EdStem 帖子合集

近期将发布

It will be released in the near future

会在近期发布

It will be released in the near future

我不知道我们是否已经激活了 Gradescope - 这取决于助教（我们周五会开会，所以我们会在下周初解决这个问题）

I don't know if we've activated Gradescope yet - that's up to the TAs (we are meeting Friday so we'll address this by start of next week)

这将由您的 P-学分助教决定...

This will be up to your P-credit TA to determine...

发现得好（或澄清得好）！这个问题旨在确定退出最终“混合”框所需的时间，但技术上图中显示了最后一根管道。如果你再加 0.01 秒，你不会扣分...

Good catch (or clarification)! The problem is meant to determine the time it takes to exit the final "Mix" box, but technically there is a final pipe shown in the figure. If you were to add another 0.01 you wouldn't lose points...

很好的发现（或澄清）！这个问题旨在确定离开最终“混合”盒所需的时间，但技术上图中显示了最后一个管道。如果你再增加 0.01 秒，你不会扣分...

Good catch (or clarification)! The problem is meant to determine the time it takes to exit the final "Mix" box, but technically there is a final pipe shown in the figure. If you were to add another 0.01 you wouldn't lose points...

发现得好（或者说澄清得好）！这个问题是为了确定离开最终“混合”盒子所需的时间，但技术上图中显示了最后一根管道。如果你再加0.01秒，你不会失分...

Good catch (or clarification)! The problem is meant to determine the time it takes to exit the final "Mix" box, but technically there is a final pipe shown in the figure. If you were to add another 0.01 you wouldn't lose points...

抓得好（或者说澄清得好）！这个问题旨在确定退出最终“混合”盒所需的时间，但技术上图中显示有最后一个管道。如果你再加 0.01 秒，你不会扣分...

Good catch (or clarification)! The problem is meant to determine the time it takes to exit the final "Mix" box, but technically there is a final pipe shown in the figure. If you were to add another 0.01 you wouldn't lose points...

算了，我在幻灯片里找到了答案。不知何故，ctrl + f 没有显示“overflow”这个词在它的定义处。这是定义：

“无论如何强调，溢出意味着结果不能在字长内表示（不一定是结果太大！）”

Never mind, I found the answer in the slides. For some reason, ctrl + f wasn't showing the word "overflow" where it was defined. Here is the definition:

"Can’t stress enough that overflow means the result cannot be represented

within the word size (not necessarily that the result is too big)!"

没关系，我在幻灯片中找到了答案。不知何故，ctrl + f 没有显示“overflow”的定义位置。这是定义：

“怎么强调都不过分，溢出意味着结果不能在字长内表示（不一定是结果太大）！”

Never mind, I found the answer in the slides. For some reason, ctrl + f wasn't showing the word "overflow" where it was defined. Here is the definition:

"Can’t stress enough that overflow means the result cannot be represented

within the word size (not necessarily that the result is too big)!"

算了，我在幻灯片中找到了答案。不知何故，Ctrl + F 没有显示定义“溢出”的地方。这是定义：

“不能再强调溢出意味着结果不能在字长内表示（不一定是结果太大）！”

Never mind, I found the answer in the slides. For some reason, ctrl + f wasn't showing the word "overflow" where it was defined. Here is the definition:

"Can’t stress enough that overflow means the result cannot be represented

within the word size (not necessarily that the result is too big)!"

符号位始终是MSB（最高有效位）。由于此问题指定的是8位架构，因此它是第7位（最左边的位）。

符号位为0并不意味着该数字被视为无符号数；它只是意味着它是一个正的有符号数。如果MSB为1，则为负数。由于您示例中的MSB为0，因此该值为正数。

The sign bit is always the MSB (Most Significant Bit). Since this problem specifies an 8-bit architecture, that is Bit 7 (the leftmost bit).

A sign bit of 0 doesn't mean the number is treated as unsigned; it just means it is a signed number that is positive. If the MSB were 1, it would be negative. Since the MSB in your example is 0, the value is positive.

符号位始终是 MSB（最高有效位）。由于此问题指定了 8 位架构，因此它是第 7 位（最左边的位）。

符号位为 0 并不意味着该数字被视为无符号数；它只是意味着它是一个正的有符号数。如果 MSB 为 1，它将是负数。由于您示例中的 MSB 为 0，因此该值为正。

The sign bit is always the MSB (Most Significant Bit). Since this problem specifies an 8-bit architecture, that is Bit 7 (the leftmost bit).

