python 传递参数

2016-09-20  本文已影响25人  bigtom
def foo(bar):
    print "inside {} with id {}".format(bar, id(bar))

bar = "bar"
print "outside {} with id {}".format(bar, id(bar))
foo(bar)

baz = [1,2]
print "outside {} with id {}".format(baz, id(baz))
foo(baz)

返回结果

outside bar with id 4334173920
inside bar with id 4334173920
outside [1, 2] with id 4333883616
inside [1, 2] with id 4333883616

显然python传参数的方式是传引用。

不可变对象

对于不可变对象,你在函数内做什么都不会改变外面的结果

def foo(bar):
    print "inside {} with id {}".format(bar, id(bar))
    bar = "baz"
    print "inside {} with id {}".format(bar, id(bar))
bar = "bar"
print "outside {} with id {}".format(bar, id(bar))
foo(bar)
print "outside {} with id {}".format(bar, id(bar))

返回结果

outside bar with id 4472704736
inside bar with id 4472704736
inside baz with id 4472704776
outside bar with id 4472704736

我们可以很明显的看到,在执行bar = "baz"之后,bar就指向了另一个字符串对象了,而在函数外面,什么都没有改变。

可变对象

对于可变对象,你可以在内部改变它,这种改变会反馈到外面。

def foo(bar):
    print "inside {} with id {}".format(bar, id(bar))
    bar.append(3)
    print "inside {} with id {}".format(bar, id(bar))

bar = [1,2]
print "outside {} with id {}".format(bar, id(bar))
foo(bar)
print "outside {} with id {}".format(bar, id(bar))

返回结果

outside [1, 2] with id 4358607072
inside [1, 2] with id 4358607072
inside [1, 2, 3] with id 4358607072
outside [1, 2, 3] with id 4358607072

我们在内部改变了bar这个list(可变对象),它的引用没有改变。

但是如果你在函数中修改了引用,那你在函数中做任何操作都不会影响到外面。

def foo(bar):
    print "inside {} with id {}".format(bar, id(bar))
    bar = [1,2,3]
    print "inside {} with id {}".format(bar, id(bar))

bar = [1,2]
print "outside {} with id {}".format(bar, id(bar))
foo(bar)
print "outside {} with id {}".format(bar, id(bar))

结果

outside [1, 2] with id 4544712928
inside [1, 2] with id 4544712928
inside [1, 2, 3] with id 4544951360
outside [1, 2] with id 4544712928

我们在函数中改变了变量bar指向的引用,它就不再指向外面那个list了,而是指向了一个新创建的list。对这个新list进行任何操作都不会影响外面的那个list

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