解析MySQL隐式转换问题

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)

root@mysqldb 22:19: [xucl]> select * from t1 where id=204027026112927603;
+--------------------+
| id   |
+--------------------+
| 204027026112927605 |
| 204027026112927603 |
+--------------------+
2 rows in set (0.00 sec)
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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;
}

{
 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);
}

/*
 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).
*/

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)

root@mysqldb 22:19: [xucl]> select * from t1 where id='204027026112927603';
+--------------------+
| id   |
+--------------------+
| 204027026112927603 |
+--------------------+
1 row in set (0.01 sec)