A sign bit of 0 doesn't mean the number is treated as unsigned; it just means it is a signed number that is positive. If the MSB were 1, it would be negative. Since the MSB in your example is 0, the value is positive.

符号位始终是MSB（最高有效位）。由于此问题指定了8位架构，因此它是第7位（最左边的位）。

符号位为0并不意味着该数字被视为无符号数；它只是意味着它是一个正的有符号数。如果MSB是1，它将是负数。由于您示例中的MSB是0，因此该值是正数。

The sign bit is always the MSB (Most Significant Bit). Since this problem specifies an 8-bit architecture, that is Bit 7 (the leftmost bit).

A sign bit of 0 doesn't mean the number is treated as unsigned; it just means it is a signed number that is positive. If the MSB were 1, it would be negative. Since the MSB in your example is 0, the value is positive.

请看 #5

See #5

我想我明白你在说什么，但你的问题措辞有点奇怪。

我们以 1e) 10001010 为例。

它将代表不同的整数值，具体取决于应用哪种表示（无符号、符号-大小、一补码、二补码）。

对于无符号数，过程相当简单。

对于其他情况，大多数人（包括我自己）在数字为负数时，最简单的方法是找到相同大小的正数。对于每种有符号表示（符号-大小、一补码、二补码），都有一个直接的过程来做到这一点。我可能会在周二的课上复习这一点。

I think I know what you are saying, but your question is worded in a funny way.

Let's take 1e) 10001010

This will represent a different integer value depending on which representation (unsigned, signed magnitude, 1's complement, 2's complement) is being applied.

For unsigned, the process is rather straightforward.

For the others, most people (including myself), when the number is negative, the easiest thing to do is to find the positive number of same magnitude. For each signed representation (signed mag, 1's c, 2's c) there is a straightforward process for doing this. I'll probably start Tuesday's class reviewing this point).

我不是助教，但在一个ed讨论帖的回复中，教授提到我们还没有学到所有可以完全完成作业1的材料。我不知道我们什么时候才能掌握作业1所需的一切，但我们还有更多的课程内容需要学习。

Not a TA, but in an ed discussion post reply the professor mentioned that we haven't learned all of the material yet to fully do HW 1. I don't know when we will have everything we need for HW 1, but we still have more material to go through in class.

我不是助教，但在一个ed讨论帖的回复中，教授提到我们还没有学完所有材料来完全完成作业1。我不知道什么时候我们能学到作业1所需的所有内容，但我们课堂上还有更多材料要学习。

Not a TA, but in an ed discussion post reply the professor mentioned that we haven't learned all of the material yet to fully do HW 1. I don't know when we will have everything we need for HW 1, but we still have more material to go through in class.

我不是助教，但在一个 ed 讨论帖的回复中，教授提到我们还没有学到所有必要的材料来完全完成硬件1。我不知道我们什么时候才能掌握硬件1所需的一切，但我们课堂上还有更多的材料要学习。

Not a TA, but in an ed discussion post reply the professor mentioned that we haven't learned all of the material yet to fully do HW 1. I don't know when we will have everything we need for HW 1, but we still have more material to go through in class.

嗨 - 我周五和助教们谈过了，他们也需要更多时间来敲定他们的日程，所以我们将把正式分配推迟一周。我会让助教们这周尽量保持联系（以防有疑问，但也可以随时在EdStem上提问）。一旦他们（和你们）的日程稳定下来，我们就会进行分配。

Hi - I spoke with the TAs on Friday and they also need more time to finalize their schedules, so we will delay the official assignment by 1 week. I'll ask the TA's to try to make themselves available this week (in case there are questions, but also feel free to ask on EdStem). Once their (and your) schedules have stabilized we will do the assignment.

是的

yup

Courseworks 上有教学大纲。你也可以查看我的 S'24 讲义（在 Courseworks 上）以更好地了解所涵盖的内容（讲义每年都会有一些变化，但不大），这里是作业发布和截止日期的暂定时间表。大约每周一次，HW 5 是例外（它很难）。

3827S26-Planning - Google Sheets.pdf

There's a syllabus available on Courseworks. You can also see my S'24 lecture notes (on Courseworks) to get an even better sense of what's being covered (the notes change a bit year-to-year but not much), and here is a tentative schedule of HW release and due date. Roughly once a week with HW 5 being the exception (it's a doozy)

3827S26-Planning - Google Sheets.pdf

请参阅 #20 (此消息发布后)

See #20 (just posted after this msg)

我建议在一个单独的笔记本或文档上完成，并复制问题以获得更多空间，因为通常要求展示过程才能获得满分。

I recommend working on a separate notebook or document and copying the questions over for more space, as people are generally expected to show work for full credit.

