发私信

三人与歌

文章 评论 浏览 30

三人与歌 2025-02-03 03:44:25

据我所知，它仅在第二个输入之后才起作用，因为此块：

const changeSearchTerm = (e) => {
    setSearchTerm(e.target.value)
    dispatch(userActions.searchByName(searchTerm));
  }

setSearchTerm 之类的状态设置实际上是异步。这意味着在您调用 setSearchTerm 之后，在下一行 searchTerm 的值仍然是旧值。

相反，您可以执行以下操作之一：

const changeSearchTerm = (e) => {
    setSearchTerm(e.target.value)
    dispatch(userActions.searchByName(e.target.value));
}

const changeSearchTerm = (e) => {
    setSearchTerm(e.target.value, () => {
      // This callback runs only after state has changed
      dispatch(userActions.searchByName(searchTerm));
    });
}

编辑：对不起，我错过了您使用了USESTATE/功能。您可以使用 useeffect ：

const changeSearchTerm = (e) => {
    setSearchTerm(e.target.value);
}

useEffect(() => {
   if (searchTerm.length > 0) {
     dispatch(userActions.searchByName(searchTerm));
   }
}, [searchTerm]);

这样，您的调度肯定会在搜索端的值时正确运行。

https://reactjs.org/docs/react-comps/react-componcon.html#html#state.html#setstate

From what I can tell, it only works after the second input because of this block:

const changeSearchTerm = (e) => {
    setSearchTerm(e.target.value)
    dispatch(userActions.searchByName(searchTerm));
  }

State setters like setSearchTerm are actually asynchronous. This means that after you call setSearchTerm, on the next line the value of searchTerm is still the old value.

You can instead do one of the following:

const changeSearchTerm = (e) => {
    setSearchTerm(e.target.value)
    dispatch(userActions.searchByName(e.target.value));
}

const changeSearchTerm = (e) => {
    setSearchTerm(e.target.value, () => {
      // This callback runs only after state has changed
      dispatch(userActions.searchByName(searchTerm));
    });
}

Edit: Sorry I missed that you used useState/functional. You can use useEffect:

const changeSearchTerm = (e) => {
    setSearchTerm(e.target.value);
}

useEffect(() => {
   if (searchTerm.length > 0) {
     dispatch(userActions.searchByName(searchTerm));
   }
}, [searchTerm]);

This way your dispatch will correctly run whenever searchTerm's value changes for sure.

https://reactjs.org/docs/react-component.html#setstate

React中的搜索栏有效，但经历非常奇怪的行为

一梦等七年七年为一梦 2025-01-27 03:44:25 浏览 1 评论 0 收藏 0
回复 原文
三人与歌 2025-02-02 14:19:51

文本 /跨度＆gt; “文本” 是标签的文本节点。因此，您必须选择[a class =“ eLLIPSIS”]，然后可以调用 .get_text（）方法以获取文本节点为文本/字符串

html='''
<html>
 <body>
  <a class="ellipsis" href="/aktier/om-aktien.html/5246/investor-a">
   <span class="flag small SE">
   </span>
   Investor A
  </a>
  <a class="ellipsis" href="/aktier/om-aktien.html/5247/investor-b">
   <span class="flag small SE">
   </span>
   Investor B
  </a>
 </body>
</html>
'''

from bs4 import BeautifulSoup

soup = BeautifulSoup(html,'lxml')

#print(soup.prettify())

for span in soup.find_all('a',class_="ellipsis"):
    txt = span.get_text(strip=True)
    print(txt)

output：

Investor A
Investor B

text behind /span> "text" are text nodes of a tag . So You have to select [a class="ellipsis" ] then you can call .get_text() method to get text nodes value as text/string

html='''
<html>
 <body>
  <a class="ellipsis" href="/aktier/om-aktien.html/5246/investor-a">
   <span class="flag small SE">
   </span>
   Investor A
  </a>
  <a class="ellipsis" href="/aktier/om-aktien.html/5247/investor-b">
   <span class="flag small SE">
   </span>
   Investor B
  </a>
 </body>
</html>
'''

from bs4 import BeautifulSoup

soup = BeautifulSoup(html,'lxml')

