We may This is symmetric initialize for two teens, two eyes.

这是对称的两个青少年 两只眼睛。

First, we initialize bonus to 0.

，我们将 bonus 为 0。

We aren't allowed to initialize anything here.

我们不允许在这里任何东西。

At first, every pixel is initialized to infinity.

，每个像素都被为无穷大。

So what it does is it initializes the UART.

所以它所做的是 UART。

First let's initialize a new variable called " bonus" and set it to 1.

让我们一个名为“bonus”的新变量并将其设置为 1。

Then we initialize the instance of the job with values.

然后我们用值作业的实例。

And then we'll initialize it down here to the create marker function.

然后我们将在此处将其为创建标记函数。

So inside of the loop definition, we're initializing var i equal to zero.

因此，在循环定义内部，我们将 var i 为零。

You need to make sure you're setting the size for the map element that you're initializing.

您需要确保设置正在的地图元素的大小。

We can initialize a matrix like so.

我们可以像这样一个矩阵。

As an example, we’ve initialized our RAM with a simple computer program that we’ll to through today.

例如，我们已经使用今天将要使用的简单计算机程序了 RAM。

Then you need to initialize the instance with some data to work with.

然后，您需要使用一些要使用的数据来实例。

As an example, we've initialized our RAM with a simple computer program that we'll to through today.

例如，我们已经使用今天将要使用的简单计算机程序了 RAM。

We'll use this to initialize our struct and then give it the data it needs to do its work.

我们将使用它来我们的结构，然后为其提供完成工作所需的数据。

To do this, we need to specify that our function “handles” this event for our button, by adding a line to our initialize function.

为此，我们需要通过向我们的函数添加一行来指定我们的函数为我们的按钮“处理”此事件。

Is this because it initializes PLLs, and these PLLs need some time to lock every time you reset? And this seems to take a different amount of time every time it does that.

是不是因为它了PLL，每次reset的时候这些PLL都需要一些时间来lock？每次这样做似乎都需要不同的时间。