The true nature of "bool"?

[Frederick Gotham]

Does "bool" behave _exactly_ like any other integer type, except for when it
comes to integer promotion? Consider the following C function whose purpose
it is to calculate the quantity of equal coordinates:

struct Coord3D {
int a,b,c;
};

typedef struct Coord3D Coord3D; /* To make it work like C++ */

unsigned QuantSame(Coord3D const *const x,Coord3D const *const y)
{
return (x->a == y->a)+(x->b == y->b)+(x->c == y->c);
}

In C, the type of the result of the equality operation would be "int", and
these int's can be added together.

In C++, however, the type of the result of the equality operation is "bool".
Am I right in thinking, that, when arithmetical operations are performed upon
a bool, that it undergoes integer promotion just like any other integer type?
(except that it _must_ become either 0 or 1).

Are there any other places in which a bool doesn't behave like it's just
another built-in integer type?

--

Frederick Gotham

[Pete Becker]

Frederick Gotham wrote:

Does "bool" behave _exactly_ like any other integer type, except for when it
comes to integer promotion? Consider the following C function whose purpose
it is to calculate the quantity of equal coordinates:

struct Coord3D {
int a,b,c;
};

typedef struct Coord3D Coord3D; /* To make it work like C++ */

unsigned QuantSame(Coord3D const *const x,Coord3D const *const y)
{
return (x->a == y->a)+(x->b == y->b)+(x->c == y->c);
}

In C, the type of the result of the equality operation would be "int", and
these int's can be added together.

In C++, however, the type of the result of the equality operation is "bool".
Am I right in thinking, that, when arithmetical operations are performed upon
a bool, that it undergoes integer promotion just like any other integer type?
(except that it _must_ become either 0 or 1).

Operands of type bool are promoted to int when used in arithmetic
expressions. false becomes 0 and true becomes 1.

Are there any other places in which a bool doesn't behave like it's just
another built-in integer type?

++ and -- are idiosyncratic.

--

-- Pete

Author of "The Standard C++ Library Extensions: a Tutorial and
Reference." For more information about this book, see
www.petebecker.com/tr1book.

[Jacek Dziedzic]

Frederick Gotham wrote:

Does "bool" behave _exactly_ like any other integer type, except for when it
comes to integer promotion? Consider the following C function whose purpose
it is to calculate the quantity of equal coordinates:

I _think_ std::vector<boolbehaves in a weird manner.

HTH,
- J.

[Steve Pope]

Jacek Dziedzic <jacek@no_spam.tygrys.no_spam.netwrote:

I _think_ std::vector<boolbehaves in a weird manner.

Yes, it is bit-packed. (Not sure if this is specified or just
the common implementation.) You almost never want to use it.

The comment about ++ and -- operators not behaving as for integer
types is also correct.

Steve

[Marcus Kwok]

Steve Pope <sp*****@speedymail.orgwrote:

Jacek Dziedzic <jacek@no_spam.tygrys.no_spam.netwrote:

> I _think_ std::vector<boolbehaves in a weird manner.

Yes, it is bit-packed. (Not sure if this is specified or just
the common implementation.) You almost never want to use it.

Also, many of the operations in std::vector<boolreturn a proxy object
instead of a reference to bool; see:

http://www.gotw.ca/publications/mill09.htm
http://www.gotw.ca/publications/N1185.pdf
http://www.gotw.ca/publications/N1211.pdf

--
Marcus Kwok
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