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And then hash character takes in any string or character, single character, gives me back a number. Notice what I do.
然后哈希接受任何字符串或字母，单个字符,返回给我一个数字，注意我要做什么。
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It is going to give me back a tuple a collection of two things, and so check out the syntax.
它将给我返回一个数组或者系列的一对值,然后检查一下语法。
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I would think that it would approach inifinity, and I would need to think about it and get back to you in terms of why we don't actually hit it and see something with an infinite wavelength.
我会认为波长会接近无穷大，而且我需要考虑一下,然后返回来告诉你,为什么我们没有说对,为什么我们没有看到,无穷大波长的东西。
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Something that takes two inputs, a left operant - and a right operant and it just returns a Boolean value is-- or rather it returns a left to right value.
它有两个输入，一个是左边的作用物,一个是右边的作用物，然后返回一个布尔值-,或者更确切地说，返回一个从左到右的值。
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Literally, return the control from this function, and take the value of the next expression, and return that as the value of the whole computation.
正如字面意义上说的，从这个函数返回，然后取得下一个表达式的值,并把这个值作为整个计算的结果返回。
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And it's just gathering together the multiplications while counting down the exponent. And you can see it when we get down to the end test here, we're going to pop out of there and we're going to return the answer.
这个方法就是通过乘法,来一个一个的减小指数,可以看到,最后面的结果测试，我们会在这里退出,然后返回答案。
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Well, this allows us to try to go back and explain some phenomena that over the years, mounting evidence was building that couldn't be explained.
好了，这个可以使得我们,返回去然后去解释一些，以前逐年积累的未能解释的现象,我们下节课将讨论。
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But I got a couple of other of these strange looking things in there with underbars to them.
我可以在这片代码中看到这些,然后我需要一些可以给我,返回这些信息的东西,但是这里还有一些看起来。
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I can use that to return values, which I can then use elsewhere, which I did-- and if I just come back and highlight this-- inside of that computation.
我可以用它来返回值，然后我可以在别的地方使用这些返回值，正是我做的--如果我回过头来,在高亮一下这个--在这部分计算内部。
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So the end result, because someone wrote this function years ago is that printf takes this thing, takes this thing, David plops David inside the middle of that formatted string and then renders the whole result.
基于这是某人多年以前写的程序,最终的结果是打印出这个，这个，在格式化字符串中间的,然后返回结果。
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And then I'm going to just return the bigger of the two. Little bit complicated, but it's basically just implementing this decision tree.
我会计算总价值,然后我会返回两者中的较大值，有点小复杂,但是它就是按照这个决策树在运行。
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And then the only other difference is obviously, it's a function I need to return a value.
然后还剩下一点明显的区别，对于方法我要有返回值。
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And if we type directly into the interpreter, it essentially does an eval and a print.
如果我们直接输入到解释器里面，它本质上会运算然后返回结果。
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And then to create the object, I'll simply do a set of inserts.
我将此返回做为我的集合，然后创建对象，我会一系统简单的插入操作。
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They'll keep trying to get it going, but maybe we should move on with our lives here while this is happening, and we'll click it back at the end, and if we have a nice set up at any point, I'll just stop and we'll go back and we'll look at it again.
它们会一直让它保持运动，但是或许我们应该,在这个过程中离开一会儿，我们在最后点击它，如果我们在每一个点,都有一个好的装置，我就会停止然后,返回再去看看它。
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What that basically says is the following: in an interpreted language, you take what's called the source code, the thing you write, it may go through a simple checker but it basically goes to the interpreter, that thing inside the machine that's going to control the flow of going through each one of the instructions, and give you an output.
也就说，如果是解释语言的话：,你要写一些叫做，源码的东西，你写的东西会经过一个简单的过滤器,然后解释器会处理你的源码，解释器会产生一个,逐条读取你的源码的,控制流，然后返回一个输出结果。
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s1 So for the fourth period, now we're into the 4 s 1 3d for potassium here. And what we notice when we get to the third element in 4s2 and the fourth period is 3d that we go 4 s 2 and then we're back to the 3 d's.
对于第四周期到现在我们来到钾的1,然后我们返回到,我们注意到当我们看到第三个元素,第四周期我们来到,然后我们返回到。
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And given those types of input, will get back output.
然后鉴于输入类型返回结果输出。
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So, I said, okay, I didn't want to assume so we clicked senior and frankly it was then a dead end because apparently to get some senior discount you need to have some special pass or something like this that we didn't have and yet that certainly wasn't obvious here so now we together and this person in particular had to figure out how you go back and then restart this process.
我说，好的，我不希望揣测，所以我点击了老人票,然后问题就来了，因为似乎,需要特殊的号码之类的东西来购买老年票,但是我们没有,明显我们不应该走这步,现在我们一起必须找出,怎样返回，然后重新开始这个步骤。
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I return immediately.
然后立即返回。
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Then if the weight of i is less than the available weight, I can return the value of i.
然后如果i的重量,小于剩下的重量，我将返回i的价值。
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I'm going to call it down here with search, which is simply going to call it, and then print an answer out.
然后返回答案，在二分法搜索中，其实有个挺美妙的名称。
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Do one comparison and return one of two possible orders on it, but I need to decide that.
虽然可以做一次比较,然后返回两个可能的顺序中的一个为结果，但是这需要我来做决定。
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It'd still give me an answer. It just would not be the answer I'm looking for.
然后运行程序，程序还是会返回一个答案,只是不是我想要的答案。
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This looks fine, right, I'm doing the right thing.
然后返回这个数组，如果我去运行这个程序。
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memo And then I'll return it.
然后我再返回。
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And then I'll call fast Fib and it returns the result it has.
然后我会调用快速Fib函数,然后它会返回它的结果。
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It says, if I want to print out something I built in Cartesian form up here, says, again, I'm going to pass it in a pointer to the instance, that self thing, and then I'm going to return a string that I combine together with an open self and close paren, a comma in the middle, and getting the x-value and the y-value and converting them into strings before I put the whole thing together.
这不仅仅是个列表,它是怎么来做的？,流程是：如果我想要返回,一些已经在笛卡尔模式下建好的值，好，再说一遍，我首先要传入一个,指向实例的指针,也就是，然后我会返回一个,由开括号，闭括号,中间的一个逗号,以及提前转换为字符串格式的。
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f Well, call toupper pass this lowercase F F to this function called toupper it's going to return capital F and so what do I assign to s2 bracket zero?
调用toupper函数，传递这个小写,然后返回的是大写的,那么我对s2【0】赋值多少？
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It says, well I'm going to print out first and last just so you can see it, and then I say, gee 2 if last minus first is less than 2, that is, if there's no more than two elements left in the list, then I can just check those two elements and return the answer.
然后它计算了尾点和开始点的差,如果小于2的话,也就是说数组中的元素小于等于，我对这两个元素进行比较,然后返回结果就可以了，否则的话，我们就去寻找中值点，注意它是怎么实现的,首先这个指向一个列表的开头。
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