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java中int类型所能表示的最大值和最小值是怎么来的?

疯狂的小佳
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11回答

牛津小学徒

你可以了解一下计算机二进制部分的内容,原码,反码,补码等,看完基本就懂了

慕莱坞4347929

暗黑风离开你来看我.............................................................................................................

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成都加米谷大数据

建议复习一下二进制的相关知识

troyoqczaq

你好a

Steve杨

int 是占4个字节, 32个0`1 的表示能力就是 2^32 (4,294,967,296) 个整数. 可以想象一条数据, 我们把整数 0, 1, 2, ... 一直到2^32 - 1  (4,294,967,295) 都表示出来了.但是int 是带正负号的,  我们生活中需要用到负200 负1000 的概率要远大于  30亿, 35亿的情况, 说白了,我们不能只考虑正数, 所以把这2^32 左移动了一半, 这样我们的数轴就变成了-2147483648     ~~ 2147483647 这样我们就得到了int类型的最大值了最小值.类似的, 如果有一天int 变成6个节字或8个字节, 我们也能快速推导出新的上界下界了.备注: 我觉得以下2个数字大概有个印象, 2^32 = 4,294,967,296       约43亿2^31 = 4,294,967,296 / 2 =  2147483648     约21.5亿

宝慕林8457364

int max =Integer.MAX_VALUE;//最大值 int min =Integer.MIN_VALUE;//最小值 再用打印输出语句输出结果即可!

翔仔

同学您好,int类型是8个字节,32位,最大值用二进制表示就是, 0111...(总共31个1)。为什么第一位是0? 二进制里,最高位(第一位)表示符号0表示正,1表示负。所以将二进制01111...(31个1)转换成十进制,就是:2^0*1 + 2^1*1 + 2^2*1...+2^n*1就变成了2的n次方求和, 这里的n就是30,因为第一位是0,即2^31 * 02的n次方求和(n=30),由等比数列可知, 2^0 + 2^1 + 2^2...+2^n=2^(n+1)-1 = 2^31-1同理最小值的二进制表示就是,1000...(总共31个0),(负数二进制转十进制,是用补码(反码加1)转的。)1000(31个0)的补码还是本身。 所以最小值就是, 2^0*0 + 2^1*0 + 2^2*0...+ 2^31*1=2^31= -2^31

Dusdii

......

qq_Kero_1

。。。 
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