关于IEEE745浮点数单精度的学习
关于5.75 161.875 -0.0234375的单精度浮点数转化.
| 原始数字 | 5.75 | 161.875 | -0.0234375 |
| 转化为二进制 | 101.11 | 10100001.111 | -0.0000011 |
| 标准化 | 1.0111*2^2 | 1.0100001111*2^7 | -1.1*2^（6） |
| 转化 | 0 10000001 0111 0000 0000 0000 0000 000 | 0 10000110 0100 0011 1100 0000 0000 000 | 1 01111001 1000 0000 0000 0000 0000 000 |
利用网站检验
利用Python转化参考https://blog.csdn.net/qq_40890756/article/details/83111431
def ConvertFixedIntegerToComplement(fixedInterger) :#浮点数整数部分转换成补码(整数全部为正) return bin(fixedInterger)[2:]
def ConvertFixedDecimalToComplement(fixedDecimal) :#浮点数小数部分转换成补码
fixedpoint = int(fixedDecimal) / (10.0**len(fixedDecimal))
s = ''
while fixedDecimal != 1.0 and len(s) < 23 :
fixedpoint = fixedpoint * 2.0
s += str(fixedpoint)[0]
fixedpoint = fixedpoint if str(fixedpoint)[0] == '0' else fixedpoint - 1.0
return s
def ConvertToExponentMarker(number) : #阶码生成
return bin(number + 127)[2:].zfill(8)
def ConvertToFloat(floatingPoint) :#转换成IEEE754标准的数
floatingPointString = str(floatingPoint)
if floatingPointString.find('-') != -1 :#判断符号位
sign = '1'
floatingPointString = floatingPointString[1:]
else :
sign = '0'
l = floatingPointString.split('.')#将整数和小数分离
front = ConvertFixedIntegerToComplement(int(l[0]))#返回整数补码
rear = ConvertFixedDecimalToComplement(l[1])#返回小数补码
floatingPointString = front + '.' + rear #整合
relativePos = floatingPointString.find('.') - floatingPointString.find('1')#获得字符1的开始位置
if relativePos > 0 :#若小数点在第一个1之后
exponet = ConvertToExponentMarker(relativePos-1)#获得阶码
mantissa = floatingPointString[floatingPointString.find('1')+1 : floatingPointString.find('.')] + floatingPointString[floatingPointString.find('.') + 1 :] # 获得尾数
else :
exponet = ConvertToExponentMarker(relativePos)#获得阶码
mantissa = floatingPointString[floatingPointString.find('1') + 1: ] # 获得尾数
mantissa = mantissa[:23] + '0' * (23 - len(mantissa))
floatingPointString = '0b' + sign + exponet + mantissa
print(floatingPointString)
return hex( int( floatingPointString , 2 ) )
print(ConvertToFloat(-5.56))
