My code works perfectly like this.
But I want to remove "org.apache.commons.lang3.*" (Whole address) and just write * so it can get all files in the current folder and subdirectories. I wanted to make my code work on different projects and don't want to change it again and again. I want to make it universal. I just want to point to the target folder and let it get all files in the folder and subdirectories files. So far I have tried:
( .**.* ) ( .* ) (.* / ) (.* \ )
Any clue how to do this?
This is not currently possible with the pitest Ant plugin. The globs are matched against all classes in the classpath, so using an 'everything' glob results on all classes being instrumented.
If no glob is supplied to the pitest maven plugin, it will scan the source directories and construct a filter based on the packages found there. There is no plan to port this functionality to Ant.
Related
Given I have a list of files, e.g foo/src/main.cpp, foo/src/bar.cpp, foo/README.md is it possible to determine which of those files are part of a bazel package?
In my example, the output would e.g. be foo/src/main.cpp, foo/src/bar.cpp since the README.md would not be part of the build.
One way to do this would be to call bazel query on each file and see if it results in an output, but that is quite inefficient and so I was wondering if there is an easier way.
Background: I am trying to determine if a changes in a set of files have an impact on a target, and I want to use bazel query somepath(//some/target, set($FILES)) for that, but this will fail if any of the files in $FILES is not part of a BUILD file.
How about flipping it around and querying for all the source files of the target with:
bazel query 'kind("source file", deps(//some:target))'
and then checking if the result has any of the files in the set
please take a look at the bin-win target in my repository here:
https://github.com/thinlizzy/bazelexample/blob/master/demo/BUILD#L28
it seems to be properly packing the executable inside a file named bin-win.tar.gz, but I still have some questions:
1- in my machine, the file is being generated at this directory:
C:\Users\John\AppData\Local\Temp_bazel_John\aS4O8v3V\execroot__main__\bazel-out\x64_windows-fastbuild\bin\demo
which makes finding the tar.gz file a cumbersome task.
The question is how can I make my bin-win target to move the file from there to a "better location"? (perhaps defined by an environment variable or a cmd line parameter/flag)
2- how can I include more files with my executable? My actual use case is I want to supply data files and some DLLs together with the executable. Should I use a filegroup() rule and refer its name in the "srcs" attribute as well?
2a- for the DLLs, is there a way to make a filegroup() rule to interpret environment variables? (e.g: the directories of the DLLs)
Thanks!
Look for the bazel-bin and bazel-genfiles directories in your workspace. These are actually junctions (directory symlinks) that Bazel updates after every build. If you bazel build //:demo, you can access its output as bazel-bin\demo.
(a) You can also set TMP and TEMP in your environment to point to e.g. c:\tmp. Bazel will pick those up instead of C:\Users\John\AppData\Local\Temp, so the full path for the output directory (that bazel-bin points to) will be c:\tmp\aS4O8v3V\execroot\__main__\bazel-out\x64_windows-fastbuild\bin.
(b) Or you can pass the --output_user_root startup flag, e.g. bazel--output_user_root=c:\tmp build //:demo. That will have the same effect as (a).
There's currently no way to get rid of the _bazel_John\aS4O8v3V\execroot part of the path.
Yes, I think you need to put those files in pkg_tar.srcs. Whether you use a filegroup() rule is irrelevant; filegroup just lets you group files together, so you can refer to the group by name, which is useful when you need to refer to the same files in multiple rules.
2.a. I don't think so.
I've got an iOS framework that has a dependency on the (presumably Google maintained) pod called '!ProtoCompiler'. In order to build my framework I'm going to need it in the sandbox. So, I have a genrule and can try to include it with
src = glob(['Pods/!ProtoCompiler/**/*']) but I get the following error:
ERROR: BUILD:2:1: //Foo:framework-debug: invalid label 'Pods/!ProtoCompiler/google/protobuf/any.proto' in element 1118 of attribute 'srcs' in 'genrule' rule: invalid target name 'Pods/!ProtoCompiler/google/protobuf/any.proto': target names may not contain '!'.
As is, this seems like a total blocker for me using bazel to do this build. I don't have the ability to rename the pod directory as far as I can tell. As far as I can tell, the ! prohibition is supposed to be for target labels, is there any way I can specify that this is just a file, not a label? Or are those two concepts completely melded in bazel?
