python - 为什么 exec() 中的符号定义语句有时对本地符号表没有影响？

问题描述

以下代码片段按预期工作：

def test():
    print(f'local symbol table before exec : {locals()}')
    exec('a = 0')
    print(f'local symbol table after exec  : {locals()}')

test()
# printed result:
# local symbol table before exec : {}
# local symbol table after exec  : {'a': 0}

但是，一旦我在函数a = 1末尾添加符号定义语句test，该语句似乎exec对本地符号表没有影响：

def test():
    print(f'local symbol table before exec : {locals()}')
    exec('a = 0')
    print(f'local symbol table after exec  : {locals()}')
    a = 1

test()
# printed result:
# local symbol table before exec : {}
# local symbol table after exec  : {}

那么，为什么会这样呢？

这是我的猜测：在函数内部静态定义的符号将在编译时以某种方式保留，如果符号已经保留，则在 exec 函数内部动态调用的任何符号定义语句都无法修改本地符号表。

真的吗？在编译期间实际发生了什么？

额外测试 1：将 exec 参数替换为'a = 0\nprint(locals())'

def test():
    print(f'local symbol table before exec : {locals()}')
    exec('a = 0\nprint(locals())')
    print(f'local symbol table after exec  : {locals()}')


test()
# printed result:
# local symbol table before exec : {}
# {'a': 0}
# local symbol table after exec  : {'a': 0}

def test():
    print(f'local symbol table before exec : {locals()}')
    exec('a = 0\nprint(locals())')
    print(f'local symbol table after exec  : {locals()}')
    a = 1


test()
# printed result:
# local symbol table before exec : {}
# {'a': 0}
# local symbol table after exec  : {}

如我们所见，符号a在执行过程中成功添加到本地符号表中exec()，但之后它就神奇地消失了a = 1。

额外测试 2：在之前添加return语句a = 1

def test():
    print(f'local symbol table before exec : {locals()}')
    exec('a = 0\nprint(locals())')
    print(f'local symbol table after exec  : {locals()}')
    return


test()
# printed result:
# local symbol table before exec : {}
# {'a': 0}
# local symbol table after exec  : {'a': 0}

def test():
    print(f'local symbol table before exec : {locals()}')
    exec('a = 0\nprint(locals())')
    print(f'local symbol table after exec  : {locals()}')
    return
    a = 1


test()
# printed result:
# local symbol table before exec : {}
# {'a': 0}
# local symbol table after exec  : {}

在第二个函数a = 1中是无法访问的test()，但它仍然会影响exec().

即使是模块中的dis()函数dis也无法区分这两个test()函数。输出完全相同，如下所示：

  5           0 LOAD_GLOBAL              0 (print)
              2 LOAD_CONST               1 ('local symbol table before exec : ')
              4 LOAD_GLOBAL              1 (locals)
              6 CALL_FUNCTION            0
              8 FORMAT_VALUE             0
             10 BUILD_STRING             2
             12 CALL_FUNCTION            1
             14 POP_TOP

  6          16 LOAD_GLOBAL              2 (exec)
             18 LOAD_CONST               2 ('a = 0\nprint(locals())')
             20 CALL_FUNCTION            1
             22 POP_TOP

  7          24 LOAD_GLOBAL              0 (print)
             26 LOAD_CONST               3 ('local symbol table after exec  : ')
             28 LOAD_GLOBAL              1 (locals)
             30 CALL_FUNCTION            0
             32 FORMAT_VALUE             0
             34 BUILD_STRING             2
             36 CALL_FUNCTION            1
             38 POP_TOP

  8          40 LOAD_CONST               0 (None)
             42 RETURN_VALUE

标签： pythonpython-3.xexeclocal-variablessymbol-table

python - 为什么 exec() 中的符号定义语句有时对本地符号表没有影响？

问题描述

解决方案

推荐阅读