This adds a timeout of 60 minutes for the Cppcheck analysis.
Timed out results do not count as crash but they are uploaded and
marked with "TO!" in the list of the latest results. No "diff" is
generated for timed out results so they do not add wrong entries to
the "Diff report".
In test-my-pr.py the timed out results are listed separately just like
the crashes.
donate-cpu-server.py: Add timeout report
For whatever reason, pygments removes leading empty lines by default.
This results in the error messages being in the wrong line, shifted by
how many empty lines were removed.
Adding the option "stripnl=False" as an option to the lexer constructor
avoids this issue.
Follow-up to bca5d0d820
The style for inconclusive messages has to be fixed too. Otherwise
the border adds 2 extra pixels resulting in improper line height.
* rename JS functions
* Use the native font stack
* Update cppcheck-htmlreport
* Use a class for hiding content and toggle that instead. This should improve performance with huge reports.
* reorder selectors
* use classes for header and footer
* remove unneeded clear property (we don't have floats anymore)
* fix wrong `font-size` and `margin` properties
* remove `dir=rtl`
Related ticket: https://trac.cppcheck.net/ticket/9574
* Try to use double quotes consistently
* minor CSS consistency changes
* fix HTML errors
* fix a few JS issues and switch to `textContent`
* use `addEventListener` instead of the onload event
* use `querySelector` and `querySelectorAll`
Co-authored-by: XhmikosR <xhmikosr@gmail.com>
Using .tar.xz packages adds about 4500 additional packages that can be
tested and changes many existing packages where a more recent version
can be used now that is only available as .tar.xz file.
Related ticket: https://trac.cppcheck.net/ticket/9508
donate-cpu.py: Require at least Python 3.4
xz support was added with 3.3.
Ensure bifurcate() does not recurse endlessly where a variable is
initialized recursively, or a variable is initialized as x(0) or x{0}
followed by a recursive assignment (for example int x(0); x = x / 1;).
The first case is solved by bailing out if there initialization is done
using x(0) or x{0}, the second by adding a missing depth argument to a
recursive call.
versat
Aleksandr Serbin
Thomas Niederberger
Daniel Marjamäki
Dmitry-Me
XhmikosR
Simon Martin
Rikard Falkeborn