(Also, if I get this to work I'm worried about the fact that this produces a .framework directory and it seems like rules are expected to produces files only. Maybe I'll zip it up and then unzip it as part of the build of the test harness.)
As far as I can tell, the ! prohibition is supposed to be for target
labels, is there any way I can specify that this is just a file, not a
label? Or are those two concepts completely melded in bazel?
They are mostly molded.
Bazel associates a label with all source files in a package that appear in BUILD files, so that you can write srcs=["foo.cc", "//bar:baz.cc"] in a build rule and it'll work regardless of foo.cc and baz.cc being a source file, a generated file, or a build rule's name that produces files suitable for this particular srcs attribute.
That said you can of course have any file in the package, but if the name won't allow Bazel to derive a label from it, then you can't reference them in the BUILD file. Since glob is evaluated during loading and is expanded to a list of labels, using glob won't work around this limitation.
(...) it seems like rules are expected to produces files only. Maybe
I'll zip it up and then unzip it as part of the build of the test
harness.
Yes, that's the usual approach.
I have C++ header file dependencies that I specify in my waf script with the includes=... parameter to bld.program().
I know the waf build configuration sees the includes because my program compiles correctly.
However, when I change a header file, waf does not detect the change. That is, when I run waf build after changing the contents of an included header, nothing gets recompiled.
Isn't waf supposed to determine #include "..." dependencies automatically?
How can I troubleshoot this?
I have looked in the build/c4che directory to see if I could make sense of the configuration files stored there. Mention of "include" in the waf generated .py files is suspiciously absent.
I am using waf version 1.9.0.
I have also tried this with waf 1.8.19 and got the same result.
EDIT: I replaced my original complicated wscript with the much simpler one listed below, and I still get the same behavior.
Here is my wscript:
top = '.'
out = 'build'
CXXFLAGS = ['-fopenmp', '-Wall', '-Werror', '-std=c++11', '-Wl,--no-as-needed']
def options(ctx):
ctx.load('compiler_cxx')
def configure(ctx):
ctx.load('compiler_cxx')
ctx.env.CXXFLAGS = CXXFLAGS
def build(ctx):
ctx.program(source="test_config_parser.cpp", target="test_config_parser", includes=["../include"], lib=['pthread', 'gomp'])
Your problem is that you use includes out of the project's directory. By default waf does not use external includes as dependencies (like system includes) to speed up things. Solutions I know of :
1/
Organize your project to have your include directory under the waf top directory :
top_dir/
wscript
include/
myinclude.h
sources/
mysource.cpp
2/
Change top directory. I think top = .. should work (not tested).
3/
Tell waf to go absolute by adding this lines at the beginning of build():
waflib.Tools.c_preproc.go_absolute=True
waflib.Tools.c_preproc.standard_includes=[]
4/
Use gcc dependencies by loading the gccdeps waf module.
Solution 1/ is probably the best.
By the way I prefer to have my build directory out of the source tree. Use out = ../build in your wscript, if you want to build out of the source tree.
my2c
I have a couple of job in Jenkins that archive artifact from the source tree for another job (some unit tests or alike). I have the current situation :
top_dir
\scripts_dir
\some_files
\dir1
\dir2
\dir3
\other_dir
I would like to archive all that is in "top_dir" including the files in "scripts_dir", but not the subdirectories "dir1, dir2,...", which I do not know the name, that are in "scripts_dir". These subdirs are actually Windows directory joints that point to other places on the disk, and I do not want them to be copied.
How do I achieve this with the inculde/excludes pattern of Jenkins ?
I already tried, having include=top_dir/ , exclude=
**/scripts_dir/*/
**/scripts_dir/*/**
**/scripts_dir/**/*
but it always exculdes the whole "scripts_dir" folder.
Finally, by using brute force, I found that the following expression does exclude all the files in the subdirectories of scripts_dir (whatever symlink or not), then removing these subdirs, while keeping the files directly in scripts_dir :
**/scripts_dir/**/*/*/
Thanks for the help anyway.
Reading the ANT manual, there an followsymlinks attribute that defaults to true. You said those things you want to exclude are symlinks (although i am not sure if this will work with Windows joints). Try adding followsymlinks=false
Another solution: if all your files under scripts_dir have a set number of characters in the extension, you can put that into your include statement. This will only pickup files with extensions of 3 characters:
**/scripts_dir/*.???
More on this here