我不是助教，但这是我的理解。-0和+0在数学上被认为是相同的。然而，它们在技术上拥有不同的位模式，所以如果你想比较位模式，那么它们是不同的。如果x的编码是1000，y的编码是0000，那么x=y，但是当你想用位操作比较它们时，x=/=y。它们是同一个数字0的不同位模式。我相信当你想“定义”除以零，例如在除以正零和负零时有不同的结果时，它们可以用于计算。换句话说，你根据你想如何使用零，给零赋予了两种不同的“定义”。

这类似于int x = 0和double y = 0是同一个数字的不同“定义”（类型）。这并不是说+0和-0在Java中必然是不同的类型，但它们确实存在细微的差别。

I am not a TA, but here is my understanding. -0 and +0 are treated as mathematically the same. However, they technically have different bit patterns, so if you want to compare bit patterns, then they are distinct. If x has encoding 1000 and y has encoding 0000, then x = y, but x =/= y when you want to compare them using bitwise operations. They are different bit patterns of the same number 0. I believe they can be used for computations when you want to "define" dividing by zero such as for having distinct outcomes for when you divide by positive versus negative zero. In other words, you're giving two different "definitions" to zero depending on how you want to use it.

It's analogous to how int x = 0 and double y = 0 are different "definitions" (types) for the same number. That's not to say that +0 and -0 are different types necessarily like in java, but they do have subtle differences.

我不是助教，但这是我的理解。-0和+0在数学上被认为是相同的。然而，它们在技术上拥有不同的位模式，所以如果你想比较位模式，那么它们是不同的。如果x的编码是1000，y的编码是0000，那么x=y，但是当你想用位操作比较它们时，x=/=y。它们是同一个数字0的不同位模式。我相信当你想“定义”除以零，例如在除以正零和负零时有不同的结果时，它们可以用于计算。换句话说，你根据你想如何使用零，给零赋予了两种不同的“定义”。

这类似于int x = 0和double y = 0是同一个数字的不同“定义”（类型）。这并不是说+0和-0在Java中必然是不同的类型，但它们确实存在细微的差别。

I am not a TA, but here is my understanding. -0 and +0 are treated as mathematically the same. However, they technically have different bit patterns, so if you want to compare bit patterns, then they are distinct. If x has encoding 1000 and y has encoding 0000, then x = y, but x =/= y when you want to compare them using bitwise operations. They are different bit patterns of the same number 0. I believe they can be used for computations when you want to "define" dividing by zero such as for having distinct outcomes for when you divide by positive versus negative zero. In other words, you're giving two different "definitions" to zero depending on how you want to use it.

It's analogous to how int x = 0 and double y = 0 are different "definitions" (types) for the same number. That's not to say that +0 and -0 are different types necessarily like in java, but they do have subtle differences.

我不是助教，乔纳的理解与我的相似，但有一点补充。问题是关于-0在有符号位表示法中的应用，如果我的理解没错，-0没有实际应用。它是有符号系统的一个副产品，通常需要被解决/规范化，据我所知，没有真正的情况是你想要将0表示为-0的。

Not a TA, and Jonah’s understanding is similar to mine with one slight addition. The question asks about application of -0 in signed magnitude, if my understanding is correct, there is no practical application of -0. It is a byproduct of the signed system and is generally something that has to be worked around/normalized, afaik there isn’t really a situation where you would want to represent 0 as -0.

