本文深入探讨了Redis中Util.h文件包含的各种实用工具函数，如字符串模式匹配、内存单位转换、数字转换及路径处理等，并提供了详细的代码分析。
Redis将很多的公用转换函数独立了出来，放入了Util.h中，包括字符串对比，内存转换，字符串数字转换，获取路径等，Redis的作者都是自己实现的，在这里我将这些函数认真学习，观察是在哪里高效并可以在之后的工作中使用。
因为Util的函数都是完全独立逻辑的，所以我在这里只看.c文件
Util.c
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <limits.h>
#include <math.h>
#include <unistd.h>
#include <sys/time.h>
#include <float.h>
#include <stdint.h>
#include <errno.h>
#include "sha1.h"//Sha1校验算法的实现文件
/* Glob-style pattern matching. 全局风格的模式对比*/
int stringmatchlen(const char *pattern, int patternLen,
const char *string, int stringLen, int nocase)
{
while(patternLen) {
switch(pattern[0]) {
case '*'://当模式的第一个字符是*时
while (pattern[1] == '*') {
pattern++;//遍历模式字符串，查看模式到哪里不为*
patternLen--;
}
if (patternLen == 1)
return 1; /* match 如果直到最后都是*的话，那么就是完全匹配，*相当于通配符*/
while(stringLen) {
if (stringmatchlen(pattern+1, patternLen-1,
string, stringLen, nocase))
return 1; /* match 这里递归自己来判断后面的字符，666*/
string++;//字符串的下一位
stringLen--;
}
return 0; /* no match 如果到这里了，就没有匹配到，因为如果匹配的话前面和下面已经都匹配了*/
break;
case '?':
if (stringLen == 0)
return 0; /* no match 如果字符串长度是0，那么就不匹配了，因为模式是？*/
string++;否则到string的下一个
stringLen--;
break;
case '[':
{
int not, match;
pattern++;//模式的下一位
patternLen--;
not = pattern[0] == '^';//这里用int来表示bool值，因为在C里没有bool
if (not) {
pattern++;//如果模式是^，匹配模式的下一位
patternLen--;
}
match = 0;
while(1) {
if (pattern[0] == '\\') {//模式是\
pattern++;//匹配下一个
patternLen--;
if (pattern[0] == string[0])//如果模式等于字符，就是匹配
match = 1;
} else if (pattern[0] == ']') {
break;
} else if (patternLen == 0) {
pattern--;
patternLen++;
break;
} else if (pattern[1] == '-' && patternLen >= 3) {
int start = pattern[0];
int end = pattern[2];
int c = string[0];
if (start > end) {
int t = start;
start = end;
end = t;
}
if (nocase) {
start = tolower(start);//全部变成小写字符
end = tolower(end);
c = tolower(c);
}
pattern += 2;
patternLen -= 2;
if (c >= start && c <= end)
match = 1;
} else {
if (!nocase) {
if (pattern[0] == string[0])
match = 1;
} else {
if (tolower((int)pattern[0]) == tolower((int)string[0]))
match = 1;
}
}
pattern++;
patternLen--;
}
if (not)
match = !match;
if (!match)
return 0; /* no match */
string++;
stringLen--;
break;
}
case '\\':
if (patternLen >= 2) {
pattern++;
patternLen--;
}
/* fall through 接着执行下面的*/
default:
if (!nocase) {
if (pattern[0] != string[0])
return 0; /* no match */
} else {
if (tolower((int)pattern[0]) != tolower((int)string[0]))
return 0; /* no match */
}
string++;
stringLen--;
break;
}
pattern++;
patternLen--;
