l41.1a

<B>Digital的Unix操作系统VAX 4.2源码</B>

#print
The problem is to produce a function
	bitct(x)
which examines the bits in x, returning a count of
the number of 1-bits.  There are various ways of doing
this job: here are two.
(1) a sane way.  Shift the word x right 16 times (you are
on UNIX) and check the rightmost bit each time, counting
the number of times it is '1'.
(2) a machine-independent (sort of) way.  The logical
bitwise AND of x and x-1 contains one fewer one bit than x itself.
Loop anding x and x-1 until you get zero.
Program either algorithm.  Compile and test it.  Leave it on
a file bitct.c and type "ready".
#once #create tzaqc.c
main()
{
	int x;
	x=23069;
	if (bitct(x) != goodct(x)) 
		return(1);
	x=0;
	if (bitct(x) != goodct(x)) 
		return(1);
	x=16384;
	if (bitct(x) != goodct(x)) 
		return(1);
	x = -1;
	if (bitct(x) != goodct(x)) 
		return(1);
	x= -200;
	if (bitct(x) != goodct(x)) 
		return(1);
	return(0);
}
goodct(x)
{
	int k, i;
	for(k=i=0; i<16; i++)
		{
		k =+ (x&1);
		x= x>>1;
		}
	return(k);
}
#user
cc tzaqc.c bitct.o
a.out
#succeed
/*  a possible solution */
bitct(x)
{
	int k, i;

	for(i=k=0; i<16; i++) {
		if (x&1)
			k++;
		x >>= 1;
	}
	return(k);
}
/* by the way, if you really care about
this problem a table lookup by whole bytes
is faster */
#log
#next
42.1a 10