FXCop 10 was Shipped with Windows 7.1 SDK

| by Jonathan Allen Follow 500 Followers on Jun 21, 2010. Estimated reading time: 1 minute |

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For .NET developers who want the rigor of code analysis without the expense of Visual Studio Premium, FXCop is the tool for choice. But with FXCop 1.36 pulled from Microsoft Downloads without warning, many developers were left wondering what happened. Fortunately this tool is still available if you know where to look.

To start, developers need to install the .NET tools from Microsoft Windows SDK for Windows 7 and .NET Framework 4. Don’t bother looking for FXCop in the installer, you won’t find it mentioned there or in any of the usual pre-installation documentation. But a careful review of the post-install release notes will reveal this passage:

6.3.6 FXCop Setup is Now Located Under the Window SDK “\Bin” Directory.
The installer for FXCop, fxcopsetup.exe, is now located in [Program Files]\Microsoft SDKs\Windows\v7.1\Bin\FXCop.

Running that installer will setup FXCop 10.0. Like previous editions, this version is missing many of the rules found in Visual Studio. Also there are some changes to how rules are grouped such as splitting up the Security group into “Security” and “Security Transparency”. Fortunately FXCop uses the same numbering scheme for rules, so if you can’t find a rule by name try looking for its CAxxxx ID.

Visual Studio FXCop
Design Design
Globalization Globalization
Interoperability Interoperability
Mobility Mobility
Naming Naming
Performance Performance
Portability Portability
Security Security & Security Transparency
Usage Usage

This version of FXCop supports both the CLR 2.0 and 4.0 runtimes.

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Minimum install set for FxCop? by Kevin McFarlane

What's the minimum set of options to select in the web installer to get FxCop? If I choose just Tools in the .NET Development set is that sufficient?

Thanks by Alexandre Victoor

Thanks a lot for this very useful news. I was searching to download FxCop 1.36 in vain ...

Re: Minimum install set for FxCop? by Jonathan Allen

I think so, but since it isn't marked I'm not really certain.

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