if (stringLen == 0) {
while(*pattern == '*') {
pattern++;
patternLen--;
}
break;
}
}
if (patternLen == 0 && stringLen == 0)
return 1;
return 0;
}
int stringmatch(const char *pattern, const char *string, int nocase) {
return stringmatchlen(pattern,strlen(pattern),string,strlen(string),nocase);
}
/* Convert a string representing an amount of memory into the number of bytes, so for instance memtoll("1Gb") will return 1073741824 that is (1024*1024*1024).转换一个字符串表示的内存到数字
On parsing error, if *err is not NULL, it's set to 1, otherwise it's set to 0. On error the function return value is 0, regardless of the fact 'err' is NULL or not. */
//这里是针对redis自己表示内存的方式来转换的，
long long memtoll(const char *p, int *err) {
const char *u;
char buf[128];
long mul; /* unit multiplier */
long long val;
unsigned int digits;
if (err) *err = 0;
/* Search the first non digit character. 找到第一个非数字的符号*/
u = p;
if (*u == '-') u++;
while(*u && isdigit(*u)) u++;
if (*u == '\0' || !strcasecmp(u,"b")) {
mul = 1;
} else if (!strcasecmp(u,"k")) {
mul = 1000;
} else if (!strcasecmp(u,"kb")) {
mul = 1024;
} else if (!strcasecmp(u,"m")) {
mul = 1000*1000;
} else if (!strcasecmp(u,"mb")) {
mul = 1024*1024;
} else if (!strcasecmp(u,"g")) {
mul = 1000L*1000*1000;
} else if (!strcasecmp(u,"gb")) {
mul = 1024L*1024*1024;
} else {
if (err) *err = 1;
return 0;
}
/* Copy the digits into a buffer, we'll use strtoll() to convert the digit (without the unit) into a number. 将数字的字符复制到一个buffer中*/
digits = u-p;
if (digits >= sizeof(buf)) {
if (err) *err = 1;
return 0;
}
memcpy(buf,p,digits);
buf[digits] = '\0';
char *endptr;
errno = 0;
val = strtoll(buf,&endptr,10);
if ((val == 0 && errno == EINVAL) || *endptr != '\0') {
if (err) *err = 1;
return 0;
}
return val*mul;
}
/* Return the number of digits of 'v' when converted to string in radix 10. See ll2string() for more information. 当以10进制转换为字符串时返回v*/
uint32_t digits10(uint64_t v) {
if (v < 10) return 1;
if (v < 100) return 2;
if (v < 1000) return 3;
if (v < 1000000000000UL) {
if (v < 100000000UL) {
if (v < 1000000) {
if (v < 10000) return 4;
return 5 + (v >= 100000);
}
return 7 + (v >= 10000000UL);
}
if (v < 10000000000UL) {
return 9 + (v >= 1000000000UL);
}
return 11 + (v >= 100000000000UL);
}
return 12 + digits10(v / 1000000000000UL);
}
/* Like digits10() but for signed values. 类似于上面的函数但是支队标识值*/
uint32_t sdigits10(int64_t v) {
if (v < 0) {
/* Abs value of LLONG_MIN requires special handling. */
uint64_t uv = (v != LLONG_MIN) ?
