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# 2022q1 Homework5 (quiz8)
## 測驗一
```c=
unsigned long mask = d << 8 | d;
mask = mask << 16 | mask;
for (unsigned int i = 32; i < LBLOCKSIZE * 8; i <<= 1)
mask = (mask << i) | mask;
```
先以 long 為 8 bytes 的狀況下進行討論
以老師給的測驗題為例
d 的初始值為 `0x000000000000002E` (10進位值是 46)
而 mask 的初始值為 `0x0000000000002E2E`
到第二行時 , mask為`0x000000002E2E2E2E`
以 long 為 8 bytes 的狀況下時 , mask 最後會是 `0x2E2E2E2E2E2E2E2E`
**DETECT_CHAR(X, MASK) (DETECT_NULL(X ^ MASK))**
1 byte 是 8 個 bits , 所以用 16 進制來看會是 2 個 數字為 1 單位
所以 **DETECT_CHAR(X, MASK)** 是用來偵測下面8個 byte中有沒有我們所要查詢的字元
以兩個數字為單位 ,如果 src 恰巧有 1 byte 的值 為 2E , 則表示在這 8 個 byte 中有找到相對應的字元
我們使用 XOR , 如果有 byte 的值恰好為 2E , 則跟 mask XOR 之後 , 則相對應的 byte 的值會變為 0
我們觀察 DETECT_NULL(X)
```c
DETECT_NULL(X) (((X)-0x0101010101010101) & ~(X)&0x8080808080808080)
```
我們一樣拆分成 用 byte 看
((x-1) 跟 ~x 的最高位 bit ) 做 and
x 分成兩種情況
* x 為 0 時 , ((x-1) 跟 ~x 的最高位 bit ) 做 and 結果為 1
* x 不為 0 時 , 其結果為 0
也就是說當有 byte 為 1 時 ,
=> 也就是原本的 input XOR mask 有 byte 為 0
=> 也就是原本的 input 中有我們要找的字元
所以當 DETECT_CHAR 為 0 時 , 表示有 8 byte 不是我們要找的字元
```c=
while (length >= LBLOCKSIZE) {
/* if DETECT_CHAR equal to zero ,
* which means the char we want to search is not in next 8 bytes
* But if DETECT_CHAR is not equal to zero
* we must know the char we want to search is exist in next 8 bytes
* so we need to break this loop to find the accurate position of the char
*/
if (DETECT_CHAR(*asrc, mask)) {
break;
}
/* change the pointer asrc to point the next eight char's address
* However, we use unsigned long to declare asrc , so we just need to add 1 (because unsigned long is eight bytes)
*
*/
asrc = asrc + 1;
length -= LBLOCKSIZE;
}
```
完整程式碼如下
```c
#include <limits.h>
#include <stddef.h>
#include <stdint.h>
#include <stdio.h>
#include <string.h>
/* Nonzero if either X or Y is not aligned on a "long" boundary */
#define UNALIGNED(X) ((long)X & (sizeof(long) - 1))
/* How many bytes are loaded each iteration of the word copy loop */
#define LBLOCKSIZE (sizeof(long))
/* Threshhold for punting to the bytewise iterator */
#define TOO_SMALL(LEN) ((LEN) < LBLOCKSIZE)
/* Different Machine may have 4 bytes long or 8 bytes long int */
#if LONG_MAX == 2147483647L
#define DETECT_NULL(X) (((X)-0x01010101) & ~(X)&0x80808080)
#else
#if LONG_MAX == 9223372036854775807L
/* Nonzero if X (a long int) contains a NULL byte. */
#define DETECT_NULL(X) (((X)-0x0101010101010101) & ~(X)&0x8080808080808080)
#else
#error long int is not a 32bit or 64bit type.
#endif
#endif
/* @return nonzero if (long)X contains the byte used to fill MASK. */
#define DETECT_CHAR(X, MASK) (DETECT_NULL(X ^ MASK))
/* @c means the char's ASCII code value*/
void *memchr_opt(const void *src_void, int c, size_t length) {
const unsigned char *src = (const unsigned char *)src_void;
unsigned char d = c;
/* Deal with the src is not word-aligned */
while (UNALIGNED(src)) {
if (!length--)
return NULL;
if (*src == d)
return (void *)src;
src++;
}
if (!TOO_SMALL(length)) {
/* If we get this far, we know that length is large and
* src is word-aligned.
*/
/* The fast code reads the source one word at a time and only performs
* the bytewise search on word-sized segments if they contain the search
* character, which is detected by XORing the word-sized segment with a
* word-sized block of the search character and then detecting for the
* presence of NULL in the result.
*/
unsigned long *asrc = (unsigned long *)src;
unsigned long mask = d << 8 | d;
mask = mask << 16 | mask;
for (unsigned int i = 32; i < LBLOCKSIZE * 8; i <<= 1)
mask = (mask << i) | mask;
while (length >= LBLOCKSIZE) {
/* if DETECT_CHAR equal to zero ,
* which means the char we want to search is not in next 8 bytes
* But if DETECT_CHAR is not equal to zero
* we must know the char we want to search is exist in next 8 bytes
* so we need to break this loop to find the accurate position of the char
*/
if (DETECT_CHAR(*asrc, mask)) {
break;
}
/* change the pointer asrc to point the next eight char's address
* However , we use unsigned long to declare asrc , so the address
* we just add 1 (because unsigned long is eight bytes)
*
*/
asrc = asrc + 1;
length -= LBLOCKSIZE;
}
/* If there are fewer than LBLOCKSIZE characters left, then we resort to
* the bytewise loop.
*/
src = (unsigned char *)asrc;
}
while (length--) {
if (*src == d)
return (void *)src;
src++;
}
return NULL;
}
int main() {
const char str[] = "http://wiki.csie.ncku.edu.tw";
const char ch = '.';
char *ret = memchr_opt(str, ch, strlen(str));
printf("String after |%c| is - |%s|\n", ch, ret);
return 0;
}
```