I can do these steps to get coverage information in covsrc
covselect -f test.cov -d
covselect -f test.cov -a dir_i_want/
covsrc -f test.cov
But when I use covxml or bullshtml, the settings i did above don't work.
So, how to exclude a directory or only include directories I want?
What I found worked for me was to use a import/export file.
covselect --file file.cov --import <importfile>
"folder" seem to need to end with a forward slash or they are ignored
exclude folder path/to/dir/
include source path/to/src.cpp
exclude function path/to/src.cpp:name::space::class::function(param& p)
Related
I'm trying to use Make to ... make modular Dockerfiles. Long story short, I want to centralize certain elements and make the composable and reusable, like classes and functions really, but the Dockerfile syntax does not - and according to the developers, will not - offer any facilities in the image of C's #include or similar composability solutions. Not to worry, #include and friends to the rescue!
Except...
I have the following Makefile in my project:
BUILD_DIR := ${CI_PROJECT_DIR}/build
TEMPLATE_FILES := $(shell find ${CI_PROJECT_DIR} -name '*.build')
TEMPLATE_FILENAMES := $(foreach file,$(TEMPLATE_FILES),$(BUILD_DIR)/$(notdir $(file)).built)
BUILT_TEMPLATES := $(TEMPLATE_FILENAMES:.build.built=.built)
DOCKER_FILES := $(shell find ${CI_PROJECT_DIR} -name '*.Dockerfile')
DOCKER_OBJS := $(foreach file,$(DOCKER_FILES),$(BUILD_DIR)/$(notdir $(file)))
all: $(BUILT_TEMPLATES) $(DOCKER_OBJS)
$(BUILD_DIR)/%.built: $(TEMPLATE_FILES) $(BUILD_DIR) # build any templated Dockerfiles
cpp -E -P -o $(BUILD_DIR)/$(notdir $#) -I ${CI_PROJECT_DIR}/modules $<
sed -i 's/__NL__ /\n/g' $(BUILD_DIR)/$(notdir $#)
$(BUILD_DIR)/%.Dockerfile: $(DOCKER_FILES) $(BUILD_DIR)
cp $< $(BUILD_DIR)/$(notdir $(#))
$(BUILD_DIR):
mkdir -p $(BUILD_DIR)
.PHONY: clean
clean:
-rm -r $(BUILD_DIR)
The objective is to run the templated Dockerfiles through GCC to compile the #includes in them into proper Docker instructions, and just copy the rest of the files. Sounds simple enough.
Except that it looks like all the target files are "offset" from their sources - like the file names are correct, but the contents are from a file elsewhere in the list, and with no discernible order either.
One thing that I'm fairly sure is wrong - but even more wrong otherwise - is the line
$(BUILD_DIR)/%.built: $(TEMPLATE_FILES) $(BUILD_DIR) # build any templated Dockerfiles
By all manuals and documentation, it ought to be
$(BUILD_DIR)/%.built: %.build $(BUILD_DIR) # build any templated Dockerfiles
but that's even worse, because then Make just says make: *** No rule to make target '/docker/build/runner-dart-2-18-firebase.built', needed by 'all'. Stop.
I'm out of ideas here, along with my limited knowledge of Make. What am I missing to make Make make - sorry - my Dockerfiles?
This line:
$(BUILD_DIR)/%.built: $(TEMPLATE_FILES) $(BUILD_DIR)
Says that if make wants to build a target that matches that pattern, and it can find all the prerequisites, then the pattern rule matches and the recipe can be used. Let's ignore BUILD_DIR (note that it's always a bad idea to list a directory as a prerequisite, but that's not causing this problem). Suppose TEMPLATE_FILES is set to the value ./foo/foo.build ./bar/bar.build. Now the above rule expands to:
./build/%.built: ./foo/foo.build ./foo/bar.build ./build
What is the recipe?
cpp -E -P -o $(BUILD_DIR)/$(notdir $#) -I ${CI_PROJECT_DIR}/modules $<
First it's always wrong to create a file that is not exactly $# so you should use just $# not $(BUILD_DIR)/$(notdir $#). But more importantly, what will $< be set to? It is always set to the first prerequisite, and the first prerequisite is always ./foo/foo.build. So every time you run this recipe, regardless of which .built file you're trying to create, you will always be preprocessing the first .build file.
Your idea that you want this instead:
$(BUILD_DIR)/%.built: %.build $(BUILD_DIR)
is correct, in general. Why do you get the error? Because if you are trying to build the target ./build/foo.built, then the stem (part that matches %) is foo. Then make will look to see if the prerequisite foo.build exists or can be created, because you said the prerequisite is %.build. That file does NOT exist and CANNOT be created (make doesn't know how to create it), because the file is ./foo/foo.build not foo.build which is a totally different file.
You have three options. You can either write separate rules for each source directory:
$(BUILD_DIR)/%.built: foo/%.build
...
$(BUILD_DIR)/%.built: bar/%.build
...