#print(soup.prettify())

for span in soup.find_all('a',class_="ellipsis"):
    txt = span.get_text(strip=True)
    print(txt)

Output:

Investor A
Investor B

可以解析出背后的文本在美丽的小组

独闯女儿国 2025-01-26 14:19:51 浏览 1 评论 0 收藏 0
回复 原文
三人与歌 2025-02-02 02:54:39

您的图形大致具有正确的形状：微小的阵列应适合L1缓存，因此获得非常高的性能。兆字节左右的阵列在L2中的阵列并获得较低的性能，除此之外，您应该从内存中流式传输并获得低性能。因此，问题大小和运行时之间的关系确实应该随着大小的增加而变得更陡峭。但是，当您击中连续的缓存边界时，结果图（BTW，OPS/SEC比单纯的运行时更常见）应具有逐步结构。我会说您没有足够的数据点来证明这一点。

另外，通常您会重复几次“实验”到1。甚至统计打ic和2。确保数据确实在缓存中。

由于您标记了此“ OpenMP”，因此您还应该探索采用给定的数组尺寸，并改变核心计数。然后，您应该获得或多或少的线性提高性能，直到处理器没有足够的带宽来维持所有核心。

评论者提出了强/弱缩放的概念。强大的缩放费用是：考虑到一定的问题大小，使用越来越多的核心。这应该使您的性能不断提高，但是随着开销开始占主导地位，回报的降低。较弱的缩放意味着：每个过程/线程/任何常数保持问题大小，并增加处理元素的数量。正如我所指出的，这应该使您几乎可以提高性能，直到您用尽带宽为止。您似乎要做的实际上都不是：您正在做“乐观的缩放”：增加问题大小，而其他一切都恒定。除了我所指出的那样，这应该给您带来越来越更好的性能。

因此，如果您想说“此代码缩放”，则必须在哪种情况下决定。就其价值而言，您的200GB/sec数字是合理的。这取决于您的体系结构的详细信息，但是对于最近的Intel节点来说，听起来很合理。

Your graph has roughly the right shape: tiny arrays should fit in the L1 cache, and therefore get very high performance. Arrays of a megabyte or so fit in L2 and get lower performance, beyond that you should stream from memory and get low performance. So the relation between problem size and runtime should indeed get steeper with increasing size. However, the resulting graph (btw, ops/sec is more common than mere runtime) should have a stepwise structure as you hit successive cache boundaries. I'd say you don't have enough data points to demonstrate this.

Also, typically you would repeat each "experiment" several times to 1. even out statistical hiccups and 2. make sure that data is indeed in cache.

Since you tagged this "openmp" you should also explore taking a given array size, and varying the core count. You should then get a more or less linear increase in performance, until the processor does not have enough bandwidth to sustain all the cores.

A commenter brought up the concepts of strong/weak scaling. Strong scaling means: given a certain problem size, use more and more cores. That should give you increasing performance, but with diminishing returns as overhead starts to dominate. Weak scaling means: keep the problem size per process/thread/whatever constant, and increase the number of processing elements. That should give you almost linear increasing performance, until -- as I indicated -- you run out of bandwidth. What you seem to do is actually neither of these: you're doing "optimistic scaling": increase the problem size, with everything else constant. That should give you better and better performance, except for cache effects as I indicated.

So if you want to say "this code scales" you have to decide under what scenario. For what it's worth, your figure of 200Gb/sec is plausible. It depends on details of your architecture, but for a fairly recent Intel node that sounds reasonable.

