引子
前几天 QQ 群里一位好友提出来一个问题: "整型（有正有负）除以2的指数结果四舍五入, 应该如何优化呢", 当时做了答, 发表在这里, 并做些许修订, 希望对大家有用.
符号约记
下取整函数的定义: (leftlfloor x ight floor = max left{ {z in mathbb{Z}:z leqslant x} ight})
上取整函数的定义: (leftlceil x ight ceil = min left{ {z in mathbb{Z}:z geqslant x} ight})
四舍五入取整函数的定义: (left[kern-0.15emleft[ x ight]kern-0.15em ight] = left{ {egin{array}{*{20}{c}} {leftlfloor {x + 0.5} ight floor }&{x > 0} \ {leftlceil {x - 0.5} ight ceil }&{x leqslant 0} end{array}} ight.)
正文
问题: 整型（有正有负）除以2的指数结果四舍五入， 应该如何优化呢?
问题转化为: 将 (left[kern-0.15emleft[ {frac{m}{2^k}} ight]kern-0.15em ight]) 转化为向下取整 (以便能够使用移位运算) 的形式，其中 (m) 为整数, (k) 为非负整数.
答: 因为 (left[kern-0.15emleft[ {frac{m}{n}} ight]kern-0.15em ight] = left{ {egin{array}{*{20}{c}} {leftlfloor {frac{m}{n} + frac{1}{2}} ight floor = leftlfloor {frac{1}{n}left( {m + leftlfloor {frac{n}{2}} ight floor } ight)} ight floor }&{mn geqslant 0} \ {leftlceil {frac{m}{n} - frac{1}{2}} ight ceil = leftlceil {frac{1}{n}left( {m - leftlfloor {frac{n}{2}} ight floor } ight)} ight ceil }&{mn < 0} end{array}} ight.), 其中 (m, n) 均为整数, (n eq 0).
特别地, 设 (n = 2^k) ( (k) 为非负整数) 时, 当 (k=0) 时, (left[kern-0.15emleft[ m
ight]kern-0.15em
ight] = m);
当 (k>0) 时, (left[kern-0.15emleft[ {frac{m}{{{2^k}}}}
ight]kern-0.15em
ight] = left{ {egin{array}{*{20}{c}}
{leftlfloor {frac{m}{{{2^k}}} + frac{1}{2}}
ight
floor = leftlfloor {frac{1}{{{2^k}}}left( {m + leftlfloor {frac{{{2^k}}}{2}}
ight
floor }
ight)}
ight
floor = leftlfloor {frac{{m + {2^{k - 1}}}}{{{2^k}}}}
ight
floor }&{m geqslant 0} \
{leftlceil {frac{m}{{{2^k}}} - frac{1}{2}}
ight
ceil = leftlceil {frac{1}{{{2^k}}}left( {m - leftlfloor {frac{{{2^k}}}{2}}
ight
floor }
ight)}
ight
ceil = leftlceil {frac{{m - {2^{k - 1}}}}{{{2^k}}}}
ight
ceil }&{m < 0}
end{array}}
ight.).
又因为 (leftlceil {frac{m}{n}} ight ceil = left{ {egin{array}{*{20}{c}} {leftlfloor {frac{{m + n - 1}}{n}} ight floor = leftlfloor {frac{{m - 1}}{n}} ight floor + 1}&{n > 0} \ {leftlfloor {frac{{m + n + 1}}{n}} ight floor = leftlfloor {frac{{m + 1}}{n}} ight floor + 1}&{n < 0} end{array}} ight.),
所以 (leftlceil {frac{{m - {2^{k - 1}}}}{{{2^k}}}} ight ceil = leftlfloor {frac{{m - {2^{k - 1}} + {2^k} - 1}}{{{2^k}}}} ight floor = leftlfloor {frac{{m + {2^{k - 1}} - 1}}{{{2^k}}}} ight floor).
综上, (left[kern-0.15emleft[ {frac{m}{{{2^k}}}} ight]kern-0.15em ight] = left{ {egin{array}{*{20}{c}} m&{k = 0} \ {leftlfloor {frac{{m + {2^{k - 1}}}}{{{2^k}}}} ight floor }&{k > 0,m geqslant 0} \ {leftlfloor {frac{{m + {2^{k - 1}} - 1}}{{{2^k}}}} ight floor }&{k > 0,m < 0} end{array}} ight.).
