在s2中存储什么呢？
C2 be the right-most character of s2.
C2是S2最右侧的字符。
S2：他们是六个男孩。
For example, schema_map ((s1, t1), (s2, t2)).
例如，schema_map ((s 1,t1)，(s 2,t2))。
The base case is whenever S1 or S2 is a zero-length string.
（基线条件（basecase）是S1或S2为长度为零的字符串的情形。
The effect is that the value of s2 changes from "/usr" to "/tmp".
结果是s2的值从“ /usr ”变成了“ /tmp ”。
In this case, the LCS of S1 and S2 is clearly a zero-length string.
在这种情况下，S1和S2的LCS显然是长度为零的字符串。
These effects are represented by a rightward shift in supply to S2.
这些效应可以供应曲线右移到S2来代表。
Put your phone into downloading mode by powering your Galaxy S2 off.
放入您的手机下载您的银河s2断开供电模式。
So that is actually making a copy of the address and putting it in s2.
这实际上是复制一个地址，然后放置在s2中。
To resolve the conflict you can merge S2 and S3 and create a new state S4.
要解决冲突，您可以将S2和S3进行合并，并且创建新的状态s4。
In that case, your S1 or S2 breakout stop would get you out with a profit.
如果是那样，用系统1或系统2的突破止损能帮助你兑现利润。
And actually, I will choose another surface, S2, that maybe looks like that.
实际上，我将选择另一个曲面s2，它看起来应该是这样的。
Well, let's think about comparing the flux integral for S1 and the flux integral for S2.
让我们考虑关于S1和S2的通量积分的比较。
Since you suspended the change to S2, you have version S1 selected again in your workspace.
由于您悬挂了到S2的变更，所以您的工作区中再次选择了版本s1。
Next, you need to obtain the actual alignment strings - S1 'and S2' - and the alignment score.
接下来，需要得到实际的比对字符串—s1'和S2 '—以及比对的得分。
Resuming the suspended change causes a conflict because Jazz doesn't know which version to select: S2 or S3?
恢复悬挂的变更导致了冲突，因为Jazz不知道选择哪个版本：S2或者S3 ？
Subjects were required to judge whether numbers of S1 and S2 were the same, and whether they were greater than 5.
要求被试判断数字对（S1、S2）是否相同，并对S2数字做“与5相比”判断大小的任务。
The existing DB2 system has 3 DPF partitions (S1, S2, and S3), and the new system requires only 2 partitions (T1 and T2).
现有的DB 2系统有3个DPF分区(S1、S2和S3)，而新系统只需要2个分区(T1和T2)。
Definition: a set S1 is a superset of another set S2 if every element in S2 is in S1. S1 may have elements which are not in S2.
定义：如果一个集合s 2中的每一个元素都在集合s 1中，且集合s1中可能包含S2中没有的元素，则集合s1就是S2的一个超集。
Malloc It returns the address of this first byte so really the address of the first char here and so what gets stored in s2 now?
返回的是什么？，What, does,malloc, return ？，它返回的是第一个字符，第一个字节的地址，现在在s2中存储的是什么？
I could choose, for example, s1 a half sphere if I want or I can choose, let's call that s1, s2 I don't know, a pointy thing, s2.
比如说，我可以选择，一个半球面，或者我可以选一个…，先叫它，诶…，一个尖的曲面。
At this point in the story both s1 and s2 are literally pointing at the same location in memory so that's now what the story looks like.
在这里，s1和s2指向的是，内存中同一个地方，这里看起来像什么？
You initially had version S1 of the file in your workspace, you made a change creating version S2 of the file, then suspended the change.
最初，您的工作区中是文件的版本s1您进行变更，创建了文件的版本s2，然后悬挂变更。
You initially had version S1 of the file in your workspace, you made a change creating version S2 of the file, then suspended the change.
最初，您的工作区中是文件的版本s1您进行变更，创建了文件的版本s2，然后悬挂变更。
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