如何测试代码的问题大小缩放性能

末が日狂欢 2025-01-26 02:54:39 浏览 6 评论 0 收藏 0
回复 原文
三人与歌 2025-02-01 14:00:27

@barmar的答案是正确的，正是您要做的。但是，一种更具智能的熊猫的方法是不要为 loop使用，而是使用 apply ：

cols2 = ['col A', 'col B', 'col C']
df[pd.Index(cols2) + '_clean'] = df[cols2].apply(lambda col: col.str.replace(r"\(|\)|,", ""))

当您调用时，请访问而无需指定 axis ，它将默认为 axis = 0 ，这意味着它将为每列调用lambda函数。

@Barmar's answer is correct and exactly what you're trying to do. However, a more idomatic pandas way to do would be to not use a for-loop and instead use apply:

cols2 = ['col A', 'col B', 'col C']
df[pd.Index(cols2) + '_clean'] = df[cols2].apply(lambda col: col.str.replace(r"\(|\)|,", ""))

When you call apply without specifying axis, it'll default to axis=0, which means it'll call the lambda function for each column.

为什么我不能在熊猫分配方法中使用F弦？

独享拥抱 2025-01-25 14:00:27 浏览 2 评论 0 收藏 0
回复 原文
三人与歌 2025-02-01 01:15:44

您可以使用 loc plt.legend（）：

plt.legend(['Group A','Group B','Group C'], loc=(1.04, 1))

output： “在此处输入图像描述”

You can use the loc keyword for plt.legend():

plt.legend(['Group A','Group B','Group C'], loc=(1.04, 1))

Output:enter image description here

传奇格式 - 大胆边框或更大 /不同的字体

抚你发端 2025-01-25 01:15:44 浏览 3 评论 0 收藏 0
回复 原文
三人与歌 2025-01-31 21:25:29 
import java.util.*;
import java.util.stream.*;

class Main {
  static final String ALLOWED_CHARS = "01";
  
  public static void main(String[] args) {
    char[] valid = {'0','0','1','0','0','0','1','1','0','1'};
    char[] invalid = {'0','0','1','0','0','A','1','1','0','1'};
    
    Set<Character> set = ALLOWED_CHARS.chars()
                                      .mapToObj(c -> (char) c)
                                      .collect(Collectors.toSet());

    isValid(set, valid);
    isValid(set, invalid);
  }

  public static void isValid(Set<Character> set, char[] array) {
    for (char c: array) {
      if (!set.contains(c)) {
        System.out.println("Invalid");
        return;
      }
    }
    System.out.println("Valid");
  }
}

另一种方法（以下是正则） - 将 werse_chars 添加到 set ，并简单地检查数组中的任何元素是否在 set set 中不存在。

import java.util.*;
import java.util.stream.*;

class Main {
  static final String ALLOWED_CHARS = "01";
  
  public static void main(String[] args) {
    char[] valid = {'0','0','1','0','0','0','1','1','0','1'};
    char[] invalid = {'0','0','1','0','0','A','1','1','0','1'};
    
    Set<Character> set = ALLOWED_CHARS.chars()
                                      .mapToObj(c -> (char) c)
                                      .collect(Collectors.toSet());

    isValid(set, valid);
    isValid(set, invalid);
  }

  public static void isValid(Set<Character> set, char[] array) {
    for (char c: array) {
      if (!set.contains(c)) {
        System.out.println("Invalid");
        return;
      }
    }
    System.out.println("Valid");
  }
}

Another approach (other than regex) — add the ALLOWED_CHARS to a Set and simple check if any element in the array is not present in the Set.

Java如何使用字符串验证字符？

中性美 2025-01-24 21:25:29 浏览 0 评论 0 收藏 0
回复 原文
三人与歌 2025-01-31 14:36:54

我宁愿您建议您尝试 BetterPlayer 。在许多情况下，它具有更多的功能，并且效果很好。

我自己与几个玩家一起工作，包括Chewie，BetterPlayer，FlickVideOplayer，Videoplayer等。在我发现最好的玩法者中，最好用字幕＆amp;最好地处理视频文件。音轨选择等。

Instead of Chewie, I'd rather suggest you to give a try with BetterPlayer. It has got more functionalities and works great in many scenarios.

