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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