0

引用数组

工作正常！

在 PowerShell 中通过引用传递数组的正常方法似乎工作正常：

Function Swap-Array ($theArray, $theArrayB, [int]$indexToSwap) {
  $temp = $theArrayA[$indexToSwap];
  $theArrayA[$indexToSwap] = $theArrayB[$indexToSwap];
  $theArrayB[$indexToSwap] = $temp;
} 

$a = @(1,2,3,4)
$b = @(3,2,4,1)
$a
$b
Swap-Array $a, $b, 2
$a
$b

输出：

a
-
1
2
3
4

b
-
3
2
4
1

a
-
1
2
4
3

b
-
3
2
3
1

问题

添加对象

当引用数组是非静态的 PSObjects 容器并且我正在尝试添加新记录时，就会出现问题。修改现有记录似乎没问题！

Function Swap-Apples($objectA, $objectB, $indexToSwap) {
  $temp = $objectA[$indexToSwap].Apples;
  $objectA[$indexToSwap].Apples = $objectB[$indexToSwap].Apples;
  $objectB[$indexToSwap].Apples = $temp;
}

Function Swap-Oranges($objectA, $objectB, $indexToSwap) {
  $temp = $objectA[$indexToSwap].Oranges;
  $objectA[$indexToSwap].Oranges = $objectB[$indexToSwap].Oranges;
  $objectB[$indexToSwap].Oranges = $temp;
}

<# heres the problematic bit #>
Function Add-Fruit ($object, [int]$howManyApples, [int]$howManyOranges) {
  $hAdd = @{
    Apples=$howManyApples
    Oranges=$howManyOranges
  }
  $hToAdd = New-Object -TypeName PSObject -Property $hAdd;
  $object += $hToAdd;
}

$a = @();
$b = @();
$a1 = @{
  Apples=3
  Oranges=2
}
$b1 = @{
  Apples=5
  Oranges=7
}
$a2 = @{
  Apples=6
  Oranges=3
}
$b2 = @{
  Apples=1
  Oranges=5
}
$aObject1 = New-Object -TypeName PSObject -Property $a1;
$bObject1 = New-Object -TypeName PSObject -Property $b1;
$aObject2 = New-Object -TypeName PSObject -Property $a2;
$bObject2 = New-Object -TypeName PSObject -Property $b2;
$a += $aObject1; $a += $aObject2;
$b += $bObject1; $b += $aObject2;

Write-Host "Values of A";
$a | Format-List

Write-Host "Values of B";
$b | Format-List

Write-Host "Now lets make a trade`!";
Swap-Apples $a $b 0
Swap-Oranges $a $b 1

Write-Host "Values of A";
$a | Format-List

Write-Host "Values of B";
$b | Format-List

Write-Host "Hey, I brought more fruit for A`!";
Add-Fruit -object $a -howManyApples 5 -howManyOranges 2

Write-Host "Values of A";
$a | Format-List

Write-Host "I brought more fruit for B too`!";
Add-Fruit -object $b -howManyApples 5 -howManyOranges 3

Write-Host "Values of B";
$b | Format-List

输出

Values of A


Oranges : 2
Apples  : 3

Oranges : 3
Apples  : 6



Values of B


Oranges : 7
Apples  : 5

Oranges : 3
Apples  : 6



Now lets make a trade!
Values of A


Oranges : 2
Apples  : 5

Oranges : 3
Apples  : 6



Values of B


Oranges : 7
Apples  : 3

Oranges : 3
Apples  : 6



Hey, I brought more fruit for A!
Values of A


Oranges : 2
Apples  : 5

Oranges : 3
Apples  : 6



I brought more fruit for B too!
Values of B


Oranges : 7
Apples  : 3

Oranges : 3
Apples  : 6

和函数似乎工作正常Swap-Apples。Swap-Oranges该程序在最后一段崩溃了，试图给 A 和 B 更多的果实！否则这通常会在本地范围内工作。我觉得这由于引用传递而分崩离析。

在这个程序结束时我将如何解决这个问题？

4

2 回答 2

0

解决方案

动态数组列表

在 PowerShell 中，您可以创建固定大小和动态分配的数组。如果我想给 A 和 B 更多的对象，我必须告诉 PowerShell 这是一个 ArrayList 而不是典型的标准 Array。