I've worked with couple of players myself including Chewie, BetterPlayer, FlickVideoPlayer, VideoPlayer etc. Among all I found BetterPlayer best to handle video files better way with subtitles & audio track selection etc.

颤音：视频在Chewie视频播放器组件中未正确显示

噩梦成真你也成魔 2025-01-24 14:36:54 浏览 1 评论 0 收藏 0
回复 原文
三人与歌 2025-01-31 11:33:38

您应该将Frameworks用作Microsoft.netcore.app

packages从Nuget软件包管理器

  1. Microsoft.entityframeworkcore
  2. microsoft.entityframeworkcore.sqlserver

现在您可以运行命令：

脚手架dbContext“ server = 10.10.10.123; database = testdb;用户id = sa;
密码= SA1234“ Microsoft.EntityFrameWorkCore.sqlServer -Outputdir
模型-Context testdbcontext -tables雇员

You should use Frameworks as Microsoft.NETCore.App

Packages from NuGet Package Manager

  1. Microsoft.EntityFrameworkCore
  2. Microsoft.EntityFrameworkCore.SqlServer

Now you can run the command:

Scaffold-DbContext "Server=10.10.10.123;Database=Testdb; user id=sa;
password=Sa1234" Microsoft.EntityFrameworkCore.SqlServer -OutputDir
Models -Context TestDBContext -Tables employee

如何在.NET Core 6中使用DB的第一种方法？

孤独患者 2025-01-24 11:33:38 浏览 3 评论 0 收藏 0
回复 原文
三人与歌 2025-01-31 07:59:57

来自： https：// https：//kubernetes.io/docs/conpepts/concepts/concepts/concepts/concepts/concepts/concepts/concepts/concepts/ /services-networking/indress/＃什么is-ingress

入口不会暴露任意端口或协议。将http和https以外的其他服务曝光到Internet通常使用类型服务的服务。Type= nodePort或service.type = loadBalancer。

NodePort可能是您想要的。这里记录了更多信息和选项：

From: https://kubernetes.io/docs/concepts/services-networking/ingress/#what-is-ingress

An Ingress does not expose arbitrary ports or protocols. Exposing services other than HTTP and HTTPS to the internet typically uses a service of type Service.Type=NodePort or Service.Type=LoadBalancer.

NodePort might be what you are looking for. More information and options are documented here: https://kubernetes.io/docs/concepts/services-networking/service/#publishing-services-service-types

如何更改NGINX入口端口号？

坚持沉默 2025-01-24 07:59:57 浏览 1 评论 0 收藏 0
回复 原文
三人与歌 2025-01-31 02:48:55

对于产品选择上的颜色选择，例如上衣或鞋子，则颜色名称可能很重要。

那里的是颜色名称的标准。它在CSS文档中。

这些CSS颜色名称可能与公司的名称不符，例如“ Dusty Sunset”，它可能是衬衫颜色或油漆颜色。

RGB值也可能非常有用。如果我是设计师或编码器，您可能会想到Gibberish，＃D1ED7E，对我来说意义重大。

视觉显示颜色名称以及色板也对某些用户也有帮助。您不一定需要为屏幕阅读器用户“隐藏”颜色名称。

这是鞋网站的示例。

当前的颜色显示为“海军”，如果我将鼠标与另一种颜色悬停在鼠标上，则显示“ tan”。工具提示并不是这样做的最佳方法，因为它们通常不会在移动设备上显示，而工具提示通常仅适用于鼠标悬停而不是键盘焦点，但它使您有一个显示颜色名称的想法。

如果我在设计系统中，并且想选择一种颜色，那么我可能想要确切的RGB或HSV值，因此显示颜色代码会有所帮助。

For color choices on a product selection, say a blouse or shoes, then color names might be important.

