Re: size of a variable

Greg Martin <>
Wed, 18 Feb 2015 14:53:32 GMT
On 2015-02-18, David Brown <> wrote:

On 17/02/15 23:51, Barry Schwarz wrote:

On Tue, 17 Feb 2015 20:36:40 +0100, David Brown
<> wrote:

On 17/02/15 19:51, Barry Schwarz wrote:

On Tue, 17 Feb 2015 14:36:20 +0100, David Brown
<> wrote:

As far as I can understand the details of the standard, there is no
(standards compliant) way to have a uint8_t type unless chars are 8-bit
on the given platform. In theory, a compiler could provide another
independent 8-bit unsigned type for uint8_t that is different from
"unsigned char" - but as it must be an integral multiple of "char" in
size, and char must be at least 8-bit, then char has to be /exactly/
8-bit if "uint8_t" exists on a platform.

While the standard talks about uintN_t with N = 8, 16, 32, and 64, it
does not limit the compiler to just those values. The compiler is
free to provide a uint9_t type. That such a type would require two
bytes on an 8-bit system with some unused bits is just a detail the
code generator would need to deal with. (Note that the unused bits
are not padding, which is prohibited, since they are not part of the

A compiler can provide any uintN_t and intN_t types, as long as those
types follow the rules - it can provide a uint9_t as long as the type
has the same sort of wrap on overflow semantics as other unsigned
integer types, and as long as there is a corresponding int9_t with two's
complement arithmetic.

But that does not change the fact that if the compiler provides a valid
uint8_t, then CHAR_BIT must be 8 (even though weird things are allowed,
if unlikely, such as "signed char" being 8-bit one's complement while
"int8_t" is 8-bit two's complement).

Similarly, what in the standard prevents CHAR_BIT from being 16 and
the compiler using any 8 bits in the byte to represent the value of a

The rules of C - a "char" is the smallest addressable unit of memory.

That doesn't stop sub-byte bit fields.

Bit-fields cannot be accessed except as part of their containing struct.
 In particular, you cannot take the address of a bitfield (?? of
N1570, if you want chapter and verse).

Ah, but if it walks like a duck and talks like a duck is it a duck?
note: I can't think of a single reason why you'd do this.

#include <iostream>

class uint8 {
    friend std::ostream& operator<< (std::ostream& out, uint8& ui);
    struct {
        unsigned char d : 8;
    uint8 (const unsigned int i) : d(i) {}
    uint8 () : d(0) {}
    uint8 (const uint8& o) : d(o.d) {}
    uint8 (uint8&& o) : d(o.d) { o.d = 0; }
    ~uint8 () {}

    uint8& operator= (const uint8& o) {
        d = o.d;
        return *this;

    uint8& operator= (uint8&& o) {
        d = o.d;
        o.d = 0;
        return *this;

    uint8& operator= (const unsigned int i) {
        d = i;
        return *this;

    bool operator== (const unsigned int i) {
        return d == i;

    bool operator== (const uint8& o) {
        return d == o.d;


std::ostream& operator<< (std::ostream& out, uint8& ui) {
    out << (unsigned int) ui.d;
    return out;

int main () {
    uint8 u8 = 255;

    std::cout << u8 << "\n";
    std::cout << (u8 == 255) << "\n";

    uint8 u8_2 = u8 = 0xFFFE;

    std::cout << u8 << "\n";
    std::cout << (u8 == 255) << "\n";
    std::cout << (u8 == u8_2) << "\n";

    std::cout << sizeof u8 << "\n";

    uint8* pu8 = &u8;

    std::cout << *pu8 << "\n";

    return 0;


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