背景

今天出现了一个bug，在数据库中我们将订单表中的order_no从之前的bigint(20)改成varchar(20)后，原有的代码逻辑在进行时查询时，之前是以Long类型传参查询的。

select * from order_main where order_no=16541913435669023

debug时的时候发现这条sql语句查询出来两条数据，另外一条毫不相关的订单也被查出来了。但是同样的sql我们放到数据库中时确是只能查到一条数据。

select * from order_main where order_no='16541913435669023'

仔细观察后发现，得到正确结果的Sql，是加了引号的，代码中的sql是没有加引号的数字类型。

根源

mysql5.7 查询varchar类型的数据时，不加引号，触发隐式转换导致的查询结果错误。

dev.mysql.com/doc/refman/…

源码解释

堆栈调用关系如下所示：

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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。

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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

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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）

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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:
 
   1. We only require IEEE (not IEEE double-extended).
   2. 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.
   3. 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.
   4. 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)