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Programs to communicate with low-level hardware interfaces need to define bit fields laid out to match the hardware data. This section explains how to do that.
Consecutive bit fields are packed together, but each bit field must fit within a single object of its specified type. In this example,
unsigned short a : 3, b : 3, c : 3, d : 3, e : 3;
all five fields fit consecutively into one two-byte short
.
They need 15 bits, and one short
provides 16. By contrast,
unsigned char a : 3, b : 3, c : 3, d : 3, e : 3;
needs three bytes. It fits a
and b
into one
char
, but c
won’t fit in that char
(they would
add up to 9 bits). So c
and d
go into a second
char
, leaving a gap of two bits between b
and c
.
Then e
needs a third char
. By contrast,
unsigned char a : 3, b : 3; unsigned int c : 3; unsigned char d : 3, e : 3;
needs only two bytes: the type unsigned int
allows c
to straddle bytes that are in the same word.
You can leave a gap of a specified number of bits by defining a
nameless bit field. This looks like type : nbits;
.
It is allocated space in the structure just as a named bit field would
be allocated.
You can force the following bit field to advance to the following
aligned memory object with type : 0;
.
Both of these constructs can syntactically share type with ordinary bit fields. This example illustrates both:
unsigned int a : 5, : 3, b : 5, : 0, c : 5, : 3, d : 5;
It puts a
and b
into one int
, with a 3-bit gap
between them. Then : 0
advances to the next int
,
so c
and d
fit into that one.
These rules for packing bit fields apply to most target platforms, including all the usual real computers. A few embedded controllers have special layout rules.
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