How to declare an unsigned 32-bit integer?

Question

Is there a way to declare a 32-bit unsigned integer in PowerShell?

I'm trying to add an unsigned (begin with 0xf) i.e. 0xff000000 + 0xAA, but it turned out to be a negative number whereas I want it to be 0xff0000AA.

Solution 1

0xff00000AA is too large for a 32 bit unsigned integer, use uint64

PS> [uint64]0xff00000AA
68451041450

Solution 2

The code from the currently accepted answer causes the following error on my system:

This is because PowerShell always tries to convert hex values to int first so even if you cast to uint64, the environment will complain if the number has a negative int value. You can see this from the following example:

PS C:\> [uint64] (-1)
Cannot convert value "-1" to type "System.UInt64". Error: "Value was either too large or too small for a UInt64."
At line:1 char:26
+ Invoke-Expression $ENUS; [uint64] (-1)
+                          ~~~~~~~~~~~~~
    + CategoryInfo          : InvalidArgument: (:) [], RuntimeException
    + FullyQualifiedErrorId : InvalidCastIConvertible

You need to cast the string representation of the number to uint64 to avoid this:

[uint64]"0xff000000" + [uint64]"0xAA"

Solution 3

The simplest way is to cast to [uint32] from a long literal (with the L suffix). You don't need to do the math in [uint64] or cast from a string

[uint32]0xff000000L + 0xAA

In PowerShell 6.2 a series of new suffixes have been added including a u suffix fof uint32 so this can be done even simpler

0xff000000u + 0xAA

Comments

[SpellingD]

I do not get the error you described, and the answer @Shay gave works just fine for me. Using 64bit PSv2. However, just entering 0xff00000AA does auto-convert to 68451041450 without the need for casting.

[Emperor XLII]

@SpellingD: The difference is Pencho used 0xFF0000AA (four zeros) to get a signed Int32 negative value (which fails to convert to an unsigned value), where the question uses 0xFF00000AA (five zeros) to get a signed Int64 positive value (which converts successfully).

[Bob Reynolds]

This does not answer the question actually asked, i.e., how to declare an unsigned 32-bit integer, with the remainder of the text implying the high bit is set. The answer provided by Pencho does provide a means to do this. Given the question asked, it seems reasonable to assume the example provided just has more hex digits than intended and not that it is "too large" for a 32 bit integer (signed or otherwise).

[jpmc26]

Do not use images to post code or errors. Copy them into the answer.

[phuclv]

0xff000000 + 0xAA = 0xff0000AA, not 0xff00000AA. All of them fit in uint32