是否可以在lambda表达式的字典中添加值?
那就是实现一个lambda,它具有与下面方法类似的功能.
def add_value(dict_x):
dict_x['a+b'] = dict_x['a'] + dict_x['b']
return dict_x
Run Code Online (Sandbox Code Playgroud) struct item
{
int a;
};
int main()
{
item *a = (item *)malloc(sizeof(item));
item *b = (item *)malloc(sizeof(item));
short *c = (short *)b;
c += 3;
memcpy(a, c, sizeof(int));
free(a);
free(b);
return 0;
}
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为什么valgrind回显“大小为 2 的无效读取”?我觉得应该是4号。
来自 Valgrind 的示例消息:
==19134== Invalid read of size 2
==19134== at 0x4C2F7E0: memcpy@@GLIBC_2.14 (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==19134== by 0x400625: main (main.cpp:19)
==19134== Address 0x51fd096 is 2 bytes after a block of size 4 alloc'd
==19134== at 0x4C2AB80: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==19134== …Run Code Online (Sandbox Code Playgroud) 这是我的桌子t1; 它有一百万行.
CREATE TABLE `t1` (
`a` varchar(10) NOT NULL,
`b` varchar(10) DEFAULT NULL,
`c` varchar(10) DEFAULT NULL,
`d` varchar(10) DEFAULT NULL,
`e` varchar(10) DEFAULT NULL,
`f` varchar(10) DEFAULT NULL,
`g` varchar(10) DEFAULT NULL,
`h` varchar(10) DEFAULT NULL,
PRIMARY KEY (`a`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 |
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结果:
mysql> select * from t1 where a=10000000;
Empty set (1.42 sec)
mysql> select * from t1 where b=10000000;
Empty set (1.41 sec)
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为什么选择主键和普通字段一样快?
我想知道在mysql中使用"select*from TABLE where primary key = x"时的时间复杂度; 答案是O(log(n))?(n是记录号)