view mercurial/lsprofcalltree.py @ 11562:efbc09fdefd8 stable

test-log: Add test for "hg log -pf" (issue647) To test f786fc4b8764 we analyze this repo: 0 | 1 | 2 | 3 | 4 a ------------> b \ b -> dir/b -------> e d We follow dir/b and expects rev 0, 1 and 2. A log following b will surprisingly report rev 0 and 1. That's because the content and the parents are the same in rev 1 and rev 3, and the revlog entry from rev 1 is thus reused in rev 3 even though the revlink is incorrect. Note that if we follow e then we will get all revs, including rev 3, because b gets added to the set of interesting filenames we are following through all revisions. That might be surprising, but that's (currently) how it is. Original test case by Joel Rosdahl <joel@rosdahl.net> Original test reviewed by Nicolas Dumazet <nicdumz@gmail.com>
author Joel Rosdahl <joel@rosdahl.net>
date Thu, 08 Jul 2010 22:44:15 +0200
parents beae42f3d93b
children 071af8d385a9
line wrap: on
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"""
lsprofcalltree.py - lsprof output which is readable by kcachegrind

Authors:
    * David Allouche <david <at> allouche.net>
    * Jp Calderone & Itamar Shtull-Trauring
    * Johan Dahlin

This software may be used and distributed according to the terms
of the GNU General Public License, incorporated herein by reference.
"""

def label(code):
    if isinstance(code, str):
        return '~' + code    # built-in functions ('~' sorts at the end)
    else:
        return '%s %s:%d' % (code.co_name,
                             code.co_filename,
                             code.co_firstlineno)

class KCacheGrind(object):
    def __init__(self, profiler):
        self.data = profiler.getstats()
        self.out_file = None

    def output(self, out_file):
        self.out_file = out_file
        print >> out_file, 'events: Ticks'
        self._print_summary()
        for entry in self.data:
            self._entry(entry)

    def _print_summary(self):
        max_cost = 0
        for entry in self.data:
            totaltime = int(entry.totaltime * 1000)
            max_cost = max(max_cost, totaltime)
        print >> self.out_file, 'summary: %d' % (max_cost,)

    def _entry(self, entry):
        out_file = self.out_file

        code = entry.code
        #print >> out_file, 'ob=%s' % (code.co_filename,)
        if isinstance(code, str):
            print >> out_file, 'fi=~'
        else:
            print >> out_file, 'fi=%s' % (code.co_filename,)
        print >> out_file, 'fn=%s' % (label(code),)

        inlinetime = int(entry.inlinetime * 1000)
        if isinstance(code, str):
            print >> out_file, '0 ', inlinetime
        else:
            print >> out_file, '%d %d' % (code.co_firstlineno, inlinetime)

        # recursive calls are counted in entry.calls
        if entry.calls:
            calls = entry.calls
        else:
            calls = []

        if isinstance(code, str):
            lineno = 0
        else:
            lineno = code.co_firstlineno

        for subentry in calls:
            self._subentry(lineno, subentry)
        print >> out_file

    def _subentry(self, lineno, subentry):
        out_file = self.out_file
        code = subentry.code
        #print >> out_file, 'cob=%s' % (code.co_filename,)
        print >> out_file, 'cfn=%s' % (label(code),)
        if isinstance(code, str):
            print >> out_file, 'cfi=~'
            print >> out_file, 'calls=%d 0' % (subentry.callcount,)
        else:
            print >> out_file, 'cfi=%s' % (code.co_filename,)
            print >> out_file, 'calls=%d %d' % (
                subentry.callcount, code.co_firstlineno)

        totaltime = int(subentry.totaltime * 1000)
        print >> out_file, '%d %d' % (lineno, totaltime)