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Now I got one just simple thing, and I simply have isolated that module inside of that function.What about abstraction?
而现在我只要做一件简单的事，那就是我只需把这个模块,加到函数中去？
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In that case point p 1 doesn't correspond to this point, it actually corresponds to the point of radius 2 and angle 1, which is about here.
基本上也就是说这是第一个点1,这是第二个点，把它们的值加到一起,然后我就得到了目标点，好，这听起来挺不错的。
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Maybe I will add this on as a problem you should do in your homework.
也许我会把这个作为,练习题加到你们的作业中去
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What happens if you add them to water?
把它们加到水里会发生什么呢
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But what looks like this orange block as of now in Scratch, add orange to your so called inventory, well, it's gonna take at least two lines of code in C to actually create the inventory array with this first line and then put something like the word orange inside of it.
就好比Scratch里面橘色的这一块，把“桔子“加到所谓的“清单“里面去,在C语言中至少需要两行代码来实现,第一行代码创建一个货单数组,第二行就需要把“桔子“放进去。
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It's going to take this list and add to it something else.
这个代码有什么作用呢？,我将会把一些东西加到这个数组里面去。
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I, somehow if, I want to walk through some sequence of data structures, gathering up or doing the same thing, adding ages in until I get a total age and then divide by the number faculty.
我想，以某种方法，来遍历,一些数据结构，把它们相同的属性加到一起，就是一直的把年龄加到一起一直到,得到了一个年龄总数,然后除以员工的数目。
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And so I'm going to go back to this idea of a class, and let's build the first of these, and that is shown right here on this piece of your handout.
你把半径加到一起,你不会把角度加到一起，你需要再笛卡尔坐标下，进行这个操作,那么我这里在讲的就是。
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This equation is begging you to be added to this equation.
把这个方程加到这个方程上去
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But there's an important thing going on here, if I in fact wanted to combine those into a string I should have told the machine to do that, and I can do that, by explicitly saying, take that, which is a number, then add it to that string.
但是这里有个很重要的事情，实际上想要把这些加到一个字符串里面去,我应该告诉计算机去做这个，我应该这么做，通过明确的输入，也就是把这个对象，一个数字，然后把它加到一个字符串里面去。
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And this just is another list, I mainly wanted you to have it in your notes of diseases that are important around the world, for which there are very active programs to develop either vaccines or more effective vaccines.
这是另一个表,我希望你们做笔记时把它加到,世界上这些严重的传染疾病上,因为现在有一些项目致力于开发,针对这些疾病的疫苗或者改进式疫苗
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So I've appended this whole object.
因此我把整个对象都加到空数组里面去了。
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And when I do that, I'll append it to history.
当我这么做的时候，我同时会把它加到历史中去。
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So this first little piece of code right here says, ok you give me 2 points, I'll create another 1 of these lists and I'll simply take the x, sorry I shouldn't say x, I'm going to assume it's the x, the x-values are the two points, add them together, just right there, the y-values, add them together and return that list.
好，为了来认识到这一点,让我们来看一个简单的小例子，在你们的课堂手册上，你可以看到我写了一个小程序,它假设我得到了，这些点中的一些，我想对它们做一些操作，例如我想把它们加到一起，那么这里的第一小片，代码的意思是，好给我两个点，我会再创建一个数组。
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That means starting at the first one, I'm going to do something to it. And what am I'm going to do? I'm going to take that character, convert it back into an integer, and add it into some digits. And I've done a little short hand here, which is I should have said some digits is equal to some digits plus this.
这意味着从第一个字符开始，我要对他们进行一些操作,我要去做什么呢？我要取得这个字符，然后把它转换为整数,然后加到某些数上面去,我在这里用了一些缩写,我本来应该写一个数字等于这个数字。
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But that little short hand there is doing exactly the same thing. It is adding that value into some digits and putting it back or signing it back into some digits. And I'll walk through that loop and when I'm done I can print out the total thing does. And if I do that, I get out what I would expect.
加上得到的这个数的,但是这个缩写声明其实是进行了同样的操作,它把我们得到的这个数加到一个数上面去,然后用和对这个数进行了重新赋值,在循环中会去遍历字符串，当完成循环后,程序会显示数字的总和，如果我运行，这个程序的话，我会得到我期待的结果。
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first So what do I do? I find the midpoint by taking last minus first, divide by 2, and add to start.
要怎么做？我把last减去,除以二，加到起点上去。
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It says, take value of y and value of x, add them together, y and create that as the new value of y.
把x和y的值，加到一起，然后把和赋值给。
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Another way of saying it is, I've actually said, what did I use there, and , It's basically saying there is the first point, there's the second point, add them together and I get that point.
我要去做的是-对不起，再运行一次-好，你可以看到我已经把，这两个点的坐标值加到一块儿了，然后返回了r的值,我会让大家看到我们，实际上得到的是什么，这看起来不错，好,我做了正确的操作。
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To get any location in memory, to get to any value of the list, I simply have to say which element do I want to get, I know that these things are stored in a particular size multiply that index by 4, add it to start, and then it's in a constant amount of time I can go to that location and get out the cell.
取得列表中的任何一个值,简单说来，想要取得列表中的任何元素,我知道这些元素存在特定的大小中,把下标乘以4，加到start上,然后定位到内存单元，并取出值就是固定的时间了,好的，如果元素以固定大小存储。
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