Need to write down something

面对多个工作机会,作者综合考虑了薪资、工作地点、公司行业地位及职业发展前景等因素,难以抉择。最终接受了Realtek公司的职位,该公司主要产品包括计算机和终端的网络接口卡。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

I have been finding a  job for around one month, with several offers in hand. It is really hard to decide which job I should take. I have to take a lot of elements to account, like salary, working place, position of the company in the IT area, and career prospect. Moreover, suggestions from my family also need to be  regarded.


Last night, I chatted with one of my best friends in college and asked him for some advice. He got a position in CISCO, a great company. We had participated in the Electronic Design Contest together during undergraduate study. After graduation, we entered different universities for further education. However, we still can't get the final decision after a long discussion. Last, we agreed to deal with it by flipping a coin. Anyway, I really appreciated his help and our friendship.


Now I accept the occupation offered by Realtek Corp. One of main products of this company is NIC for computers and terminals.




Now, we have a way to perceive that the user is pressing a key. We know when we want to move up, down, left, and right. We know at each iteration of the main loop exactly, what the user wants; we just have to update the circle with a new position depending on this input. This method gives us a great advantage. So finally we can write something in our update() function, namely, the movement of our player. We check which of the four Boolean member variables is true, and determine the movement accordingly. By using += (instead of =) and if (instead of else if), we implicitly handle the case where two opposite keys, such as right and left are pressed at the same time—the movement stays zero. The update() function is shown in the following code snippet: void Game::update() { sf::Vector2f movement(0.f, 0.f); if (mIsMovingUp) movement.y -= 1.f; if (mIsMovingDown) movement.y += 1.f; if (mIsMovingLeft) movement.x -= 1.f; if (mIsMovingRight) movement.x += 1.f; mPlayer.move(movement); } We introduce two new things here: a vector and the move() function on the circle shape. The move() function does what its name says, it moves the shape by the amount we provide it. Vector algebra Vectors are an important part of algebraic mathematics. They imply lots of rules and definitions, which go beyond the scope of our book. However, SFML's sf::Vector2 class template is way more practical, both in concept and functionality. To be as simple as we could possibly be, we know that a coordinate in a two-dimensional Cartesian system would need two components: x and y. Because in graphics all coordinates are expressed with the decimal float data type, sf::Vector2 is instantiated as sf::Vector2<float>, which conveniently has a typedef named sf::Vector2f. Such an object is made to contain two member variables, x and y. This makes our life simpler, because now we don't need to pass two variables to functions, as we can fit both in a single sf::Vector2f object. sf::Vector2f also defines common vector operations, such as additions and subtractions with other vectors, or multiplications and divisions with scalars (single values), effectively shortening our code.翻译,要求每行中英文对照
03-08
资源下载链接为: https://pan.quark.cn/s/d9ef5828b597 在本文中,我们将探讨如何通过 Vue.js 实现一个带有动画效果的“回到顶部”功能。Vue.js 是一款用于构建用户界面的流行 JavaScript 框架,其组件化和响应式设计让实现这种交互功能变得十分便捷。 首先,我们来分析 HTML 代码。在这个示例中,存在一个 ID 为 back-to-top 的 div 元素,其中包含两个 span 标签,分别显示“回到”和“顶部”文字。该 div 元素绑定了 Vue.js 的 @click 事件处理器 backToTop,用于处理点击事件,同时还绑定了 v-show 指令来控制按钮的显示与隐藏。v-cloak 指令的作用是在 Vue 实例渲染完成之前隐藏该元素,避免出现闪烁现象。 CSS 部分(backTop.css)主要负责样式设计。它首先清除了一些默认的边距和填充,对 html 和 body 进行了全屏布局,并设置了相对定位。.back-to-top 类则定义了“回到顶部”按钮的样式,包括其位置、圆角、阴影、填充以及悬停时背景颜色的变化。此外,与 v-cloak 相关的 CSS 确保在 Vue 实例加载过程中隐藏该元素。每个 .page 类代表一个页面,每个页面的高度设置为 400px,用于模拟多页面的滚动效果。 接下来是 JavaScript 部分(backTop.js)。在这里,我们创建了一个 Vue 实例。实例的 el 属性指定 Vue 将挂载到的 DOM 元素(#back-to-top)。data 对象中包含三个属性:backTopShow 用于控制按钮的显示状态;backTopAllow 用于防止用户快速连续点击;backSeconds 定义了回到顶部所需的时间;showPx 则规定了滚动多少像素后显示“回到顶部”按钮。 在 V
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值