I'm using waf to build a C program. I'd like to check for the existence of a particular header file during the configuration phase. Is there a way to do that?
Ah, a bit of googling found the answer to my question: You can use the check method on Configuration objects, like so:
def configure(conf):
conf.check(header_name="stdbool.h")
Related
I'm writing some validation code for a bazel build rule and I need to do some path validation. I need to check that a certain file exists in the same directory as the BUILD file. I notice that there's a context attribute build_file_path which points to the BUILD file. I'd like to extract the parent directory from this.
It looks like I can't create a new path object - I don't see a constructor/initializer. It also seems like Starlark doesn't support os.path like python because imports aren't supported.
What's the canonical way to get the parent directory of a string object representing a path in Starlark?
I can't answer your final question, but hopefully the following will help with the initial problem:
You could use the Label of the target for which this instance of the rule is being built and find its package. This will give you a string representing the parent directory of the BUILD file.
i.e. ctx.label.package
load("#bazel_skylib//lib:paths.bzl", "paths")
paths.dirname(path_str)
See https://github.com/bazelbuild/bazel-skylib/blob/main/docs/paths_doc.md
I like using .fsi signature files to control visibility. However, if I have both Foo.fsi and Foo.fs files in my solution, and #load "Foo.fs" in a script, it doesn't seem like the corresponding signature file gets used. If I do:
#load "Foo.fsi"
#load "Foo.fs"
... then the desired visibility control happens. Is this the recommended way to achieve this, or is there a better way to do it? In a perfect world, one would like to see the signature file automatically loaded, too.
Not a final answer, but a better way.
From reading Expert F# 4.0 one can do
#load "Foo.fsi" "Foo.fs" "Foo.fsx"
All three loads are on one line.
TL;DR
The link to the book is via WolrdCat just put in a zip code and it will show you locations near there where the book can be found.
I have a script which saves some files at a given location. It works fine but when I send this code to someone else, he has to change the paths in the code. It's not comfortable for someone who does not know what is in that code and for me to explain every time where and how the code should be changed.
I want to get this path in a variable which will be taken from the configuration file. So it will be easier for everyone to change just this config file and nothing in my code. But I have never done this before and could not find any information on how I can do this in the internet.
PS: I do not have any code and I ask about an ultimate solution but it is really difficult to find something good in the internet about dxl, especially since I'm new with that. Maybe someone of you already does that or has an idea how it could be done?
DXL has a perm to read the complete context of a file into a variable: string readFile (string) (or Buffer readFile (string))
you can split the output by \n and then use regular expressions to find all lines that match the pattern
^\s*([^;#].*)\s*=\s*(.*)\s*$
(i.e. key = value - where comment lines start with ; or #)
But in DOORS I prefer using DOORS modules as configuration modules. Object Heading can be the key, Object Text can be the value.
Hardcode the full name of the configuration module into your DXL file and the user can modify the behaviour of the application.
The advantage over a file is that you need not make assumptions on where the config file is to be stored on the file system.
It really depends on your situation. You are going to need to be a little more specific about what you mean by "they need to change the paths in the code". What are these paths to? Are they DOORS module paths, are they paths to local/network files, or are the something else entirely?
Like user3329561 said, you COULD use a DOORS module as a configuration file. I wouldn't recommend it though, simply because that is not what DOORS modules were designed for. DOORS is fully capable of reading system files in one line at a time as well as all at once, but I can't recommend that option either until I know what types of paths you want to load and why.
I suspect that there is a better solution for your problem that will present itself once more information is provided.
I had the same problem, I needed to specify the path of my configuration file used in my dxl script.
I solved this issue passing the directory path as a parameter to DOORS.exe as follow:
"...\DOORS\9.3\bin\doors.exe" -dxl "string myVar = \"Hello Word\"
then in my dxl script, the variable myVar is a global variable.
I need to create a symlink to a sub-directory using Ant. The issue is that I don't know where the target sub-directory is.
To create a symlink with ant I do this:
<symlink link="${parent.dir}/FOO/linkname" resource="${parent.dir}/BAR/target"/>
But I don't know what BAR is called in advance so I need to do a search for "target" under parent.dir and then pass the one result into the resource.
Is this possible using fileset? Or another way?
It might be possible to use a fileset but that might give you several symlinks or none.
A much better approach is to define the path to BAR in a property. If there is a dynamic part in this path, change the code so that Ant evaluates the dynamic part and everyone else uses Ant's value.
The typical example here is that the path contains a version or timestamp. Define those in your build file so you can use them everywhere. If a Java process needs the values, pass them to the process as a system property (-D...).
The include directive is usually used for a .hrl file at the top of an .erl file.
But, I would like to use include from the Erlang console directly.
I am trying to use some functions in a module. I have compiled the erl file from the console. But, the functions I want to use do not work without access to the hrl file.
Any suggestions?
"But, the functions I want to use do not work without access to the hrl file."
This can't be true, but from this I'll take a shot at guessing that you want access to records in the hrl file that you don't (normally) have in the shell.
If you do rr(MODULE) you will load all records defined in MODULE(including those defined in an include file included by MODULE).
Then you can do everything you need to from the shell.
(Another thing you may possibly want for testing is to add the line -compile(export_all) to your erl file. Ugly, but good sometimes for testing.)
Have you tried the compile:file option? You can pass a list of modules to be included thus:
compile:file("myfile.erl", [{i, "/path/1/"}, {i, "/path/2/"}])
It's worth nothing that jsonerl.hrl doesn't contain any functions. It contains macros. As far as I know, macros are a compile-time-only construct in Erlang.
The easiest way to make them available would be to create a .erl file yourself that actually declares functions that are implemented in terms of the macro. Maybe something like this:
-module(jsonerl_helpers).
-include("jsonerl.hrl").
record_to_struct_f(RecordName, Record) ->
?record_to_struct(RecordName, Record).
... which, after you compile, you could call as:
jsonerl_helpers:record_to_struct_f(RecordName, Record)
I don't know why the author chose to implement those as macros; it seems odd, but I'm sure he had his reasons.