这意味着我不能像这样声明我的数组：

$a = @();
$b = @();

.NET 中的这种类型被调用System.Collections.ArrayList，并且可以在 PowerShell 程序中通过引用传递，如下所示：

$a = New-Object -TypeName 'System.Collections.ArrayList';
$b = New-Object -TypeName 'System.Collections.ArrayList';

现在它不是固定大小的，所以我可以随意在我的程序中的任何地方添加记录，甚至通过引用！

这是解决方案：

Function Swap-Apples($objectA, $objectB, $indexToSwap) {
  $temp = $objectA[$indexToSwap].Apples;
  $objectA[$indexToSwap].Apples = $objectB[$indexToSwap].Apples;
  $objectB[$indexToSwap].Apples = $temp;
}

Function Swap-Oranges($objectA, $objectB, $indexToSwap) {
  $temp = $objectA[$indexToSwap].Oranges;
  $objectA[$indexToSwap].Oranges = $objectB[$indexToSwap].Oranges;
  $objectB[$indexToSwap].Oranges = $temp;
}

<# ArrayList! #>
Function Add-Fruit ([System.Collections.ArrayList]$object, [int]$howManyApples, [int]$howManyOranges) {
  $hAdd = @{
    Apples=$howManyApples
    Oranges=$howManyOranges
  }
  $hToAdd = New-Object -TypeName PSObject -Property $hAdd;
  <# We have to call the ArrayList Add method to add to our dynamic object #>
  $object.Add($hToAdd);
}

$a = New-Object -TypeName 'System.Collections.ArrayList';
$b = New-Object -TypeName 'System.Collections.ArrayList';
$a1 = @{
  Apples=3
  Oranges=2
}
$b1 = @{
  Apples=5
  Oranges=7
}
$a2 = @{
  Apples=6
  Oranges=3
}
$b2 = @{
  Apples=1
  Oranges=5
}
$aObject1 = New-Object -TypeName PSObject -Property $a1;
$bObject1 = New-Object -TypeName PSObject -Property $b1;
$aObject2 = New-Object -TypeName PSObject -Property $a2;
$bObject2 = New-Object -TypeName PSObject -Property $b2;
<# Here we call the ArrayList Add method #>
$a.Add($aObject1); $a.Add($aObject2);
$b.Add($bObject1); $b.Add($aObject2);

Write-Host "Values of A";
$a | Format-List

Write-Host "Values of B";
$b | Format-List

Write-Host "Now lets make a trade`!";
Swap-Apples $a $b 0
Swap-Oranges $a $b 1

Write-Host "Values of A";
$a | Format-List

Write-Host "Values of B";
$b | Format-List

Write-Host "Hey, I brought more fruit for A`!";
Add-Fruit -object $a -howManyApples 5 -howManyOranges 2

Write-Host "Values of A";
$a | Format-List

Write-Host "I brought more fruit for B too`!";
Add-Fruit -object $b -howManyApples 5 -howManyOranges 3

Write-Host "Values of B";
$b | Format-List

以及（有点）预期的输出（见下文）：

0
1
0
1
Values of A


Oranges : 2
Apples  : 3

Oranges : 3
Apples  : 6



Values of B


Oranges : 7
Apples  : 5

Oranges : 3
Apples  : 6



Now lets make a trade!
Values of A


Oranges : 2
Apples  : 5

Oranges : 3
Apples  : 6



Values of B


Oranges : 7
Apples  : 3

Oranges : 3
Apples  : 6



Hey, I brought more fruit for A!
Values of A


Oranges : 2
Apples  : 5

Oranges : 3
Apples  : 6

Oranges : 2
Apples  : 5



I brought more fruit for B too!
Values of B


Oranges : 7
Apples  : 3

Oranges : 3
Apples  : 6

Oranges : 3
Apples  : 5

额外输出？

如果您在输出中观察到：

0
1
0
1

ArrayList 的 Add 方法返回您添加的记录的索引。由于它返回此值，因此它会从您的管道中退出到 Std-Out，因此请确保相应地引导它。

如果您的程序中不需要此输出，就像我在这里不需要的那样，请将其通过管道传输到 null 设备，如下所示：

$a.Add($aObject1) | Out-Null; $a.Add($aObject2) | Out-Null;
$b.Add($bObject1) | Out-Null; $b.Add($aObject2) | Out-Null;

