lz | an old science experiment | Functional Programming library

Thanks to Tsyvarev comment the solution was to remove the -lOBD2 from the target_link_options. As per Tsyvarev suggestion now all the linker options are linked through target_link_libraries.

The solution to the CMakeLists.txt

You're very close. There were a few minor issues that I spotted with your YAML/PowerShell:

1. You forgot the semicolon after the variable name in "##vso[task.setvariable variable=JsonLZ]$json", it should be: "##vso[task.setvariable variable=JsonLZ;]$json"
2. You should be using $(JsonLZ) instead of ${{ variables.JsonLZ }}. The former will be evaluated at runtime, the latter at compile-time. Here's a link to the MS Docs: Understand variable syntax

Give this a try to see a working example:

If you see errors about a specific library missing, you can use dnf itself to find out the name. For example, on Fedora 35:

You're using a pre-warmed image

Here's how I got DBD::mysql installed on macOS:

As you can see in strace, the execve command executes as: execve("/bin//nc", ["/bin//nc", "/bin//nc-e //bin/bash -nvlp 4455"], NULL) = 0 It seems to be taking the whole /bin//nc-e //bin/bash -nvlp 4455 as a single argument and thus thinks it's a hostname. In order to get around that, the three argv[] needed for execve() is pushed seperately. argv[]=["/bin/nc", "-e/bin/bash", "-nvlp4455"] These arguments are each pushed into edx, ecx, and ebx. since ebx needs to be /bin/nc, which was already done in the original code. we just needed to push 2nd and 3rd argv[] into ecx and edx and push it into stack. After that we just copy the whole stack into ecx, and then xor edx,edx to set edx as NULL.

Here is the correct solution:

The issue was caused by WORKDIR clause before copy instruction.

Fixed Dockerfile:

Why -z add 0a instead of 00?What does output mean in Treat input and output data as sequences of lines, each terminated by a zero byte (the ASCII NUL character) instead of a newline.?

The -z option is about the data that grep is matching against the given pattern. It has two effects:

1. The data will be interpreted as having null-terminated lines instead of newline-terminated lines.

2. When grep echos input data to its output (normally a line matching the pattern, but if -v is in effect then a line not matching the pattern) then it terminates that line with a null character, thus preserving that characteristic of the input. This is what "output" means in the documentation for the -z option.

Neither of those is relevant to your particular data and grep -rlz command.

In the first place, the contents of file /tmp/target/it is a test.txt is just test -- no null characters in sight. grep therefore treats the entire contents of the file as one line, though that's no different than if -z were not in effect, there being no newline either.

In the second place, the -l option is in effect, so instead of printing any matched lines (with null terminators), grep prints the name of the file. It appends a newline because that is its default, and you have not overridden it -- filename printing is what the -Z option is about, not the -z option.

Note also that -Z is effective on filename printing even when -l is not in effect. When -r is in effect or multiple filenames are given on the grep command line, grep will normally precede each line of input data it prints (that is, each output line) with the corresponding file name and a colon. When -Z is in effect, it instead precedes each line with the file name and a null character.

Ok I think I've found out your issue. Let's download a fresh file to use as an example and work with that. I'm working with a small snippet of a gray square.