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From: Janek Kozicki (janek_listy_at_[hidden])
Date: 2006-10-12 15:28:56
Douglas Gregor said: (by the date of Thu, 12 Oct 2006 12:00:48 -0400)
> 75 failures in 13 libraries
> filesystem (2)
> io (1)
> iostreams (8)
> mpl (6)
> parameter (6)
> program_options (9)
> python (9)
> random (1)
> rational (5)
> serialization (18)
> spirit (7)
> test (2)
> utility (1)
Hi Doug, after 10 minutes (no more), including some googling I produced this:
75 failures in 13 libraries (37 are from non-broken platforms)
non-broken broken
filesystem 0 2
io 1
iostreams 7 8
mpl 5 6
parameter 2 6
program_options 0 9
python 8 9
random 1
rational 3 5
serialization 4 18
spirit 4 7
test 1 2
utility 1
The key function here is string.ljust().
If you like this, here is the diff -u
@@ -409,18 +409,17 @@
str(len(self.libraries)) + ' libraries')
if any_broken_platforms:
message += ' (' + str(self.numReportableFailures()) + ' are from non-broken platforms)'
- message += '\n'
+ message += '\n\n non-broken broken\n'
# Display the number of failures per library
for k in sorted_keys( self.libraries ):
library = self.libraries[k]
num_failures = library.numFailures()
- message += (' ' + library.name + ' ('
+ message += (' ' + library.name.ljust(22) + ' '
+ str(library.numReportableFailures()))
if library.numReportableFailures() < num_failures:
- message += (' of ' + str(num_failures)
- + ' failures are from non-broken platforms')
- message += ')\n'
+ message += (' ' + str(num_failures))
+ message += '\n'
pass
# If we have any broken platforms, tell the user how we're
-- Janek Kozicki |
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