[build] Provide a default $(Configuration) value

[jonpryor]

When Directory.Build.props is imported by MSBuild, the
$(Configuration) property will only be defined if it was excplitily
provided on the command line. As such, when building the repo when
using make(1), it builds:

    $ make prepare
    $ make all
    # works!

After make all, if we attempt to manually build a single .csproj,
it may fail:

    $ msbuild tests/Java.Interop-Tests/Java.Interop-Tests.csproj
    ...
    BuildInteropTestJar:
      "" -source 1.6 -target 1.6 -bootclasspath "" -d "obj/Debug/it-classes" -classpath …
    …: error MSB3073: The command """ -source 1.6 -target 1.6 …" exited with code 127.

Building a single .csproj works if the Configuration property is
explicitly provided:

    $ msbuild tests/Java.Interop-Tests/Java.Interop-Tests.csproj /p:Configuration=Debug
    # no error

Update Directory.Build.props so that if no $(Configuration) is
provided we default to the Debug configuration. This allows all the
relevant $(Configuration)-dependent properties to have correct and
consistent values, which allows projects to build as intended:

    $ msbuild tests/Java.Interop-Tests/Java.Interop-Tests.csproj
    # no errors

[jonathanpeppers]

I was hitting this locally yesterday, and I just changed Directory.Build.props:

-    <_Configuration Condition=" '$(Configuration)' == 'Gendarme' ">Debug</_Configuration>
+    <_Configuration Condition=" '$(Configuration)' == 'Gendarme' Or '$(Configuration)' == '' ">Debug</_Configuration>

I don't know if this has another issue though ^^

[jpobst]

Anything with Gendarme can be killed, as we no longer use it. I think Peppers may be on to something, and we could just add the standard setter at the top of Directory.Build.props:

<Configuration Condition=" '$(Configuration)' == '' ">Debug</Configuration>
<Platform Condition=" '$(Platform)' == '' ">AnyCPU</Platform>

[jonathanpeppers]

👍 LTGM if it builds