Will a `char` always-always-always have 8 bits?

Solution 1:

1. Yes, char and byte are pretty much the same. A byte is the smallest addressable amount of memory, and so is a char in C. char always has size 1.

   From the spec, section 3.6 byte:

   byte

   addressable unit of data storage large enough to hold any member of the basic character set of the execution environment

   And section 3.7.1 character:

   character

   single-byte character
   <C> bit representation that fits in a byte

2. A char has CHAR_BIT bits. It could be any number (well, 8 or greater according to the spec), but is definitely most often 8. There are real machines with 16- and 32-bit char types, though. CHAR_BIT is defined in limits.h.

   From the spec, section 5.2.4.2.1 Sizes of integer types <limits.h>:

   The values given below shall be replaced by constant expressions suitable for use in #if preprocessing directives. Moreover, except for CHAR_BIT and MB_LEN_MAX, the following shall be replaced by expressions that have the same type as would an expression that is an object of the corresponding type converted according to the integer promotions. Their implementation-defined values shall be equal or greater in magnitude (absolute value) to those shown, with the same sign.

   — number of bits for smallest object that is not a bit-field (byte)
       CHAR_BIT                               8

3. sizeof(char) == 1. Always.

   From the spec, section 6.5.3.4 The sizeof operator, paragraph 3:

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

4. You can allocate as much memory as your system will let you allocate - there's nothing in the standard that defines how much that might be. You could imagine, for example, a computer with a cloud-storage backed memory allocation system - your allocatable memory might be practically infinite.

   Here's the complete spec section 7.20.3.3 The malloc function:

   Synopsis

   1 #include <stdlib.h>
   void *malloc(size_t size);

   Description

   2 The malloc function allocates space for an object whose size is specified by size and whose value is indeterminate.

   Returns

   3 The malloc function returns either a null pointer or a pointer to the allocated space.

   That's the entirety of the specification, so there's not really any limit you can rely on.

Solution 2:

sizeof(char) is always 1 byte. A byte is not always one octet, however: The Texas Instruments TI C55x, for example, is a DSP with a 16-bit byte.

Solution 3:

sizeof(char) is defined to always be 1. From C99:

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

It is not however guaranteed to be 8 bits. In practice, on the vast majority of platforms out there, it will be, but no, you cannot technically count on that to always be the case (nor should it matter as you should be using sizeof anyway).

Solution 4:

Concretely, some architectures, especially in the DSP field have char:s larger than 8 bits. In practice, they sacrifice memory space for speed.

Solution 5:

Traditionally, a byte is not necessarily 8 bits, but merely a smallish region of memory, usually suitable for storing one character. The C Standard follows this usage, so the bytes used by malloc and sizeof can be more than 8 bits. [footnote] (The Standard does not allow them to be less.)

But sizeof(char) is always 1.

Memorizing the C FAQ is a career-enhancing move.