相关疑难解决方法(0)

我想了解如何使用dis(Python字节码的解析器).具体来说,如何解释dis.dis(或dis.disassemble)的输出？

.

这是一个非常具体的例子(在Python 2.7.3中):

dis.dis("heapq.nsmallest(d,3)")

      0 BUILD_SET             24933
      3 JUMP_IF_TRUE_OR_POP   11889
      6 JUMP_FORWARD          28019 (to 28028)
      9 STORE_GLOBAL          27756 (27756)
     12 LOAD_NAME             29811 (29811)
     15 STORE_SLICE+0  
     16 LOAD_CONST            13100 (13100)
     19 STORE_SLICE+1

我看到JUMP_IF_TRUE_OR_POP等等是字节码指令(虽然有趣的是,BUILD_SET这个列表中没有出现,但我希望它可以工作BUILD_TUPLE).我认为右边的数字是内存分配,左边的数字是转到数字......我注意到它们每次几乎增加3(但不完全).

如果我dis.dis("heapq.nsmallest(d,3)")在函数内部换行:

def f_heapq_nsmallest(d,n):
    return heapq.nsmallest(d,n)

dis.dis("f_heapq(d,3)")

      0 BUILD_TUPLE            26719
      3 LOAD_NAME              28769 (28769)
      6 JUMP_ABSOLUTE          25640
      9 <44>                                      # what is <44> ?  
     10 DELETE_SLICE+1 
     11 STORE_SLICE+1 

我对python如何实现闭包感兴趣？

为了举例，请考虑这个

def closure_test():
    x = 1
    def closure():
        nonlocal x
        x = 2
        print(x)
    return closure

closure_test()()

这里，该函数closure_test有一个x嵌套函数closure捕获的局部变量。

当我运行该程序时，我得到以下输出

2

当我看到函数的反汇编时closure_test，

  2           0 LOAD_CONST               1 (1)
              2 STORE_DEREF              0 (x)

  3           4 LOAD_CLOSURE             0 (x)
              6 BUILD_TUPLE              1
              8 LOAD_CONST               2 (<code object closure at 0x7f14ac3b9500, file "<string>", line 3>)
             10 LOAD_CONST               3 ('closure_test.<locals>.closure')
             12 MAKE_FUNCTION            8 (closure)
             14 STORE_FAST               0 (closure)

  7          16 LOAD_FAST                0 (closure)
             18 RETURN_VALUE

Disassembly of …

来自Google样式指南的词法范围:

嵌套的Python函数可以引用封闭函数中定义的变量,但不能赋值给它们.

这两个似乎最初都要检查:

# Reference
def toplevel():
    a = 5
    def nested():
        print(a + 2)
    nested()
    return a
toplevel()
7
Out[]: 5

# Assignment
def toplevel():
    a = 5
    def nested():
        a = 7 # a is still 5, can't modify enclosing scope variable
    nested()
    return a
toplevel()
Out[]: 5

那么,为什么嵌套函数中的引用和赋值的组合会导致异常呢？

# Reference and assignment
def toplevel():
    a = 5
    def nested():
        print(a + 2)
        a = 7
    nested()
    return a
toplevel()
# UnboundLocalError: local variable 'a' referenced …