问题 {#heading}

我在Get-FileHash函数中发现了以下语句（来源：Microsoft.PowerShell.Utility.psm1）：

$hasherType = "System.Security.Cryptography.${Algorithm}CryptoServiceProvider" -as [Type]

我的问题是：

写"-as [Type]"和简单写"[Type]$xxxx"之间有什么区别（如果有的话）？
${Algorithm}也可以定义为：$($Algorithm) 吗？

英文:

I found in the Get-FileHash function (source: Microsoft.PowerShell.Utility.psm1) this statement:

$hasherType = &quot;System.Security.Cryptography.${Algorithm}CryptoServiceProvider&quot; -as [Type]

my questions are:

what is the difference (if any) writing "-as [Type]" or just simply "[Type]$xxxx?
${Algorithm} could be defined as well: $($Algorithm)?

答案1 {#1}

得分: 3

使用-as [Type]或者只是简单地使用[Type]$xxxx有什么区别（如果有的话）？

-as操作符受到了C#中的所谓无检查转换操作符的启发（对应于C#关键字as） - 在转换失败的情况下，该操作将简单地评估为$null。这种行为与常规的[type]$value转换不同，后者在失败时会引发异常。

测试$null通常比捕获和处理异常更快，因此在预期某些值不符合给定类型的情况下，这可能会有用：

$bunchOfMixedValues | ForEach-Object {
  $thing = $_ -as [TargetType]
  if ($null -ne $thing) {
    # 处理 $thing
  }
  else {
    # 无需关心无效输入，什么都不做
  }
}

${Algorithm} 也可以这样定义：$($Algorithm)吗？

最终结果是相同的，是的。

${...}表示与$...完全相同的元素 - 变量路径文字表达式。您可以将{和}括号视为变量表达式的_引号_，换句话说，它们_限定了_变量名称的边界 - 因此PowerShell知道不要尝试展开$AlgorithmCryptoServiceProvider。因此，在生成代码时使用${...}语法更安全，这就是为什么在许多由Microsoft编写的模块中看到它的原因。英文:

> what is the difference (if any) writing -as [Type] or just simply [Type]$xxxx?

The -as operator is heavily inspired by the so-called unchecked conversion operator from C# (corresponding to the C# keyword as) - the operation will simply evaluate to $null in case of conversion failure. This behavior differs significantly from a regular [type]$value cast, which will throw an exception on failure.

Testing for $null is usually faster than capturing and handling an exception, so this can be useful in situations where you expect some values not to conform to a given type:

$bunchOfMixedValues |ForEach-Object {
  $thing = $_ -as [TargetType]
  if ($null -ne $thing) {
    # process $thing
  }
  else {
    # no need to care about invalid input, do nothing
  }
}

> ${Algorithm} could be defined as well: $($Algorithm)?

The end result would be the same, yes.

${...} represents the exact same element as $... - a variable path literal expression. You can think of the { and } brackets as quotation marks for variable expressions, in other words they qualify the boundary of the variable name - so PowerShell knows not to attempt to expand $AlgorithmCryptoServiceProvider instead.

It's therefore safer to use the ${...} syntax when generating code for example, which is why you'll see it in lots of Microsoft-authored modules.