(uint64_t)-v : ((uint64_t) LLONG_MAX)+1;
return digits10(uv)+1; /* +1 for the minus. */
} else {
return digits10(v);
}
}ll2string
/* Convert a long long into a string. Returns the number of characters needed to represent the number. If the buffer is not big enough to store the string, 0 is returned.将长整型转换为一个字符串，返回表示使用的字符串，如果缓冲区不足，返回0
Based on the following article (that apparently does not provide a novel approach but only publicizes an already used technique):
https://www.facebook.com/notes/facebook-engineering/three-optimization-tips-for-c/10151361643253920 Modified in order to handle signed integers since the original code was designed for unsigned integers. */
int ll2string(char* dst, size_t dstlen, long long svalue) {
static const char digits[201] =
"0001020304050607080910111213141516171819"
"2021222324252627282930313233343536373839"
"4041424344454647484950515253545556575859"
"6061626364656667686970717273747576777879"
"8081828384858687888990919293949596979899";
int negative;
unsigned long long value;
/* The main loop works with 64bit unsigned integers for simplicity, so we convert the number here and remember if it is negative. 为了简单起见，主循环与64位无符号整数一起使用，所以我们在这里转换数字并记住它是否为负数*/
if (svalue < 0) {
if (svalue != LLONG_MIN) {
value = -svalue;
} else {
value = ((unsigned long long) LLONG_MAX)+1;
}
negative = 1;
} else {
value = svalue;
negative = 0;
}
/* Check length. */
uint32_t const length = digits10(value)+negative;
if (length >= dstlen) return 0;
/* Null term. */
uint32_t next = length;
dst[next] = '\0';
next--;
while (value >= 100) {
int const i = (value % 100) * 2;
value /= 100;
dst[next] = digits[i + 1];
dst[next - 1] = digits[i];
next -= 2;
}
/* Handle last 1-2 digits. */
if (value < 10) {
dst[next] = '0' + (uint32_t) value;
} else {
int i = (uint32_t) value * 2;
dst[next] = digits[i + 1];
dst[next - 1] = digits[i];
}
/* Add sign. */
if (negative) dst[0] = '-';
return length;
}string2ll
/* Convert a string into a long long. Returns 1 if the string could be parsed
* into a (non-overflowing) long long, 0 otherwise. The value will be set to
* the parsed value when appropriate. */
int string2ll(const char *s, size_t slen, long long *value) {
const char *p = s;
size_t plen = 0;
int negative = 0;
unsigned long long v;
if (plen == slen)
return 0;
/* Special case: first and only digit is 0. */
if (slen == 1 && p[0] == '0') {
if (value != NULL) *value = 0;
return 1;
}
if (p[0] == '-') {
negative = 1;
p++; plen++;
/* Abort on only a negative sign. */
if (plen == slen)
return 0;
}
/* First digit should be 1-9, otherwise the string should just be 0. */
if (p[0] >= '1' && p[0] <= '9') {
v = p[0]-'0';
p++; plen++;
} else if (p[0] == '0' && slen == 1) {
*value = 0;
return 1;
} else {
return 0;
}
while (plen < slen && p[0] >= '0' && p[0] <= '9') {
if (v > (ULLONG_MAX / 10)) /* Overflow. */
return 0;
v *= 10;
if (v > (ULLONG_MAX - (p[0]-'0'))) /* Overflow. */
return 0;
v += p[0]-'0';
p++; plen++;
}
/* Return if not all bytes were used. */
if (plen < slen)
return 0;
if (negative) {
if (v > ((unsigned long long)(-(LLONG_MIN+1))+1)) /* Overflow. */
return 0;
if (value != NULL) *value = -v;
} else {
if (v > LLONG_MAX) /* Overflow. */
return 0;
if (value != NULL) *value = v;
}
return 1;
}getrandomhexchar
/* Generate the Redis "Run ID", a SHA1-sized random number that identifies a given execution of Redis, so that if you are talking with an instance having run_id == A, and you reconnect and it has run_id == B, you can be sure that it is either a different instance or it was restarted. 生成redis运行id，这是一个sha1大小的随机数，用于标识给定的Redis执行情况*/
void getRandomHexChars(char *p, unsigned int len) {
char *charset = "0123456789abcdef";
unsigned int j;
/* Global state. */
static int seed_initialized = 0;
