提交 0ce352c1 编写于 作者: A antirez

Faster ll2string() implementation.

Based on ideas documented in this blog post:

https://www.facebook.com/notes/facebook-engineering/three-optimization-tips-for-c/10151361643253920

The original code was modified to handle signed integers, reformetted to
fit inside the Redis code base, and was stress-tested with a program
in order to validate the implementation against snprintf().

Redis was measured to be measurably faster from the point of view of
clients in real-world operations because of this change, since sometimes
number to string conversion is used extensively (for example every time
a GET results into an integer encoded object to be returned to the
user).
上级 0adf4482
......@@ -214,28 +214,87 @@ long long memtoll(const char *p, int *err) {
return val*mul;
}
/* Return the number of digits of 'v' when converted to string in radix 10.
* See ll2string() for more information. */
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);
}
/* Convert a long long into a string. Returns the number of
* characters needed to represent the number, that can be shorter if passed
* buffer length is not enough to store the whole number. */
int ll2string(char *s, size_t len, long long value) {
char buf[32], *p;
unsigned long long v;
size_t l;
if (len == 0) return 0;
v = (value < 0) ? -value : value;
p = buf+31; /* point to the last character */
do {
*p-- = '0'+(v%10);
v /= 10;
} while(v);
if (value < 0) *p-- = '-';
p++;
l = 32-(p-buf);
if (l+1 > len) l = len-1; /* Make sure it fits, including the nul term */
memcpy(s,p,l);
s[l] = '\0';
return l;
* characters needed to represent the number.
* If the buffer is not big enough to store the string, 0 is returned.
*
* 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. */
if (svalue < 0) {
value = -svalue;
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;
}
/* Convert a string into a long long. Returns 1 if the string could be parsed
......
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