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dynamically determine .net baseframework version of exe or dllHi All,
Whether in a Winform application (.exe) or a .Net assembly (.dll), is there a way to determine what version of the .net baseframework the run-time module (i.e. exe or dll) was loaded with (is using)? Seems that using reflection this is possible...just can't seem to figure out what classes/instances will get me there. Thanks in advance! Tony Anthony,
>Whether in a Winform application (.exe) or a .Net assembly (.dll), is there System.Environment.Version>a way to determine what version of the .net baseframework the run-time >module (i.e. exe or dll) was loaded with (is using)? Mattias -- Mattias Sjögren [C# MVP] mattias @ mvps.org http://www.msjogren.net/dotnet/ | http://www.dotnetinterop.com Please reply only to the newsgroup. If you are looking for a non-programmatic way you can enable CLR Load
Logging which I blogged about http://pointerx.net/blogs/glozano/archive/2006/03/23/84.aspx Gabriel Lozano-Morán Show quote "Anthony LaMark" <anth***@eXcSoftware.com> wrote in message news:OiNVusYEHHA.992@TK2MSFTNGP03.phx.gbl... > Hi All, > > Whether in a Winform application (.exe) or a .Net assembly (.dll), is > there a way to determine what version of the .net baseframework the > run-time module (i.e. exe or dll) was loaded with (is using)? Seems that > using reflection this is possible...just can't seem to figure out what > classes/instances will get me there. Thanks in advance! > > Tony > Hi
The framework library, that includes .exes and .dlls, (eg :say System.dll) are .net assemblies.... So you can always query the metadata of assebmly through reflection. hope this is helpful Thanks -Cnu Anthony LaMark wrote: Show quote > Hi All, > > Whether in a Winform application (.exe) or a .Net assembly (.dll), is there > a way to determine what version of the .net baseframework the run-time > module (i.e. exe or dll) was loaded with (is using)? Seems that using > reflection this is possible...just can't seem to figure out what > classes/instances will get me there. Thanks in advance! > > Tony |
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