static unsigned char seed[20]; /* The SHA1 seed, from /dev/urandom. */
static uint64_t counter = 0; /* The counter we hash with the seed. */
if (!seed_initialized) {
/* Initialize a seed and use SHA1 in counter mode, where we hash the same seed with a progressive counter. For the goals of this function we just need non-colliding strings, there are no cryptographic security needs. 在计数器模式下初始化一个种子并使用SHA1，在那里我们用一个累进计数器对相同的种子进行哈希处理*/
FILE *fp = fopen("/dev/urandom","r");
if (fp && fread(seed,sizeof(seed),1,fp) == 1)
seed_initialized = 1;
if (fp) fclose(fp);
}
if (seed_initialized) {
while(len) {
unsigned char digest[20];
SHA1_CTX ctx;
unsigned int copylen = len > 20 ? 20 : len;
SHA1Init(&ctx);
SHA1Update(&ctx, seed, sizeof(seed));
SHA1Update(&ctx, (unsigned char*)&counter,sizeof(counter));
SHA1Final(digest, &ctx);
counter++;
memcpy(p,digest,copylen);
/* Convert to hex digits. */
for (j = 0; j < copylen; j++) p[j] = charset[p[j] & 0x0F];
len -= copylen;
p += copylen;
}
} else {
/* If we can't read from /dev/urandom, do some reasonable effort in order to create some entropy, since this function is used to generate run_id and cluster instance IDs 如果我们不能正常读取数据，创建一些熵*/
char *x = p;
unsigned int l = len;
struct timeval tv;
pid_t pid = getpid();
/* Use time and PID to fill the initial array. */
gettimeofday(&tv,NULL);
if (l >= sizeof(tv.tv_usec)) {
memcpy(x,&tv.tv_usec,sizeof(tv.tv_usec));
l -= sizeof(tv.tv_usec);
x += sizeof(tv.tv_usec);
}
if (l >= sizeof(tv.tv_sec)) {
memcpy(x,&tv.tv_sec,sizeof(tv.tv_sec));
l -= sizeof(tv.tv_sec);
x += sizeof(tv.tv_sec);
}
if (l >= sizeof(pid)) {
memcpy(x,&pid,sizeof(pid));
l -= sizeof(pid);
x += sizeof(pid);
}
/* Finally xor it with rand() output, that was already seeded with time() at startup, and convert to hex digits. */
for (j = 0; j < len; j++) {
p[j] ^= rand();
p[j] = charset[p[j] & 0x0F];
}
}
}getabsolutepath
/* Given the filename, return the absolute path as an SDS string, or NULL
* if it fails for some reason. Note that "filename" may be an absolute path
* already, this will be detected and handled correctly.
*
* The function does not try to normalize everything, but only the obvious
* case of one or more "../" appearning at the start of "filename"
* relative path. */
sds getAbsolutePath(char *filename) {
char cwd[1024];
sds abspath;
sds relpath = sdsnew(filename);
relpath = sdstrim(relpath," \r\n\t");
if (relpath[0] == '/') return relpath; /* Path is already absolute. */
/* If path is relative, join cwd and relative path. */
if (getcwd(cwd,sizeof(cwd)) == NULL) {
sdsfree(relpath);
return NULL;
}
abspath = sdsnew(cwd);
if (sdslen(abspath) && abspath[sdslen(abspath)-1] != '/')
abspath = sdscat(abspath,"/");
/* At this point we have the current path always ending with "/", and
* the trimmed relative path. Try to normalize the obvious case of
* trailing ../ elements at the start of the path.
*
* For every "../" we find in the filename, we remove it and also remove
* the last element of the cwd, unless the current cwd is "/". */
while (sdslen(relpath) >= 3 &&
relpath[0] == '.' && relpath[1] == '.' && relpath[2] == '/')
{
sdsrange(relpath,3,-1);
if (sdslen(abspath) > 1) {
char *p = abspath + sdslen(abspath)-2;
int trimlen = 1;
while(*p != '/') {
p--;
trimlen++;
}
sdsrange(abspath,0,-(trimlen+1));
}
}
/* Finally glue the two parts together. */
abspath = sdscatsds(abspath,relpath);
sdsfree(relpath);
return abspath;
}
/* Return true if the specified path is just a file basename without any
* relative or absolute path. This function just checks that no / or \
* character exists inside the specified path, that's enough in the
* environments where Redis runs. */
int pathIsBaseName(char *path) {
return strchr(path,'/') == NULL && strchr(path,'\\') == NULL;
}
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