Or, you can change your generated files so they are not all in the same directory but instead keep the source directory structure; you would change this:
TEMPLATE_FILENAMES := $(foreach file,$(TEMPLATE_FILES),$(BUILD_DIR)/$(notdir $(file)).built)
BUILT_TEMPLATES := $(TEMPLATE_FILENAMES:.build.built=.built)
to just this:
BUILT_TEMPLATES := $(patsubst %.build,$(BUILD_DIR)/%.built,$(TEMPLATE_FILES))
then create the output directory as part of the recipe:
#mkdir -p $(#D)
cpp -E -P -o $# -I ${CI_PROJECT_DIR}/modules $<
sed -i 's/__NL__ /\n/g' $#
Or finally, you could use VPATH to tell make what directories to look in to find the *.build files:
VPATH := $(sort $(dir $(TEMPLATE_FILES)))
(note, you should choose only one of these options).
I have a file file.txt with filenames ending with *.sha256, including the full paths of each file. This is a toy example:
file.txt:
/path/a/9b/x3.sha256
/path/7c/7j/y2.vcf.gz.sha256
/path/e/g/7z.sha256
Each line has a different path/file. The *.sha256 files have checksums.
I want to run the command "sha256sum -c" on each of these *.sha256 files and write the output to an output_file.txt. However, this command only accepts the name of the .sha256 file, not the name including its full path. I have tried the following:
while read in; do
sha256sum -c "$in" >> output_file.txt
done < file.txt
but I get:
"sha256sum: WARNING: 1 listed file could not be read"
which is due to the path included in the command.
Any suggestion is welcome
#!/bin/bash
while read in
do
thedir=$(dirname "$in")
thefile=$(basename "$in")
cd "$thedir"
sha256sum -c "$thefile" >>output_file.txt
done < file.txt
Modify your code to extract the directory and file parts of your in variable.
Suppose I have two tar.gz files
a1.tar.gz
a2.tar.gz
and each archive contains many files, including a file called
target.txt
How do I search for BLAH in target.txt in both of these archives using zgrep without searching all of the other files in each archive?
If I try
zgrep -a BLAH *.tar.gz
then that searches all files in each archive, and if I try
zgrep --include=target.txt -a BLAH *.tar.gz
then I get
zgrep: --include=target.txt: option not supported
You can use zgrep, however this command will not locate the file you are looking for as it searches the entire tar-formatted file for matches. This will report which tar file has a match for BLAH but this does not limit searching tar files containing the file target.txt.
There is an open source tool called ugrep to search archives (zip, tar, pax, cpio, jar) and tar.gz tarballs. Use option -z and -g target.txt to search for BLAH in target.txt in all of the tar files found recursively:
ugrep -z "BLAH" -g target.txt
To search the working directory only without recursing deeper:
ugrep -z "BLAH" -g target.txt .
Note that option -g takes a glob. Use a quoted glob like -g "*.txt" to match all .txt files.
I'm creating a tarball of a large codebase managed in ClearCase. Every directory has a sub-directory named ".CC". I'd like to exclude these from my tarball.
I've found Excluding directory when creating a .tar.gz file, but excluding that would appear to require passing each and every .CC directory on the commndline. This is impractical in my case.
Is there a way to exclude directories that meet a particular pattern?
EDIT:
I am not asking how to exclude a specific finite list of directories. I am asking how to exclude all directories that end in a particular pattern.
Instead of manually typing --exclude 'root/a/.CC' --exclude 'root/b/.CC' ... you can type $(find root -type d -name .CC -exec echo "--exclude \'{}\'" \;|xargs)
You can use whatever patterns find supports, or even use something like grep inbetween find and xargs.
The following bash script should do the trick. It uses the answer given by #Marcus Sundman.
#!/bin/bash
echo -n "Please enter the name of the tar file you wish to create with out extension "
read nam
echo -n "Please enter the path to the directories to tar "
read pathin
echo tar -czvf $nam.tar.gz
excludes=`find $pathin -iname "*.CC" -exec echo "--exclude \'{}\'" \;|xargs`
echo $pathin
echo tar -czvf $nam.tar.gz $excludes $pathin
This will print out the command you need and you can just copy and paste it back in. There is probably a more elegant way to provide it directly to the command line.
*.CC could be exchanged for any other common extension and this should still work.
I have a lot of files and I want to find where is MYVAR.
I'm sure it's in one of .yml files but I can't find in the grep manual how to specify the filetype.
grep -rn --include=*.yml "MYVAR" your_directory
please note that grep is case sensitive by default (pass -i to tell to ignore case), and accepts Regular Expressions as well as strings.
You don't give grep a filetype, just a list of files. Your shell can expand a pattern to give grep the correct list of files, though:
$ grep MYVAR *.yml
If your .yml files aren't all in one directory, it may be easier to up the ante and use find:
$ find -name '*.yml' -exec grep MYVAR {} \+
This will find, from the current directory and recursively deeper, any files ending with .yml. It then substitutes that list of files into the pair of braces {}. The trailing \+ is just a special find delimiter to say the -exec switch has finished. The result is matching a list of files and handing them to grep.
If all your .yml files are in one directory, then cd to that directory, and then ...
grep MYWAR *.yml
If all your .yml files are in multiple directories, then cd to the top of those directories, and then ...
grep MYWAR `find . -name \*.yml`
If you don't know the top of those directories where your .yml files are located and want to search the whole system ...
grep MYWAR `find / -name \*.yml`
The last option may require root privileges to read through all directories.
The ` character above is the one that is located along with the ~ key on the keyboard.
find . -name \*.yml -exec grep -Hn MYVAR {} \;