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MySQL令人大跌眼鏡的隱式轉換

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MySQL的隱式轉換

一、問題描述

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)
 
 select * from t1;
+--------------------+
| id                 |
+--------------------+
| 204027026112927605 |
| 204027026112927603 |
| 2040270261129276   |
| 2040270261129275   |
| 100                |
| 101                |
+--------------------+
6 rows in set (0.00 sec)

奇怪的現象:

select * from t1 where id=204027026112927603;
+--------------------+
| id                 |
+--------------------+
| 204027026112927605 |
| 204027026112927603 |
+--------------------+
2 rows in set (0.00 sec)


什么鬼,明明查的是204027026112927603,為什么204027026112927605也出來了

二、源碼解釋

堆棧調用關系如下所示:

其中JOIN::exec()是執行的入口,Arg_comparator::compare_real()是進行等值判斷的函數,其定義如下

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;
}

比較步驟如下圖所示,逐行讀取t1表的id列放入val1,而常量204027026112927603存在于cache中,類型為double類型(2.0402702611292762E+17),所以到這里傳值給val2后val2=2.0402702611292762E+17。

當掃描到第一行時,204027026112927605轉成doule的值為2.0402702611292762e17,等式成立,判定為符合條件的行,繼續往下掃描,同理204027026112927603也同樣符合

如何檢測string類型的數字轉成doule類型是否溢出呢?這里經過測試,當數字超過16位以后,轉成double類型就已經不準確了,例如20402702611292711會表示成20402702611292712(如圖中val1)

MySQL string轉成double的定義函數如下:

{
  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);
}


真正轉換函數my_strtod_int位置在dtoa.c(太復雜了,簡單貼個注釋吧)

/*
  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).
*/


既然是這樣,我們測試下沒有溢出的案例

select * from t1 where id=2040270261129276;
+------------------+
| id               |
+------------------+
| 2040270261129276 |
+------------------+
1 row in set (0.00 sec)
 
 select * from t1 where id=101;
+------+
| id   |
+------+
| 101  |
+------+
1 row in set (0.00 sec)

結果符合預期,而在本例中,正確的寫法應當是

select * from t1 where id='204027026112927603';
+--------------------+
| id                 |
+--------------------+
| 204027026112927603 |
+--------------------+
1 row in set (0.01 sec)

三、結論

避免發生隱式類型轉換,隱式轉換的類型主要有字段類型不一致、in參數包含多個類型、字符集類型或校對規則不一致等
隱式類型轉換可能導致無法使用索引、查詢結果不準確等,因此在使用時必須仔細甄別
數字類型的建議在字段定義時就定義為int或者bigint,表關聯時關聯字段必須保持類型、字符集、校對規則都一致

最后貼一下官網對于隱式類型轉換的說明吧

1、If one or both arguments are NULL, the result of the comparison is NULL, except for the NULL-safe
=> equality comparison operator. For NULL => NULL, the result is true. No conversion is needed.
2、If both arguments in a comparison operation are strings, they are compared as strings.
3、If both arguments are integers, they are compared as integers.
4、Hexadecimal values are treated as binary strings if not compared to a number.
5、If one of the arguments is a TIMESTAMP or DATETIME column and the other argument is a
constant, the constant is converted to a timestamp before the comparison is performed. This is
done to be more ODBC-friendly. This is not done for the arguments to IN(). To be safe, always
use complete datetime, date, or time strings when doing comparisons. For example, to achieve best
results when using BETWEEN with date or time values, use CAST() to explicitly convert the values to
the desired data type.
A single-row subquery from a table or tables is not considered a constant. For example, if a subquery
returns an integer to be compared to a DATETIME value, the comparison is done as two integers.
The integer is not converted to a temporal value. To compare the operands as DATETIME values,
use CAST() to explicitly convert the subquery value to DATETIME.
6、If one of the arguments is a decimal value, comparison depends on the other argument. The
arguments are compared as decimal values if the other argument is a decimal or integer value, or as
floating-point values if the other argument is a floating-point value.
7、In all other cases, the arguments are compared as floating-point (real) numbers.

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