这是最终的程序/输出：

代码：（Arraytest.ps1）

Function Swap-Apples($objectA, $objectB, $indexToSwap) {
  $temp = $objectA[$indexToSwap].Apples;
  $objectA[$indexToSwap].Apples = $objectB[$indexToSwap].Apples;
  $objectB[$indexToSwap].Apples = $temp;
}

Function Swap-Oranges($objectA, $objectB, $indexToSwap) {
  $temp = $objectA[$indexToSwap].Oranges;
  $objectA[$indexToSwap].Oranges = $objectB[$indexToSwap].Oranges;
  $objectB[$indexToSwap].Oranges = $temp;
}

<# ArrayList! #>
Function Add-Fruit ([System.Collections.ArrayList]$object, [int]$howManyApples, [int]$howManyOranges) {
  $hAdd = @{
    Apples=$howManyApples
    Oranges=$howManyOranges
  }
  $hToAdd = New-Object -TypeName PSObject -Property $hAdd;
  <# We have to call the ArrayList Add method to add to our dynamic object #>
  $object.Add($hToAdd) | Out-Null;
}

$a = New-Object -TypeName 'System.Collections.ArrayList';
$b = New-Object -TypeName 'System.Collections.ArrayList';
$a1 = @{
  Apples=3
  Oranges=2
}
$b1 = @{
  Apples=5
  Oranges=7
}
$a2 = @{
  Apples=6
  Oranges=3
}
$b2 = @{
  Apples=1
  Oranges=5
}
$aObject1 = New-Object -TypeName PSObject -Property $a1;
$bObject1 = New-Object -TypeName PSObject -Property $b1;
$aObject2 = New-Object -TypeName PSObject -Property $a2;
$bObject2 = New-Object -TypeName PSObject -Property $b2;
<# Here we call the ArrayList Add method #>
$a.Add($aObject1) | Out-Null; $a.Add($aObject2) | Out-Null;
$b.Add($bObject1) | Out-Null; $b.Add($aObject2) | Out-Null;

Write-Host "Values of A";
$a | Format-List

Write-Host "Values of B";
$b | Format-List

Write-Host "Now lets make a trade`!";
Swap-Apples $a $b 0
Swap-Oranges $a $b 1

Write-Host "Values of A";
$a | Format-List

Write-Host "Values of B";
$b | Format-List

Write-Host "Hey, I brought more fruit for A`!";
Add-Fruit -object $a -howManyApples 5 -howManyOranges 2

Write-Host "Values of A";
$a | Format-List

Write-Host "I brought more fruit for B too`!";
Add-Fruit -object $b -howManyApples 5 -howManyOranges 3

Write-Host "Values of B";
$b | Format-List

输出：

Values of A


Oranges : 2
Apples  : 3

Oranges : 3
Apples  : 6



Values of B


Oranges : 7
Apples  : 5

Oranges : 3
Apples  : 6



Now lets make a trade!
Values of A


Oranges : 2
Apples  : 5

Oranges : 3
Apples  : 6



Values of B


Oranges : 7
Apples  : 3

Oranges : 3
Apples  : 6



Hey, I brought more fruit for A!
Values of A


Oranges : 2
Apples  : 5

Oranges : 3
Apples  : 6

Oranges : 2
Apples  : 5



I brought more fruit for B too!
Values of B


Oranges : 7
Apples  : 3

Oranges : 3
Apples  : 6

Oranges : 3
Apples  : 5
于 2018-02-26T11:28:51.117 回答
0

它对我有用：

Function Swap-Array ($theArrayA, $theArrayB, [int]$indexToSwap) 
{
  $temp = $theArrayA[$indexToSwap]
  $theArrayA[$indexToSwap] = $theArrayB[$indexToSwap]
  $theArrayB[$indexToSwap] = $temp
} 

$a = @(1,2,3,4)
$b = @(3,2,4,1)
$a
$b
Swap-Array $a $b 2
$a
$b

只需消除昏迷。

于 2018-02-26T16:17:19.313 回答