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a = 1
函数的末尾添加了符号定义语句test
,似乎exec
语句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,为什么会发生这种情况?
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被保留,如果符号已经被保留,则在 exec 函数内部动态调用的任何符号定义语句将无法修改本地符号表.
这是真的吗?编译期间实际发生了什么?
额外测试 1:用 '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exec()
执行期间成功添加到本地符号表中,但之后随着a = 1
的存在它神奇地消失了。
额外测试 2:在 return
前添加 a = 1
语句
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 : {}
第二个 a = 1
函数中的 test()
无法访问,但它仍然影响 exec()
的行为。
即使是 dis()
模块中的 dis
函数也无法区分这两个 test()
函数之间的区别。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
解决方法
根据documentation :
注意:默认 locals 的行为如下面的函数 locals() 所述:不应尝试修改默认 locals 字典。 传递一个显式的locals 如果您需要在函数 exec() 返回后查看代码对局部变量的影响,请查看字典。
所以我认为这属于“意外行为”,但我想你可以去 exec()
的实现来真正深入了解。