将 integer 转换为 IEEE 浮点数？

Question

I am currently reading "Computer Systems: A Programmer's Perspective". In the book, big-endian is used(most significant bits first). In the context of IEEE floating point numbers, using 32-bit single-precision, here is a citation of conversion between an integer and IEEE floating point:

One useful exercise for understanding floating-point representations is to convert sample integer values into floating-point form. For example, we saw in Figure 2.15 that 12,345 has binary representation [11000000111001]. We create a normalized representation of this by shifting 13 positions to the right of a binary point, giving 12,345 = 1.10000001110012 × 2^13. To encode this in IEEE single-precision format, we construct the fraction field by dropping the leading 1 and adding 10 zeros to the end, giving binary representation [10000001110010000000000]. To construct the exponent field, we add bias 127 to 13, giving 140, which has binary representation [10001100]. We combine this with a sign bit of 0 to get the floating-point representation in binary of [01000110010000001110010000000000].

What I do not understand is "by dropping the leading 1 and adding 10 zeros to the end, giving binary representation [10000001110010000000000]." If big-endian is used, why can you add 10 zeros to the end of 1000000111001? Doesn't that lead to another value than that after the binary point? It would make sense to me if we added 10 zeros in the front since the final decimal value would still be that originally after the binary point.

Why/how can you add 10 zeros to the back without changing the value if big-endian is used?

Answer 1

This is how the number 12345 is represented as a 32-bit single-precision IEEE754 float:

                  3  2          1         0
                  1 09876543 21098765432109876543210
                  S ---E8--- ----------F23----------
          Binary: 0 10001100 10000001110010000000000
             Hex: 4640 E400
       Precision: SP
            Sign: Positive
        Exponent: 13 (Stored: 140, Bias: 127)
       Hex-float: +0x1.81c8p13
           Value: +12345.0 (NORMAL)

Since this is a NORMAL value, the fractional part is interpreted with an implicit 1-bit; that is it's 1.10000001110010000000000 . So, to fill the 23 bit mantissa you simply add 10 0's at the end as it doesn't change the value.

Endianness isn't really related to how these numbers are represented, as each bit has a fixed meaning. But in general, the most-significant-bit is to the left in both the exponent and the mantissa.