上式可以很方便地转写为只包含加减和位运算的C/C++代码.
int div_exp2(int x, unsigned char k)
{
assert( (k >= 0) && (k < INT_BITS));
if (k == 0) return x;
int tail = x >> (INT_BITS - 1);
int exp2_k_1 = 1 << (k - 1);
// 当 x >=0 时, bias = 1 << (k - 1)
// 当 x < 0 时, bias = (1 << (k - 1)) - 1
// 这时 k > 0, 所以 bias >= 0 恒成立.
int bias = tail + exp2_k_1;
// 当 (x >= 0) && (x + bias < 0) 时, 表明 x + bias 在 int 取值范围内溢出,
// 而其在 unsigned int 取值范围内不会溢出, 所以将其转化为 unsigned int 进行运算.
if ((x >= 0) && (x + bias < 0))
{
return int(((unsigned int)(x + bias)) >> k);
}
else
{
return (x + bias) >> k;
}
}
下面尝试合并 div_exp2 函数的最后的条件语句. 利用 "算术右移" 和 "逻辑右移" 的关系, 可以将 :
int(((unsigned int)(x + bias)) >> k);
转换为
int mask = ~(-1 << (INT_BITS - k));
return ((x + bias) >> k) & mask;
于是得到合并后的语句:
// 当 (x >= 0) && (x + bias < 0) 时, 由于 tail = 0, condition = -1, 所以 mask = ~(-1 << (INT_BITS - k)), 等效于上面的代码.
// 其他情况下, mask = -1, ((x + bias) >> k) & mask 等效于 (x + bias) >> k.
int condition = (x + bias) >> (INT_BITS - 1);
int mask = ~( (-1 << (INT_BITS - k) ) & ~tail & condition);
return ((x + bias) >> k) & mask;
所以, 最终得到:
int div_exp2(int x, unsigned char k)
{
assert( (k >= 0) && (k < INT_BITS));
if (k == 0) return x;
int tail = x >> (INT_BITS - 1);
int exp2_k_1 = 1 << (k - 1);
// 当 x >=0 时, bias = 1 << (k - 1)
// 当 x < 0 时, bias = (1 << (k - 1)) - 1
// 这时 k > 0, 所以 bias >= 0 恒成立.
int bias = tail + exp2_k_1;
// 当 (x >= 0) && (x + bias < 0) 时, 表明 x + bias 在 int 取值范围溢出,
// 而其在 unsigned int 取值范围内不会溢出, 所以将其转化为 unsigned int 进行运算.
int condition = (x + bias) >> (INT_BITS - 1);
int mask = ~( (-1 << (INT_BITS - k) ) & ~tail & condition);
return ((x + bias) >> k) & mask;
}
测试代码
#include <stdlib.h>
#include <time.h>
#include <stdio.h>
#include <limits.h>
#include <assert.h>
const int INT_BITS = 32;
double div_exp2_float(int x, unsigned char k)
{
assert( (k >= 0) && (k < INT_BITS));
unsigned int den = 1 << k;
return double(x) / den;
}
int div_exp2_plain(int x, unsigned char k)
{
assert( (k >= 0) && (k < INT_BITS));
unsigned int den = 1 << k;
if (x > 0)
{
return int(double(x) / den + 0.5);
}
else
{
return int(double(x) / den - 0.5);
}
}
int div_exp2(int x, unsigned char k)
{
assert( (k >= 0) && (k < INT_BITS));
if (k == 0) return x;
int tail = x >> (INT_BITS - 1);
int exp2_k_1 = 1 << (k - 1);
// 当 x >=0 时, bias = 1 << (k - 1)
// 当 x < 0 时, bias = (1 << (k - 1)) - 1
// 这时 k > 0, 所以 bias >= 0 恒成立.
int bias = tail + exp2_k_1;
// 当 (x >= 0) && (x + bias < 0) 时, 表明 x + bias 在 int 取值范围溢出,
// 而其在 unsigned int 取值范围内不会溢出. 所以将其转化为 unsigned int 进行运算.
int condition = (x + bias) >> (INT_BITS - 1);
int mask = ~( (-1 << (INT_BITS - k) ) & ~tail & condition);
return ((x + bias) >> k) & mask;
}
int random(int lower, int upper)
{
if (lower > upper)
{
int temp = lower;
lower = upper;
upper = temp;
}
return rand() % (upper - lower + 1) + lower;
}
int main()
{
int k = 5;
printf("=========================
");
for (int i = 0; i < 100; ++i)
{
int x = random(-1111, 11111);
float val = div_exp2_float(x, k);
int val1 = div_exp2_plain(x, k);
int val2 = div_exp2(x, k);
if (val1 != val2)
{
printf("%d / 2^%d: %.2f, %d, %d
", x, k, val, val1, val2);
}
else
{
// printf("Pass
");
printf("%d / 2^%d: %.2f, %d, %d
", x, k, val, val1, val2);
}
}
printf("=========================
");
printf("%f
", div_exp2_float(INT_MIN, 31));
printf("%f
", div_exp2_float(INT_MIN + 1, 31));
printf("%f
", div_exp2_float(INT_MAX, 31));
printf("%d
", div_exp2_plain(INT_MIN, 31));
printf("%d
", div_exp2_plain(INT_MIN + 1, 31));
printf("%d
", div_exp2_plain(INT_MAX, 31));
printf("%d
", div_exp2(INT_MIN, 31));
printf("%d
", div_exp2(INT_MIN + 1, 31));
printf("%d
", div_exp2(INT_MAX, 31));
return 0;
}
版权声明
版权声明：自由分享，保持署名-非商业用途-非衍生，知识共享3.0协议。
如果你对本文有疑问或建议，欢迎留言！转载请保留版权声明！
如果你觉得本文不错, 也可以用微信赞赏一下哈.
