了解Strlen实施中的代码 [英] Understanding code in strlen implementation
问题描述
关于glibc中string.h中strlen的实现，我有两个问题.
I have two questions regarding the implementation of strlen in string.h in glibc.
-
该实现使用带有'holes'的幻数.我不明白这是如何工作的.有人可以帮我了解一下此片段:
The implementation uses a magic number with 'holes'. I am not able to understand how this works. Can someone please help me understand this snippet:
size_t
strlen (const char *str)
{
const char *char_ptr;
const unsigned long int *longword_ptr;
unsigned long int longword, himagic, lomagic;
/* Handle the first few characters by reading one character at a time.
Do this until CHAR_PTR is aligned on a longword boundary. */
for (char_ptr = str; ((unsigned long int) char_ptr
& (sizeof (longword) - 1)) != 0;
++char_ptr)
if (*char_ptr == '\0')
return char_ptr - str;
/* All these elucidatory comments refer to 4-byte longwords,
but the theory applies equally well to 8-byte longwords. */
longword_ptr = (unsigned long int *) char_ptr;
/* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
the "holes." Note that there is a hole just to the left of
each byte, with an extra at the end:
bits: 01111110 11111110 11111110 11111111
bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
The 1-bits make sure that carries propagate to the next 0-bit.
The 0-bits provide holes for carries to fall into. */
himagic = 0x80808080L;
lomagic = 0x01010101L;
if (sizeof (longword) > 4)
{
/* 64-bit version of the magic. */
/* Do the shift in two steps to avoid a warning if long has 32 bits. */
himagic = ((himagic << 16) << 16) | himagic;
lomagic = ((lomagic << 16) << 16) | lomagic;
}
if (sizeof (longword) > 8)
abort ();
/* Instead of the traditional loop which tests each character,
we will test a longword at a time. The tricky part is testing
if *any of the four* bytes in the longword in question are zero. */
for (;;)
{
longword = *longword_ptr++;
if (((longword - lomagic) & ~longword & himagic) != 0)
{
/* Which of the bytes was the zero? If none of them were, it was
a misfire; continue the search. */
const char *cp = (const char *) (longword_ptr - 1);
if (cp[0] == 0)
return cp - str;
if (cp[1] == 0)
return cp - str + 1;
if (cp[2] == 0)
return cp - str + 2;
if (cp[3] == 0)
return cp - str + 3;
if (sizeof (longword) > 4)
{
if (cp[4] == 0)
return cp - str + 4;
if (cp[5] == 0)
return cp - str + 5;
if (cp[6] == 0)
return cp - str + 6;
if (cp[7] == 0)
return cp - str + 7;
}}}
什么是魔幻数字?
为什么不简单地增加指针直到NULL字符并返回计数呢?这种方法更快吗?为什么会这样?
Why not simply increment pointer until NULL character and return count? Is this approach faster? Why is it so?
推荐答案
这用于一次查看4个字节(32位)甚至8个(64位)，以检查其中之一是否为零(结束)字符串)，而不是单独检查每个字节.
This is used to look at 4 bytes (32 bits) or even 8 (64 bits) in one go, to check if one of them is zero (end of string), instead of checking each byte individually.
这里是检查空字节的一个示例:
Here is one example to check for a null byte:
unsigned int v; // 32-bit word to check if any 8-bit byte in it is 0
bool hasZeroByte = ~((((v & 0x7F7F7F7F) + 0x7F7F7F7F) | v) | 0x7F7F7F7F);
有关更多信息，请参见位扭曲的黑客.
For some more see Bit Twiddling Hacks.
这里使用的那个(32位示例):
The one used here (32-bit example):
还有一种更快的方法-使用hasless(v，1)，该方法已定义 以下;它可以进行4次操作，不需要后续操作 确认.简化为
There is yet a faster method — use hasless(v, 1), which is defined below; it works in 4 operations and requires no subsquent verification. It simplifies to
#define haszero(v) (((v) - 0x01010101UL) & ~(v) & 0x80808080UL)
子表达式(v-0x01010101UL)的计算结果为 v中对应的字节为零或大于零时的任何字节 0x80.子表达式〜v& 0x80808080UL评估为高位设置 以字节为单位，其中v的字节未设置高位(因此 字节小于0x80).最后，通过对这两个子表达式进行与"运算 结果是设置高位，其中v中的字节为零，因为 由于第一个中的值大于0x80而设置了高位 子表达式被第二个掩模遮盖了.
The subexpression (v - 0x01010101UL), evaluates to a high bit set in any byte whenever the corresponding byte in v is zero or greater than 0x80. The sub-expression ~v & 0x80808080UL evaluates to high bits set in bytes where the byte of v doesn't have its high bit set (so the byte was less than 0x80). Finally, by ANDing these two sub-expressions the result is the high bits set where the bytes in v were zero, since the high bits set due to a value greater than 0x80 in the first sub-expression are masked off by the second.
一次查看一个字节至少要花费一个完整的整数值(寄存器宽)所需的cpu周期.在此算法中，将检查完整整数以查看它们是否包含零.如果不是，则几乎不使用指令，并且可以对下一个完整整数进行 jump .如果内部的字节数为零，则会进行进一步检查以查看其确切位置.
Looking at one byte at a time costs at least as much cpu cycles as looking at a full interger value (register wide). In this algorithm, full integers are checked to see if they contain a zero. If not, little instructions are used, and a jump can be made to the next full integer. If there is a zero byte inside, a further check is done to see at what exact position it was.
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