将121.8125十进制数转化为二进制数?要求写出整数部分和小数部分的计算过程如何做
推荐回答(3个)
整数部分和小数部分一般分开进行计算。
121除以2余1,所以第1位为1,商60
60除以2余0,所以第2位为0,商30
30除以2余0,所以第3位为0,商15
15除以2余1,所以第4位为1,商7
7除以2余1,所以第5位为1,商3
3除以2余1,所以第6位为1,商1
1除以2余1,所以第7位为1
故整数部分为1111001
0.8125乘以2=1.625>1,所以第1位为1,1.625-1=0.625
0.625乘以2= 1.25>1,所以第2位为1, 1.25-1=0.25
0.25乘以2= 0.5<1,所以第3位为0
0.5乘以2= 1=1,所以第4位为1
故小数部分为0.1101
综上得到121.8125的二进制为1111001.1101
因为要参加软考了(当然也只有考试有这种魅力),我得了概浮点数转化为二进制表示这个最难的知识点(个人认为最难)。俺结合大量的从网上收集而来的资料现整理如下,希望对此知识点感兴趣的pfan有所帮助。基础知识:
十进制转十六进制;
十六进制转二进制;
IEEE制定的浮点数表示规则;
了解:
目前C/C++编译器标准都遵照IEEE制定的浮点数表示法来进行float,double运算。这种结构是一种科学计数法,用符号、指数和尾数来表示,底数定为2——即把一个浮点数表示为尾数乘以2的指数次方再添上符号。下面是具体的规格:
符号位 阶码 尾数 长度
float 1 8 23 32
double 1 11 52 64
以下通过几个例子讲解浮点数如何转换为二进制数
例一:
已知:double类型38414.4。
求:其对应的二进制表示。
分析:double类型共计64位,折合8字节。由最高到最低位分别是第63、62、61、……、0位:
最高位63位是符号位,1表示该数为负,0表示该数为正;
62-52位,一共11位是指数位;
51-0位,一共52位是尾数位。
步骤:按照IEEE浮点数表示法,下面先把38414.4转换为十六进制数。
把整数部和小数部分开处理:整数部直接化十六进制:960E。小数的处理:
0.4=0.5*0+0.25*1+0.125*1+0.0625*0+……
实际上这永远算不完!这就是著名的浮点数精度问题。所以直到加上前面的整数部分算够53位就行了。隐藏位技术:最高位的1不写入内存(最终保留下来的还是52位)。
如果你够耐心,手工算到53位那么因该是:38414.4(10)=1001011000001110.0110011001100110011001100110011001100(2)
科学记数法为:1.001011000001110 0110011001100110011001100110011001100,右移了15位,所以指数为15。或者可以如下理解:
1.001011000001110 0110011001100110011001100110011001100×2^15
于是来看阶码,按IEEE标准一共11位,可以表示范围是-1024 ~ 1023。因为指数可以为负,为了便于计算,规定都先加上1023(2^10-1),在这里,阶码:15+1023=1038。二进制表示为:100 00001110;
符号位:因为38414.4为正对应 为0;
合在一起(注:尾数二进制最高位的1不要):
01000000 11100010 11000001 110 01100 11001100 11001100 11001100 11001100
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