我不是助教，但这是我的理解。-0和+0在数学上被认为是相同的。然而，它们在技术上具有不同的位模式，所以如果你想比较位模式，它们是不同的。如果x的编码是1000，y的编码是0000，那么x = y，但是当你使用按位操作比较它们时，x =/= y。它们是相同数字0的不同位模式。我相信它们可以用于计算，当你想要“定义”除以零时，例如在除以正零和负零时产生不同的结果。换句话说，你根据你想要如何使用零来给出两种不同的“定义”。

这类似于int x = 0和double y = 0是相同数字的不同“定义”（类型）。这并不是说+0和-0在Java中必然是不同的类型，但它们确实有细微的差别。

I am not a TA, but here is my understanding. -0 and +0 are treated as mathematically the same. However, they technically have different bit patterns, so if you want to compare bit patterns, then they are distinct. If x has encoding 1000 and y has encoding 0000, then x = y, but x =/= y when you want to compare them using bitwise operations. They are different bit patterns of the same number 0. I believe they can be used for computations when you want to "define" dividing by zero such as for having distinct outcomes for when you divide by positive versus negative zero. In other words, you're giving two different "definitions" to zero depending on how you want to use it.

It's analogous to how int x = 0 and double y = 0 are different "definitions" (types) for the same number. That's not to say that +0 and -0 are different types necessarily like in java, but they do have subtle differences.

我不是助教，但这是我的理解。-0和+0在数学上被认为是相同的。然而，它们在技术上具有不同的位模式，所以如果你想比较位模式，它们是不同的。如果x的编码是1000，y的编码是0000，那么x = y，但是当你使用按位操作比较它们时，x =/= y。它们是相同数字0的不同位模式。我相信它们可以用于计算，当你想要“定义”除以零时，例如在除以正零和负零时产生不同的结果。换句话说，你根据你想要如何使用零来给出两种不同的“定义”。

这类似于int x = 0和double y = 0是相同数字的不同“定义”（类型）。这并不是说+0和-0在Java中必然是不同的类型，但它们确实有细微的差别。

I am not a TA, but here is my understanding. -0 and +0 are treated as mathematically the same. However, they technically have different bit patterns, so if you want to compare bit patterns, then they are distinct. If x has encoding 1000 and y has encoding 0000, then x = y, but x =/= y when you want to compare them using bitwise operations. They are different bit patterns of the same number 0. I believe they can be used for computations when you want to "define" dividing by zero such as for having distinct outcomes for when you divide by positive versus negative zero. In other words, you're giving two different "definitions" to zero depending on how you want to use it.

It's analogous to how int x = 0 and double y = 0 are different "definitions" (types) for the same number. That's not to say that +0 and -0 are different types necessarily like in java, but they do have subtle differences.

我不是助教，乔纳的理解与我的类似，但有一点补充。问题是关于符号位负零在带符号位大小表示法中的应用，如果我的理解正确，-0并没有实际应用。它是带符号系统的一个副产品，通常需要解决/规范化，据我所知，没有真正的情况是你希望将0表示为-0的。

Not a TA, and Jonah’s understanding is similar to mine with one slight addition. The question asks about application of -0 in signed magnitude, if my understanding is correct, there is no practical application of -0. It is a byproduct of the signed system and is generally something that has to be worked around/normalized, afaik there isn’t really a situation where you would want to represent 0 as -0.

我不是助教，但这是我的理解。-0和+0在数学上被视为相同。然而，它们在技术上具有不同的位模式，因此如果你想比较位模式，那么它们是不同的。如果x的编码是1000，y的编码是0000，那么x=y，但是当你想要使用位操作比较它们时，x=/=y。它们是相同数字0的不同位模式。我相信它们可以用于计算，当你想要“定义”除以零的情况，例如当你除以正零或负零时有不同的结果。换句话说，你根据如何使用零来给出两种不同的“定义”。

这类似于int x = 0和double y = 0是相同数字的不同“定义”（类型）。这并不是说+0和-0在java中必然是不同的类型，但它们确实存在细微的差别。

I am not a TA, but here is my understanding. -0 and +0 are treated as mathematically the same. However, they technically have different bit patterns, so if you want to compare bit patterns, then they are distinct. If x has encoding 1000 and y has encoding 0000, then x = y, but x =/= y when you want to compare them using bitwise operations. They are different bit patterns of the same number 0. I believe they can be used for computations when you want to "define" dividing by zero such as for having distinct outcomes for when you divide by positive versus negative zero. In other words, you're giving two different "definitions" to zero depending on how you want to use it.

It's analogous to how int x = 0 and double y = 0 are different "definitions" (types) for the same number. That's not to say that +0 and -0 are different types necessarily like in java, but they do have subtle differences.

不会 - 我们的CPU不会执行浮点运算。所以如果你只对其有肤浅的了解（对于本课程而言）也没关系。

No - our CPU will not perform floating point operations. So it is ok (for the purposes of this course) if you only understand it at a superficial level.

一个未优化的CPU可能只是用全0来表示绝对0（它在值上非常接近，即浮点数中的2^{-127}，双精度浮点数中的2^{-1023}）。但由于0经常出现，我认为实际的浮点实现会将0作为特殊情况处理，即

X + 0 = X

X * 0 = 0

X/0 = nan

对于这门课，我们不需要担心这个问题。

An unoptimized CPU might just use all 0's to represent absolute 0 (it's pretty close in value, i.e., 2^{-127} in floating point, 2^{-1023} in double). But since 0 comes up a lot, I would think that actual floating point implementations handle 0 as a special case, i.e.,

X + 0 = X

X * 0 = 0

X/0 = nan

For this class, we don't need to worry about it.

如果你喜欢你的助教，我会让它更容易阅读（即，分成相应的字段）。

If you like your TA, I'd make it easier to read (i.e., break into the corresponding fields).

当然——我们会给你打

π

分。

说真的，作业只占你成绩的 10%，而且有 9 份作业，每份作业大约占你成绩的 1%——但思考这个问题有助于你的理解（请注意这是以前的期中考试题目）——所以我想你主要是为了这个原因才去做它。

Sure - we will give you a grade of

π.

Seriously, the HW is only 10% of your grade and with 9 assignments, each HW is about 1% of your grade - but thinking about the problem will help with your understanding (note it was a previous midterm question) - so I'd think about doing it for that reason mostly.

任何不能用3位表示的结果都将构成溢出，所以是的，由于无法表示大于13的值，它们都将是溢出。请注意，问题明确指出操作是“加法”，因此您无需考虑减法。由于第一个操作数有8种选择，第二个操作数有8种选择，因此共有64种情况需要考虑。有些可以很快排除（例如，000是否会参与溢出？111是否永远不会参与“加法”溢出）。而且，交换律也会减少您需要处理的情况数量...

Any result that cannot be represented using the 3 bits would constitute an overflow, so yes, since there is no way to represent values over 13, they would all be overflows. Note the problem specifies the operation is "add", so you don't need to consider subtraction. Since there are 8 choices for the first term and 8 for the second, there are 64 cases to consider. Some can quickly be ruled out (e.g., can 000 ever be involved in overflow? Can 111 ever not be involved in an "add" overflow). And commutativity also cuts down the number of cases you need to work through...

不确定这是否是你说的通过移动小数点来规范化数字，但你的过程应该是：

1) 计算你需要移动小数点多少位才能使数字变成1.XXXXXXX形式。

2) 第一部分中的数字现在成为你的指数（左为正，右为负），XXXXXX成为你的尾数。

3) 如果数字是负数，不要忘记包含符号位。

你的最终答案应该是 [第3部分][127 + 第1部分][第2部分]。

希望这有帮助！

Not sure if this is what you mean by normalizing the number by moving the period, but your process should be:

1) Calculate the number of decimal places you need to move the period so the number becomes 1.XXXXXXX form.

2) The number in part 1 now becomes your exponent (left is positive, right is negative), and the XXXXXX becomes your mantissa.

3) If the number is negative, don't forget to include the sign.

Your final answer should be [Part 3][127 + Part 1][Part 2].

Hope this makes sense!

你好，

（顺便说一下，我不是助教。）他们仍在设置评分系统，所以目前还没有 Gradescope，也没有官方的作业提交方式。我认为这取决于批改你作业的助教个人决定是否通过 Gradescope 提交作业，但就目前而言，你并没有错过任何关于访问课程 Gradescope 的信息，而且他们可能最终根本不使用 Gradescope。（至少，我目前无法访问 Gradescope 上的任何内容。）

作业 1 和作业 2 在 Courseworks 上的作业文件中。作业 0 是可选的。

Hello,

(I am not a TA by the way.) They are still getting the grading situation setup, so there is no gradescope nor any official way to submit the homework yet. I think it's at the discretion of the individual TA who grades your work as to whether they want you to submit homework through gradescope, but you haven't missed anything regarding accessing gradescope for the class in general as of now and they might end up not even using gradescope at all I think. (At the very least, I don't have access to anything on gradescope as of now.)

Homework 1 & 2 is in the homework file on courseworks. Homework 0 is optional.