. And char can be 2 byte, or (may be) 7 bit?

Update3:
Ok. All machines have sizeof(char) == 1. But what machines have CHAR_BIT > 8 ?

Best Answer

It is always one in C99, section 6.5.3.4:

When applied to an operand that has type char, unsigned char, or signed char, (or a qualified version thereof) the result is 1.

Edit: not part of your question, but for interest from Harbison and Steele, 3rd ed. (pre c99) p. 148:

A storage unit is taken to be the amount of storage occupied by one character; the size of an object of type char is therefore 1.

Edit: In answer to your updated question, the following question and answer from Harbison and Steele is relevant (ibid, Ex. 4 of Ch. 6):

Is it allowable to have a C implementation in which type char can represent values ranging from -2,147,483,648 through 2,147,483,647? If so, what would be sizeof(char) under that implementation? What would be the smallest and largest ranges of type int?

Answer (ibid, p. 382):

It is permitted (if wasteful) for an implementation to use 32 bits to represent type char. Regardless of the implementation, the value of sizeof(char) is always 1.

While this does not specifically address a case where, say bytes are 8 bits and char are 4 of those bytes (actually impossible with the c99 definition, see below), the fact that sizeof(char) = 1 always is clear from the c99 standard and Harbison and Steele.

Edit: In fact (this is in response to your upd 2 question), as far as c99 is concerned sizeof(char) is in bytes, from section 6.5.3.4 again:

The sizeof operator yields the size (in bytes) of its operand

so combined with the quotation above, bytes of 8 bits and char as 4 of those bytes is impossible: for c99 a byte is the same as a char.

In answer to your mention of the possibility of a 7 bit char: this is not possible in c99. According to section 5.2.4.2.1 of the standard the minimum is 8:

Their implementation-defined values shall be equal or greater [my emphasis] in magnitude to those shown, with the same sign.

— number of bits for smallest object that is not a bit-field (byte)

 **CHAR_BIT 8**

— minimum value for an object of type signed char

**SCHAR_MIN -127//−(27−1)**

— maximum value for an object of type signed char

**SCHAR_MAX +127//27−1**

— maximum value for an object of type unsigned char

**UCHAR_MAX 255//28−1**

— minimum value for an object of type char

**CHAR_MIN**    see below

— maximum value for an object of type char

**CHAR_MAX**    see below

[...]

If the value of an object of type char is treated as a signed integer when used in an expression, the value of CHAR_MIN shall be the same as that of SCHAR_MIN and the value of CHAR_MAX shall be the same as that of SCHAR_MAX. Otherwise, the value of CHAR_MIN shall be 0 and the value of CHAR_MAX shall be the same as that of UCHAR_MAX. The value UCHAR_MAX shall equal 2^CHAR_BIT − 1.

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