python 整数越界问题详解
python 内部自带大整数运算能力,整数运算不会溢出,只要内存足够,就oK
下面的例子演示了两个32位整数加法的情况(通过位运算实现),为了模拟溢出的效果,必须人工的进行位运算,~运算符除了求反,还是二进制的补运算符,运算过后的二进制数字按照补码解释,例如 ~(0011 1100) = (1100 0011) = -61
def getSum(a, b):
"""
:type a: int
:type b: int
:rtype: int
"""
MAX = 0X7fffffff
MIN = 0X80000000
while b != 0 :
a,b = a^b,(a&b)<<1
print(" a = {0:b},b = {1:b}".format(a,b))
return a
def getSum_(a, b):
"""
:type a: int
:type b: int
:rtype: int
"""
MAX = 0x7FFFFFFF
MIN = 0x80000000
mask = 0xFFFFFFFF
while b != 0:
a, b = (a ^ b) & mask, ((a & b) << 1) & mask
print(type(a))
print(" a = {0:b},b = {1:b}".format(a,b))
return a if a <= MAX else ~(a^mask)
print(getSum_(-1,-1))
print(getSum(-1,1))
补充:python 循环内部添加多个条件判断会出现越界
1.循环遍历数组是,想添加条件修改时,只删除第一个
# -*- coding: utf-8 -*-
a=[11,22,33,44,55]
for i in a:
if i == 11 or i ==22:
a.remove(i)
for i in a:
print(i)
'''
33
55
[Finished in 0.1s]
'''
2.应该引入被删除为一个数组
# -*- coding: utf-8 -*-
a=[11,22,33,44,55]
b=[]
for i in a:
if i == 11 or i ==22:
b.append(i)
for i in b:
a.remove(i)
for i in a:
print(i)
'''
33
44
55
[Finished in 0.1s]
'''
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