What will be the value in float if I have a binary number as 1111111111111111 and the storage formats used by Intel processors is 32 bits? [closed]

问题

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Closed 7 years ago.

I have to represent a binary number in floating point number. I have a hexadecimal number as FFFF, when I am converting this hexadecimal number into Binary I am getting the corresponding binary number as 1111111111111111. The storage formats used by my Intel processor is 32 bits means 1 bit for the sign, 8 bits for the exponent, and 23 bits for the mantissa. I have some idea but quite confused. Can Anyone help me out what will be the corresponding float value for this binary number??

回答1:

32 ones

Simply try it out:

#include <stdio.h>

int main() {
  union {
    unsigned i;
    float f;
  } u;
  u.i = 0xffffffff;
  printf("%f\n", u.f);
  return 0;
}

prints -nan. This experiment assumes that you actually wanted 0xffffffff not 0xffff as your question states.

Looking at http://en.wikipedia.org/wiki/Single_precision you find that exponent 0xFF together with a non-zero significand is treated as NaN.

16 ones

If you really are after 0xFFFF only, as your question writes, then the code will print 0.000000, but changing the %f to %g you get 9.18341e-41. This is because both the integer and the float use the same endianess, i.e. you are talking about the float corresponding to the bit pattern 0x0000ffff.

There you see that you'll now have zero sign (i.e. positive), zero exponent and a non-zero significand. According to the same wikipedia article, this represents a subnormal number. So this is really 0xffff ∙ 2⁻¹⁴⁹ = 65535 ∙ 2⁻¹⁴⁹.

回答2:

Assuming IEEE-754 32-bit float, 0xFFFFFFFF (so 32 ones):

The first is the sign, so it's a negative float. Then comes the exponent, since it's the maximum value, it's a "special" exponent.

Because the significand is not zero, the result is NaN.

Source: Wikipedia.

If you just want 16 ones, then:

Assuming the remaining bits are zero and we're using little-endian, we'll have `0xFFFF0000`. This is still not enough for the significand to be zero, so the result is still `NaN`. In fact, regardless of which values you pick for the last 16 bits, the result will always be `NaN`.

Apparently, this is not the case as demonstrated by MvG.

Assuming the remaining bits are zero, the input is 0x0000FFFF. The first bit is the sign, which is a zero. So we have a positive float.
Then we have 8 bits for the exponent. Since they are all zero, we'll have (depending on the fraction) either zero or a subnormal number.
Because the fraction is not all zeroes, this float is a subnormal number.
At this point, we've checked 9 bits, so there are still 7 extra zeroes.
The fraction is made of 7 zeroes, followed by 16 ones, therefore the significand is (in binary) 0.00000001111111111111111. Multiply that by 2e-126 and you'll have your answer.
The result is, apparently, 9.18341e-41.

来源：https://stackoverflow.com/questions/13673045/what-will-be-the-value-in-float-if-i-have-a-binary-number-as-1111111111111111-an