sil*_*ado 9 python dictionary immutability
考虑我kwargs在方法中更改dict 的情况:
def print_arg(**kwargs):
print kwargs.pop('key')
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如果我pop_arg用这样的字典调用方法:
mydict = {'key':'value'}
print_arg(**mydict)
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mydict这个电话会改变吗?
我也对mydict更改或不更改的底层方法调用机制的更详细解释感兴趣.
mgi*_*son 15
不,mydict不会改变.kwargs被解压缩到一本新词典中.
考虑一下你有以下情况:
def print_arg(key=1,**kwargs):
print key
print kwargs
print_arg(**{'key':2,'foo':3,'bar':4})
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在这种情况下,显而易见的是,kwargs是一个不同于你传入的字典,因为当它被解压缩时,它会丢失'key'密钥.
@ mgilson的回答是正确的.但是你也应该知道浅拷贝.
def print_arg(**kwargs):
print kwargs.pop('key')
kwargs['list'].pop() # will affect the original
kwargs['dict'].pop('a') #will affect the original
mydict = {'key':'value', 'list':[2,3,4,5] ,'dict': { 'a':1,'b':2 }}
print_arg(**mydict)
print (mydict) # Output : {'dict': {'b': 2}, 'list': [2, 3, 4], 'key': 'value'}
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让我们来看看:
import dis
dis.dis(lambda: print_arg(**{'key': 'value'}))
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6 0 LOAD_GLOBAL 0 (print_arg)
3 BUILD_MAP 1
6 LOAD_CONST 1 ('value')
9 LOAD_CONST 2 ('key')
12 STORE_MAP
13 CALL_FUNCTION_KW 0
16 RETURN_VALUE
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让我们找到CALL_FUNCTION_KW(ceval.c):
case CALL_FUNCTION_VAR:
case CALL_FUNCTION_KW:
case CALL_FUNCTION_VAR_KW:
{
// ...
x = ext_do_call(func, &sp, flags, na, nk);
// ...
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static PyObject *
ext_do_call(PyObject *func, PyObject ***pp_stack, int flags, int na, int nk)
{
int nstar = 0;
PyObject *callargs = NULL;
PyObject *stararg = NULL;
PyObject *kwdict = NULL;
PyObject *result = NULL;
if (flags & CALL_FLAG_KW) { // if ** is involved
kwdict = EXT_POP(*pp_stack); // get the dict passed with **
if (!PyDict_Check(kwdict)) {
PyObject *d;
d = PyDict_New(); // make a NEW dict
if (d == NULL)
goto ext_call_fail;
if (PyDict_Update(d, kwdict) != 0) { // update it with old
// .. fail ..
goto ext_call_fail;
}
Py_DECREF(kwdict);
kwdict = d; // kwdict is now the new dict
}
}
....
result = PyObject_Call(func, callargs, kwdict); // call with new dict
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