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| 2002/12/3 [Computer/SW/Compilers] UID:26695 Activity:high |
12/2 What does the following struct mean?
struct { instr_t :18, _i:1, :13; } u_i;
#define i_i u_i._i
#define getI(x) (((((uint32)(x)) << 18)) >> 31)
#define putI(x) ((((uint32)(x)) << 31) >> 18)
\_ it means you should read up on bit fields.
\_ Hi paolo!
\_ Ok, I give up. Why is this funny?
\_ Especially because paolo doesn't log in to soda.
\_ Well, not in the past week perhaps, but:
pst ttyEm 63.73.217.160 Fri Nov 22 12:32 - 12:33 (00:01)
pst ttyEH 63.73.217.160 Wed Nov 20 14:22 - 14:27 (00:05)
pst ttyCi 63.73.217.160 Tue Nov 19 20:04 - 21:35 (01:31)
pst ttyBS 63.73.217.160 Mon Nov 18 10:42 - 10:46 (00:03)
\_ yes yes that's nice. Why is this funny?
\_ The struct is composed of three members, one of
type instr_t that is 18 bits, one of type _i that
is 1 bit, and I think a signed int of 13 bits.
This might help:
http://www.cs.cf.ac.uk/Dave/C/node13.html
\_ they're commas, not semicolons. _i is the field identifier,
not the type. All three fields are of type instr_t, with
the first 18 and the last 13 bits being unnamed.
\_ As far as the macros are concerned, does the first
one return the value of the last 14 bits * 62?
I have no idea what the second one does since it
seems to deal with only the last bit.
\_ I think even the order of the bits in bitfields is up to
the compiler. So getI() only works for some compilers.
Why does he need getI() anyway? "foo = u_i._i;" will do.
--- yuen
\_ I would guess that it's for cases where that 32-bit
value got read as an integer rather than as the struct.
of course, this could have been easily solved with a
union.
\_ wha--? It's not that hard to understand. you've got a
32-bit value where you're only concerned with is the 19th
bit from the left. getI gets that bit. It shifts left 18
bits then shifts right by 31 bits (hint: there's only one
bit left). putI does the opposite; it takes a value for
that bit and places it in the right spot in the returned
value.
\_ But the problem is that you don't know whether _i is
the 19th bit from the left or from the right. You don't
know whether the 18-bits or the 13-bits are the more
significant bits. It's up to the compiler. -- yuen
\_ http://www.cs.cf.ac.uk/Dave/C/node13.html
Read Portability. They prefer shifting for
portability.
\_ shifting is preferable to the bitfields, but
I think masks would have been better than
the extra shifts and would be just as portable
(if not more so, since they wouldn't necessarily
be dependent on operating on 32-bit values.)
\_ yes. sorry, I didn't mean to imply that it wasn't
compiler-dependent. there's also the assumption
that there's an appropriate underlying type for
the uint32 typedef. my assumption is that the
code is for some specified compiler and
architecture. |
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| www.cs.cf.ac.uk/Dave/C/node13.html Exercises Low Level Operators and Bit Fields We have seen how pointers give us control over low level memory operations. NOTE: The combination of pointers and bit-level operators makes C useful for many low level applications and can almost replace assembly code. The shift operators perform appropriate shift by operator on the right to the operator on the left. For example: x << 2 shifts the bits in x by 2 places to the left. So: if x = 00000010 (binary) or 2 (decimal) then: $x \gt\gt= 2 \Rightarrow x = 00000000$ or 0 (decimal) Also: if x = 00000010 (binary) or 2 (decimal) $x <<= 2 \Rightarrow x = 00001000$ or 8 (decimal) Therefore a shift left is equivalent to a multiplication by 2. Similarly a shift right is equal to division by 2 NOTE: Shifting is much faster than actual multiplication or division by 2. So if you want fast multiplications or division by 2 use shifts. To illustrate many points of bitwise operators let us write a function, Bitcount, that counts bits set to 1 in an 8 bit number (unsigned char) passed as an argument to the function. This is especially useful when memory or data storage is at a premium. Typical examples: * Packing several objects into a machine word. C lets us do this in a structure definition by putting :bit length after the variable. C automatically packs the above bit fields as compactly as possible, provided that the maximum length of the field is less than or equal to the integer word length of the computer. If this is not the case then some compilers may allow memory overlap for the fields whilst other would store the next field in the next word (see comments on bit fiels portability below). So type cannot take values larger than 15 (4 bits long). |