And there is a standard for color names. It's in the CSS documentation.

Those CSS color names might not match a company's name for a color, like "Dusty Sunset", which could be a shirt color or a paint color.

RGB values can also be very useful. What you might think of gibberish, #d1ed7e, might mean a lot to me if I'm a designer or coder.

Visually displaying the color name along with the swatch can also be helpful for some users. You don't necessarily have to "hide" the color name for just screen reader users.

Here's an example from a shoe website.

enter image description here

The current color displays as "Navy" and if I hover with the mouse over the other color, it shows "Tan". Tooltips are not the best way to do this since they're generally not displayed on mobile devices and tooltips often only work with a mouse hover and not keyboard focus, but it gives you an idea of showing color names.

If I'm in a design system and I want to choose a color, I probably want the exact RGB or HSV values so displaying the color code would be helpful.

色板是否应该为scree读者带有标签吗？

若水微香 2025-01-24 02:48:55 浏览 4 评论 0 收藏 0
回复 原文
三人与歌 2025-01-31 02:17:54

问题是 test.lastindexof（“”），如果 test 中没有空间，则返回-1。看看

您可以使用 string.contains 

var firstname = !string.IsNullOrEmpty(test) && test.Contains(" ") 
        ? test.Substring(0, test.LastIndexOf(" ")) 
        : string.IsNullOrEmpty(test) ? string.Empty : test;
var lastname = !string.IsNullOrEmpty(test) && test.Contains(" ") 
        ? test.Split(' ').Last() 
        : string.Empty;

检查 test 中是否有空间。

我建议使用而不是 test！= null 。

The problem is test.LastIndexOf(" ") which returns -1 if there is no space in test. Have a look at MSDN - String.Substring

You can use String.Contains

var firstname = !string.IsNullOrEmpty(test) && test.Contains(" ") 
        ? test.Substring(0, test.LastIndexOf(" ")) 
        : string.IsNullOrEmpty(test) ? string.Empty : test;
var lastname = !string.IsNullOrEmpty(test) && test.Contains(" ") 
        ? test.Split(' ').Last() 
        : string.Empty;

to check if there is a space in test.

I recommend using String.IsNullOrEmpty instead of test != null.

子字符串长度不能小于零

白色秋天 2025-01-24 02:17:54 浏览 2 评论 0 收藏 0
回复 原文
三人与歌 2025-01-31 01:43:39

您可以尝试 str.split ，然后使用 str [0] 访问列表的第一个元素

df['starting_hours'] = df['School_Hours'].str.split('-').str[0].str.strip('AM|PM')

print(df)

      School_Hours starting_hours
0  08:00AM-O3:00PM          08:00

You can try str.split then access the first element of list with str[0]

df['starting_hours'] = df['School_Hours'].str.split('-').str[0].str.strip('AM|PM')

print(df)

      School_Hours starting_hours
0  08:00AM-O3:00PM          08:00

＆＃x27; float＆＃x27;当使用pandas从单元格中提取数据时，对象不是可订阅的错误

以为你会在 2025-01-24 01:43:39 浏览 2 评论 0 收藏 0
回复 原文
三人与歌 2025-01-31 00:54:01

我提出了一个解决方案，这似乎可以解决问题。此功能（Typescript / TensorFlow.js）将分类头扩展到一个类，并将受过训练的权重调整到新形状：

const addClass = (model: tf.LayersModel): tf.LayersModel => {
  const formerHead = model.layers[model.layers.length - 1];
  const formerHeadWeights = formerHead.getWeights();
  const newOutput = tf.layers.dense({
    units: (formerHead as any).units + 1,
    activation: 'softmax',
    weights: [
      tf.tensor((formerHeadWeights[0].arraySync() as number[][]).map((row) => [...row, 0])),
      tf.tensor([...(formerHeadWeights[1].arraySync() as number[]), 0])
    ]
  })
    .apply(model.layers[model.layers.length - 2].output) as tf.SymbolicTensor;

  return tf.model({inputs: model.inputs, outputs: newOutput});
};

