Are there machines, where sizeof(char) != 1, or at least CHAR
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 ?
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.
There are no machines where sizeof(char) is 4. It's always 1 byte. That byte might contain 32 bits, but as far as the C compiler is concerned, it's one byte. For more details, I'm actually going to point you at the C++ FAQ 26.6. That link covers it pretty well and I'm fairly certain C++ got all of those rules from C. You can also look at comp.lang.c FAQ 8.10 for characters larger than 8 bits.
Upd2: But sizeof result is NOT a BYTES ! it is the size of CHAR. And char can be 2 byte, or (may be) 7 bit?
Yes, it is bytes. Let me say it again. sizeof(char) is 1 byte according to the C compiler. What people colloquially call a byte (8 bits) is not necessarily the same as what the C compiler calls a byte. The number of bits in a C byte varies depending on your machine architecture. It's also guaranteed to be at least 8.
PDP-10 and PDP-11 was.
Update: there like no C99 compilers for PDP-10.
Some models of Analog Devices 32-bit SHARC DSP have CHAR_BIT=32, and Texas Instruments DSP from TMS32F28xx have CHAR_BIT=16, reportedly.
Update: There is GCC 3.2 for PDP-10 with CHAR_BIT=9 (check include/limits.h in that archive).
链接地址: http://www.djcxy.com/p/79246.html上一篇: 它是堆栈还是堆？