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If I want to get out right now the versions of these things, I can ask what's the value of c p 1 x, and it returns it back out.
你可以在那里看到那些,代表笛卡尔坐标点的东西，如果我想要得到现在，这个类的版本的东西的话。
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So you're clobbering the value of X and then you can't get back X's value because it's too late.
所以你们清除了X的值,然后你们不能找回X的值了，因为太晚了。
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I going to come back in a second to how it actually does that, but it basically says, get me x value for p 1, get me the x value for p 2, compare them, just as you would normally.
是一个类的实例,我要去取的这个实例，所关联的x值，我稍后会讲讲实际上，这里是怎么实现的，但是基本上它的意思就是，给我p1的x值。
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x and y can be whatever you like, but I picked a very special example You should go back and remember what we did.
你可以任意选取 x 和 y,但我这里讲个特例,你们应该回忆一下上节课的内容
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But here, if I decide I'm going to store things not in x and y, but with some other set of names, for example, I've gotta go back into these pieces of code that use the points, and change them. So I've lost modularity.
除了要改下借口,但是这里，如果我决定，我不把值放在x和y中，而是和其他一些变量名进行绑定,例如这样的话,我就得回到使用这个点的代码，那儿去做更改了。
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Z So it would be incorrect to try to assign this to a variable X or Y or Z, because it doesn't actually give me anything back.
这个是错误的，来赋值这个给变量X或Y或,因为它的确没有返回什么给我。
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And so this time when you hand it back, this value, I store it in -- we'll say X, and now we actually have retained the value.
这次当你拿回这个值时，我把它存储在X中，现在我们把这个值保存了。
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Locally x got bound to 3, 3 I added 1 to it, whoop-dee-doo, 4 I get back a 4. But what's the value of x?
这是我想要的对吧？,在局部程序中x被赋值为，我给它加了1，我得到了？
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You go back to the same x-y plane; here is some vector A.
回到刚才的 x-y 平面上,这是某个矢量 A
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So let me go back now to this expression, x=at^/2+c+bt.
让我们再回到这个式子,x=at^/2+c+bt
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And in fact, now let me ask those two questions about this piece of code. First question is, for what values of integers-- we're going to assume integers-- but for what values of x does this code terminate? And the second question is, for what values of x does it give me back the right answer?
实际上，让我关于这块代码问两个问题,第一个问题是，对于什么样的整数值，我们会假设是整数,对于x的什么值程序,会最终终止？第二个问题是,对于x的什么值程序,会返回正确答案？
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