Python函数中闭包和延迟绑定详情

def multipliers():
return [lambda x : i*x for i in range(4)]
print ([m(2) for m in multipliers()] )
"""
[6, 6, 6, 6]
"""

def multipliers():
funcs = []
for i in range(4):
def bar(x):
return x*i
funcs.append(bar)
return funcs
print ([m(2) for m in multipliers()] )
"""
[6, 6, 6, 6]
"""

# 为了便于理解，你可以想象下multipliers内部是这样的(这个是伪代码，并不是准确的)：
def multipliers():
return [lambda x: 3 * x, lambda x: 3 * x, lambda x: 3 * x, lambda x: 3 * x]

def multipliers():
# 添加了一个默认参数i=i
return [lambda x, i=i: i*x for i in range(4)]
print ([m(2) for m in multipliers()] )
"""
[0, 2, 4, 6]
"""

def multipliers():
funcs = []
for i in range(4):
def bar(x, i=i):
return x * i
funcs.append(bar)
return funcs
print ([m(2) for m in multipliers()] )
"""
[0, 2, 4, 6]
"""

# 为了便于理解，你可以想象下multipliers内部是这样的(这个是伪代码只是为了理解)：
def multipliers():
return [lambda x,i=0: i*x, lambda x,i=1: i*x, lambda x,i=2: i*x, lambda x,i=3:i*x i=3]
# x的引用是2 所以output的结果就是：[0,2,4,6]

def multipliers():
# 添加了一个默认参数a=i
return [lambda x, a=i: x*a for i in range(4)]
print ([m(2) for m in multipliers()] )
"""
[0, 2, 4, 6]
"""