I came to a solution, which seems to do the trick. This function (typescript / tensorflow.js) extends the classification head by one class and adapts the trained weights to the new shape:

const addClass = (model: tf.LayersModel): tf.LayersModel => {
  const formerHead = model.layers[model.layers.length - 1];
  const formerHeadWeights = formerHead.getWeights();
  const newOutput = tf.layers.dense({
    units: (formerHead as any).units + 1,
    activation: 'softmax',
    weights: [
      tf.tensor((formerHeadWeights[0].arraySync() as number[][]).map((row) => [...row, 0])),
      tf.tensor([...(formerHeadWeights[1].arraySync() as number[]), 0])
    ]
  })
    .apply(model.layers[model.layers.length - 2].output) as tf.SymbolicTensor;

  return tf.model({inputs: model.inputs, outputs: newOutput});
};

扩展分类头后是否可以调整训练有素的重量的形状？

子栖 2025-01-24 00:54:01 浏览 2 评论 0 收藏 0
回复 原文
三人与歌 2025-01-30 22:49:34

我不太了解C ++，因此花了一些时间编写代码。运算符的实现是用外部.ASM文件编写的。我从creel 在这里）。

// TestTMR.cpp

#include <iostream>
struct TMR {
    double Data[4];
    inline TMR& operator = (const long long Src);
    inline TMR& operator = (const TMR& Src);
    friend inline TMR& operator + (const TMR& O1, const TMR& O2);
    friend std::ostream& operator << (std::ostream& out, const TMR& Src);
 };

TMR vD;

TMR const cA = {1.0, 2.0, 3.0, 4.0};

std::ostream& operator << (std::ostream& out, const TMR& Src)
{
    return out << Src.Data[0] << "  " << Src.Data[1] << "  " << Src.Data[2] << "  " << Src.Data[3];
};

int main()
{
    vD = 4;
    vD = cA;
    vD = cA + vD;
   std::cout << vD << std::endl;
}

;TestTMRasm.asm
.code

; inline TMR& operator = (const long long Src)
??4TMR@@QEAAAEAU0@_J@Z proc
    vcvtsi2sd xmm1, xmm7, rdx
    vbroadcastsd ymm0, xmm1
    vmovdqu ymmword ptr [rcx], ymm0
    ret
??4TMR@@QEAAAEAU0@_J@Z endp

; inline TMR& operator = (const TMR& Src);
??4TMR@@QEAAAEAU0@AEBU0@@Z proc
    vmovdqu ymm0, ymmword ptr [rdx]
    vmovdqu ymmword ptr [rcx], ymm0
    ret
??4TMR@@QEAAAEAU0@AEBU0@@Z endp

; inline TMR& operator + (const TMR& O1, const TMR& O2)
??H@YAAEAUTMR@@AEBU0@0@Z proc
    sub rsp, 20h
    vmovdqu ymm1, ymmword ptr [rcx]
    vmovdqu ymm2, ymmword ptr [rdx]
    vaddpd ymm0, ymm1, ymm2
    vmovdqu ymmword ptr [rsp], ymm0
    lea rax, qword ptr [rsp]
    add rsp, 20h
    ret
??H@YAAEAUTMR@@AEBU0@0@Z endp

end

但是，C ++编译器与Delphi一样，并不会使它们在线（请参阅拆卸代码）。也许我没有正确设置编译选项。如果有人取得积极的结果，请告诉我。事实证明，C ++比Delphi更好，因为在分配之前无需实现类型的转换操作员，并且能够返回对结果的参考。
尽管Delphi将引用对 ca 调用 tmr（CA）时，对常数的本机支持也很好，尽管Delphi只需将引用到 ca 。

