Question

我想密切监视从某个函数调用的函数调用链。

import pdb; pdb.set_trace()
res = api.InsertVideoEntry(video_entry, video)

我正在寻找一种方法，可以轻松查看调用api.insertVideoEntry(video_entry, video)的{{1}}来电foo() bar()，

这是一个非常粗略的图表来显示我的意思。我不需要这种形式，但这是我正在寻找的那种信息。

baz()

Answer 1

这是一个有趣的学习经历。也许你可以使用这里显示的代码？ This demonstration可以让您了解使用trace时可以预期的输出类型。

# Functions to trace
# ==================

def baz():
    pass

def baz2():
    pass

def bar():
    baz()
    baz2()

def log():
    pass

def foo():
    bar()
    log()

def finish():
    pass

def insertVideoEntry():
    foo()
    finish()

# Names to trace
# ==============

names = list(locals())

# Machinery for tracing
# =====================

import os
import sys

def trace(start, *names):
    def tracefunc(frame, event, arg):
        if event == 'call':
            code = frame.f_code
            name = code.co_name
            if name in names:
                level = -start
                while frame:
                    frame = frame.f_back
                    level += 1
                print('{}{}.{}()'.format(
                    '    ' * level,
                    os.path.splitext(os.path.basename(code.co_filename))[0],
                    name))
                return tracefunc
    sys.settrace(tracefunc)

# Demonstration of tracing
# ========================

trace(2, *names)

insertVideoEntry()

如果您对递归演示感兴趣，可能希望this variation带有被调参数读数：

import os
import sys

def main(discs):
    a, b, c = list(range(discs, 0, -1)), [], []
    line = '-' * len(repr(a))
    print(a, b, c, sep='\n')
    for source, destination in towers_of_hanoi(discs, a, b, c):
        destination.append(source.pop())
        print(line, a, b, c, sep='\n')

def towers_of_hanoi(count, source, via, destination):
    if count > 0:
        count -= 1
        yield from towers_of_hanoi(count, source, destination, via)
        yield source, destination
        yield from towers_of_hanoi(count, via, source, destination)

def trace(start, *names):
    def tracefunc(frame, event, arg):
        if event == 'call':
            code = frame.f_code
            name = code.co_name
            if name in names:
                level = -start
                args = ', '.join(repr(frame.f_locals[name]) for name in
                                 code.co_varnames[:code.co_argcount])
                while frame:
                    frame = frame.f_back
                    level += 1
                print('{}{}.{}({})'.format(
                    ' ' * (level * 4),
                    os.path.splitext(os.path.basename(code.co_filename))[0],
                    name, args))
                return tracefunc
    sys.settrace(tracefunc)

if __name__ == '__main__':
    trace(3, 'main', 'towers_of_hanoi')
    main(3)

我如何步骤使用python调试器来打破每个函数调用？

1 个答案: