一、问题描述
root@mysqldb 22:12: [xucl]> show create table t1\G
*************************** 1. row ***************************
Table: t1
Create Table: CREATE TABLE t1 (
id varchar(255) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8
1 row in set (0.00 sec)
root@mysqldb 22:19: [xucl]> select * from t1;
±-------------------+
| id |
±-------------------+
| 204027026112927605 |
| 204027026112927603 |
| 2040270261129276 |
| 2040270261129275 |
| 100 |
| 101 |
±-------------------+
6 rows in set (0.00 sec)
奇怪的现象：
root@mysqldb 22:19: [xucl]> select * from t1 where id=204027026112927603;
±-------------------+
| id |
±-------------------+
| 204027026112927605 |
| 204027026112927603 |
±-------------------+
2 rows in set (0.00 sec)
640?wx_fmt=jpeg
什么鬼，明明查的是204027026112927603，为什么204027026112927605也出来了
二、源码解释
堆栈调用关系如下所示：
其中JOIN::exec()是执行的入口，Arg_comparator::compare_real()是进行等值判断的函数，其定义如下
int Arg_comparator::compare_real()
{
/*
Fix yet another manifestation of Bug#2338. 'Volatile' will instruct
gcc to flush double values out of 80-bit Intel FPU registers before
performing the comparison.
*/
volatile double val1, val2;
val1= (*a)->val_real();
if (!(*a)->null_value)
{
val2= (*b)->val_real();
if (!(*b)->null_value)
{
if (set_null)
owner->null_value= 0;
if (val1 < val2) return -1;
if (val1 == val2) return 0;
return 1;
}
}
if (set_null)
owner->null_value= 1;
return -1;
}
比较步骤如下图所示，逐行读取t1表的id列放入val1，而常量204027026112927603存在于cache中，类型为double类型（2.0402702611292762E+17），所以到这里传值给val2后val2=2.0402702611292762E+17。
当扫描到第一行时，204027026112927605转成doule的值为2.0402702611292762e17，等式成立，判定为符合条件的行，继续往下扫描，同理204027026112927603也同样符合
如何检测string类型的数字转成doule类型是否溢出呢?这里经过测试，当数字超过16位以后，转成double类型就已经不准确了，例如20402702611292711会表示成20402702611292712（如图中val1）
MySQL string转成double的定义函数如下：
{
char buf[DTOA_BUFF_SIZE];
double res;
DBUG_ASSERT(end != NULL && ((str != NULL && *end != NULL) ||
(str == NULL && *end == NULL)) &&
error != NULL);
res= my_strtod_int(str, end, error, buf, sizeof(buf));
return (*error == 0) ? res : (res < 0 ? -DBL_MAX : DBL_MAX);
}
真正转换函数my_strtod_int位置在dtoa.c（太复杂了，简单贴个注释吧）
/*
strtod for IEEE–arithmetic machines.
This strtod returns a nearest machine number to the input decimal
string (or sets errno to EOVERFLOW). Ties are broken by the IEEE round-even
rule.
Inspired loosely by William D. Clinger’s paper "How to Read Floating
Point Numbers Accurately" [Proc. ACM SIGPLAN '90, pp. 92-101].
Modifications:
-
We only require IEEE (not IEEE double-extended).
-
We get by with floating-point arithmetic in a case that
Clinger missed -- when we're computing d * 10^n
for a small integer d and the integer n is not too
much larger than 22 (the maximum integer k for which
we can represent 10^k exactly), we may be able to
compute (d*10^k) * 10^(e-k) with just one roundoff.
- Rather than a bit-at-a-time adjustment of the binary
result in the hard case, we use floating-point
arithmetic to determine the adjustment to within
one bit; only in really hard cases do we need to
compute a second residual.
- Because of 3., we don’t need a large table of powers of 10
for ten-to-e (just some small tables, e.g. of 10^k
for 0 <= k <= 22).
*/
既然是这样，我们测试下没有溢出的案例
root@mysqldb 23:30: [xucl]> select * from t1 where id=2040270261129276;
±-----------------+
| id |
±-----------------+
| 2040270261129276 |
±-----------------+
1 row in set (0.00 sec)
root@mysqldb 23:30: [xucl]> select * from t1 where id=101;
±-----+
| id |
±-----+
| 101 |
±-----+
1 row in set (0.00 sec)
结果符合预期，而在本例中，正确的写法应当是
root@mysqldb 22:19: [xucl]> select * from t1 where id=‘204027026112927603’;
±-------------------+
| id |
±-------------------+
| 204027026112927603 |
±-------------------+
1 row in set (0.01 sec)
三、结论
避免发生隐式类型转换，隐式转换的类型主要有字段类型不一致、in参数包含多个类型、字符集类型或校对规则不一致等
隐式类型转换可能导致无法使用索引、查询结果不准确等，因此在使用时必须仔细甄别
数字类型的建议在字段定义时就定义为int或者bigint，表关联时关联字段必须保持类型、字符集、校对规则都一致
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