    23:     vD = 4;
00007FF7BAC710F4 BA 04 00 00 00       mov         edx,4  
00007FF7BAC710F9 48 8D 0D 28 45 00 00 lea         rcx,[vD (07FF7BAC75628h)]  
00007FF7BAC71100 E8 7B 07 00 00       call        TMR::operator= (07FF7BAC71880h)  
    24:     vD = cA;
00007FF7BAC71105 48 8D 15 C4 21 00 00 lea         rdx,[cA (07FF7BAC732D0h)]  
00007FF7BAC7110C 48 8D 0D 15 45 00 00 lea         rcx,[vD (07FF7BAC75628h)]  
00007FF7BAC71113 E8 77 07 00 00       call        TMR::operator= (07FF7BAC7188Fh)  
    25:     vD = cA + vD;
00007FF7BAC71118 48 8D 15 09 45 00 00 lea         rdx,[vD (07FF7BAC75628h)]  
00007FF7BAC7111F 48 8D 0D AA 21 00 00 lea         rcx,[cA (07FF7BAC732D0h)]  
00007FF7BAC71126 E8 6D 07 00 00       call        operator+ (07FF7BAC71898h)  
00007FF7BAC7112B 48 8B D0             mov         rdx,rax  
00007FF7BAC7112E 48 8D 0D F3 44 00 00 lea         rcx,[vD (07FF7BAC75628h)]  
00007FF7BAC71135 E8 55 07 00 00       call        TMR::operator= (07FF7BAC7188Fh)

I don't know C++ very well, so it took some time to write the code.The implementation of the operators is written in an external .asm file. I took this trick from Creel here.

// TestTMR.cpp

#include <iostream>
struct TMR {
    double Data[4];
    inline TMR& operator = (const long long Src);
    inline TMR& operator = (const TMR& Src);
    friend inline TMR& operator + (const TMR& O1, const TMR& O2);
    friend std::ostream& operator << (std::ostream& out, const TMR& Src);
 };

TMR vD;

TMR const cA = {1.0, 2.0, 3.0, 4.0};

std::ostream& operator << (std::ostream& out, const TMR& Src)
{
    return out << Src.Data[0] << "  " << Src.Data[1] << "  " << Src.Data[2] << "  " << Src.Data[3];
};

int main()
{
    vD = 4;
    vD = cA;
    vD = cA + vD;
   std::cout << vD << std::endl;
}

;TestTMRasm.asm
.code

; inline TMR& operator = (const long long Src)
??4TMR@@QEAAAEAU0@_J@Z proc
    vcvtsi2sd xmm1, xmm7, rdx
    vbroadcastsd ymm0, xmm1
    vmovdqu ymmword ptr [rcx], ymm0
    ret
??4TMR@@QEAAAEAU0@_J@Z endp

; inline TMR& operator = (const TMR& Src);
??4TMR@@QEAAAEAU0@AEBU0@@Z proc
    vmovdqu ymm0, ymmword ptr [rdx]
    vmovdqu ymmword ptr [rcx], ymm0
    ret
??4TMR@@QEAAAEAU0@AEBU0@@Z endp

; inline TMR& operator + (const TMR& O1, const TMR& O2)
??H@YAAEAUTMR@@AEBU0@0@Z proc
    sub rsp, 20h
    vmovdqu ymm1, ymmword ptr [rcx]
    vmovdqu ymm2, ymmword ptr [rdx]
    vaddpd ymm0, ymm1, ymm2
    vmovdqu ymmword ptr [rsp], ymm0
    lea rax, qword ptr [rsp]
    add rsp, 20h
    ret
??H@YAAEAUTMR@@AEBU0@0@Z endp

end

However, the C++ compiler, like Delphi, does not make them inline (see the disassembled code). Maybe I didn't set the compilation options correctly. If anyone achieves a positive result, please let me know. C++ proved to be better than Delphi in that there is no need to implement type conversion operators before assignment, as well as the ability to return a reference to the result.
Native support for constants is also good, although Delphi will simply pass the reference to cA to the statement when calling TMR(cA).

