我尝试进行一些hex-bin / bin-hex转换,并测试哪个函数会更快,但是在将“ a”添加为整数时遇到了奇怪的错误。
#include <stdio.h>
/* convert bin to hex char [0-9a-f] */
static inline unsigned char ToHex4bits1(unsigned char znak)
//unsigned char ToHex4bits1(unsigned char znak)
{
znak &= 0x0F;
switch(znak)
{
case 10: return 'a';
case 11: return 'b';
case 12: return 'c';
case 13: return 'd';
case 14: return 'e';
case 15: return 'f';
default: return znak + 48; /// 48 0x30 '0'
}
}
/* convert bin to hex char [0-9a-f] */
static inline unsigned char ToHex4bits2(unsigned char znak)
//unsigned char ToHex4bits2(unsigned char znak)
{
//unsigned char add = '0';
int add = '0'; /// [0-9]; add value of '0' (65 0x41 '0')
znak &=0x0F;
if(znak > 9) /// [a-f]; if `znak' <0x0a, 0x0f> /// just one comparison as `znak' cannot be bigger than 15 anyway (znak &=0x0F;)
{
add = 0x61; /// 'a'; // 87 0x61 'a'
}
return znak + add;
}
//-----------//
int main()
{
int i;
//char z;
int z;
printf("\nToHex4bits1(i)\n");
for(i=0; i<16; i++)
{
z = ToHex4bits1(i);
printf("%d\t%02x\t%c\n", z, z, z);
}
printf("\nToHex4bits2(i)\n");
for(i=0; i<16; i++)
{
z = ToHex4bits2(i);
printf("%d\t%02x\t%c\n", z, z, z);
}
return 0;
}
当我运行它时,$ gcc -o tohex4bits tohex4bits.c; ./tohex4bits
我得到以下结果:
ToHex4bits1(i)
48 30 0
49 31 1
(...)
57 39 9
97 61 a
98 62 b
(...)
102 66 f
48 30 0
# which is what I expected
ToHex4bits2(i)
48 30 0
49 31 1
(...)
57 39 9
107 6b k # that's where things get interesting; it's 10 too much ('k'-'a'==10)
108 6c l
109 6d m
110 6e n
111 6f o
112 70 p
# which is wrong
第二个函数实际上有什么问题ToHex4bits2()
,为什么加上'a'(97 / 0x61)使其加上'k'(107 / 0x6b)或'A'=>'K'?
原因很简单。如果znak
为10,则您想返回'a'
,但是您要返回'a'+10
。因此,请返回znak+add-10
。
但是,这给自己带来了极大的困难。处处都有魔术常数,完成一项简单任务的代码极其复杂。这样可以:
{
znak &= 0x0F;
if(znak > 9)
return znak + 'a' - 10;
else
return znak + '0';
}
或者,如果您想更紧凑。您显然不怕复杂的代码:
{
znak &= 0x0F;
return znak > 9 ? znak + 'a' - 10 : znak + '0';
}
您提到您正在尝试优化此代码。我很难看到您如何才能做到这一点。您可能最好对更大的块进行优化,以查看算法是否存在问题。但是我们可以对第一个做些小事情,就是这样:
#define likely(x) __builtin_expect(!!(x), 1)
#define unlikely(x) __builtin_expect(!!(x), 0)
static inline unsigned char ToHex4bits1(unsigned char znak)
{
{
znak &= 0x0F;
// Hint the compiler that the first branch is less likely, which
// improves branch prediction
if(unlikely(znak > 9))
return znak + 'a' - 10;
else
return znak + '0';
}
}
在这里阅读有关内容:https : //www.geeksforgeeks.org/branch-prediction-macros-in-gcc/
但是我认为最快的方法是这样的:
static inline unsigned char ToHex4bits1(unsigned char znak)
{
const unsigned char ret[] = { '0', '1', '2', '3', '4', '5', '6', '7',
'8', '9', 'a', 'b', 'c', 'd', 'e', 'f' };
return ret[znak & 0x0F];
}
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