DDlog is a programming language for incremental computation. It is well suited for writing programs that continuously update their output in response to input changes. A DDlog programmer does not write incremental algorithms; instead they specify the desired input-output mapping in a declarative manner.
-o DIR --output-dir=DIR Output directory (default based on program name).
To me, this means that the output directory is based on the program name by default. Sure enough, if I run ddlog -i foo.dl I end up with output in foo_ddlog. Cool.
So, if I run ddlog -i foo.dl -o bar, I'll get an output directory not based on the program name but named bar instead, right? No. Actually it's still based on the program name, just nested inside the bar directory:
[blp@sigxcpu]$ ddlog -i foo.dl -o bar
Finished compiling "foo.dl" in 0.76s
[blp@sigxcpu]$ ls bar/
foo_ddlog
[blp@sigxcpu]$
This doesn't do what I expect. I don't know whether the implementation is wrong (so that the code should be changed to do what I expect) or the usage message is wrong (so that the usage message should be changed to something like "output base directory (default is .)".
ddlog --help
says the following about-o
:To me, this means that the output directory is based on the program name by default. Sure enough, if I run
ddlog -i foo.dl
I end up with output infoo_ddlog
. Cool.So, if I run
ddlog -i foo.dl -o bar
, I'll get an output directory not based on the program name but namedbar
instead, right? No. Actually it's still based on the program name, just nested inside thebar
directory:This doesn't do what I expect. I don't know whether the implementation is wrong (so that the code should be changed to do what I expect) or the usage message is wrong (so that the usage message should be changed to something like "output base directory (default is .)".