    23:     vD = 4;
00007FF7BAC710F4 BA 04 00 00 00       mov         edx,4  
00007FF7BAC710F9 48 8D 0D 28 45 00 00 lea         rcx,[vD (07FF7BAC75628h)]  
00007FF7BAC71100 E8 7B 07 00 00       call        TMR::operator= (07FF7BAC71880h)  
    24:     vD = cA;
00007FF7BAC71105 48 8D 15 C4 21 00 00 lea         rdx,[cA (07FF7BAC732D0h)]  
00007FF7BAC7110C 48 8D 0D 15 45 00 00 lea         rcx,[vD (07FF7BAC75628h)]  
00007FF7BAC71113 E8 77 07 00 00       call        TMR::operator= (07FF7BAC7188Fh)  
    25:     vD = cA + vD;
00007FF7BAC71118 48 8D 15 09 45 00 00 lea         rdx,[vD (07FF7BAC75628h)]  
00007FF7BAC7111F 48 8D 0D AA 21 00 00 lea         rcx,[cA (07FF7BAC732D0h)]  
00007FF7BAC71126 E8 6D 07 00 00       call        operator+ (07FF7BAC71898h)  
00007FF7BAC7112B 48 8B D0             mov         rdx,rax  
00007FF7BAC7112E 48 8D 0D F3 44 00 00 lea         rcx,[vD (07FF7BAC75628h)]  
00007FF7BAC71135 E8 55 07 00 00       call        TMR::operator= (07FF7BAC7188Fh)

＆＃x41d;用覆盖方法定义恒定的托管记录

夜雨飘雪 2025-01-23 22:49:34 浏览 1 评论 0 收藏 0
回复 原文
三人与歌 2025-01-30 22:48:15

我认为您的政策可能是正确的，只需创建并使用它即可。

IAM可能正在抛出此错误，在我的情况下，这是这样做的：

RedShift-Serverless：IAM无法识别此服务。该服务可能包括错字或可能是预览或自定义服务。

由于Redshift Server无预览仍在预览中，因此IAM尚未识别它，因此不要被IAM抛弃，而没有识别该服务。

来源：

预览服务 - 预览中的服务不支持视觉编辑器。如果您正在参加预览，则可以忽略警告并继续，尽管您必须手动输入操作和资源来完成您的策略。另外，您可以选择键入JSON选项卡或粘贴JSON策略文档。

I think that your policy might be correct, just create it and use it.

IAM might be throwing this error which in does in my case:

redshift-serverless: IAM does not recognize this service. The service might include a typo or might be a previewed or custom service.

As Redshift serverless is still in preview, it is not yet recognized by iam, don't be thrown off by IAM not recognizing the service.

Source: https://docs.aws.amazon.com/IAM/latest/UserGuide/troubleshoot_policies.html?icmpid=docs_iam_console#troubleshoot_policies-unrecognized-visual

Preview service – Services that are in preview do not support the visual editor. If you are participating in the preview, you can ignore the warning and continue, though you must manually type the actions and resource ARNs to complete your policy. Alternatively, you can choose the JSON tab to type or paste a JSON policy document.

IAM用户访问Amazon RedShift无服务器

望笑 2025-01-23 22:48:15 浏览 4 评论 0 收藏 0
回复 原文
更多

热门标签

操作系统 程序设计 IT运维 Linux系统管理 JavaScript 服务器应用 solaris C/C++ PHP Shell BSD Vue.js aix Oracle Python HTML 系统管理 HTML5 CSS 前端
更多

推荐作者

櫻之舞

文章 0 评论 0

弥枳

文章 0 评论 0

m2429

文章 0 评论 0

寻找一个思念的角度

文章 0 评论 0

野却迷人

文章 0 评论 0

我怀念的。

文章 0 评论 0

更多

友情链接

文江博客
    我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站，即表示您同意使用 Cookies 和您的相关数据。
    原文