问题描述
如何添加和减去16位浮点半精度数字?
How do I add and subtract 16 bit floating point half precision numbers?
说我需要加或减:
1 10000 0000000000
1 10000 0000000000
1 01111 1111100000
1 01111 1111100000
2的补码形式.
推荐答案
假设您使用的是与IEEE单/双精度类似的非规范化表示形式，只需计算符号=(-1)^ S，尾数为1如果E！= 0则为.M，如果E == 0则为0.M，并且指数= E-2 ^(n-1)，则对这些自然表示进行运算，然后转换回16位格式.
Assuming you are using a denormalized representation similar to that of IEEE single/double precision, just compute the sign = (-1)^S, the mantissa as 1.M if E != 0 and 0.M if E == 0, and the exponent = E - 2^(n-1), operate on these natural representations, and convert back to the 16-bit format.
sign1 = -1尾数1 = 1.0指数1 = 1
sign1 = -1mantissa1 = 1.0exponent1 = 1
sign2 = -1尾数2 = 1.11111exponent2 = 0
sign2 = -1mantissa2 = 1.11111exponent2 = 0
总和:符号= -1尾数= 1.111111指数= 1
sum:sign = -1mantissa = 1.111111exponent = 1
表示形式:1 10000 1111110000
Representation: 1 10000 1111110000
自然地，这假设对指数进行过多编码.
Naturally, this assumes excess encoding of the exponent.
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