单精度浮点数在vc++6.0中内存格式研究

 

#include <iostream.h>

int main()
{
    float a=2;  /*这里分别让a等于2，-2，4，6，1，0.75，2.5，0.1，0  */
    char *p=(char*)&a;
    p+=3; /*将地址由高到低*/
    int line=0,temp=0;
    while(temp<4)
    {
        for(int i=7;i>=0;i--)
        {
            if(*p&(1<<i)) cout<<'1'; /*作与运算，如果和1与为1，说明该二进制是1，否则为0，其中左移运算让1不断往高位移动*/
            else cout<<'0';
            line++;
            if(line%4==0) cout<<' ';
        }
        temp++;

        p--;
    }
    cout<<endl;
    return 0;
}

　

　

测试结果及结果说明，给出MSDN Library上的解释（英文原版，原汁原味）：

The format, then, for the various sizes is as follows:

Format BYTE 1 BYTE 2 BYTE 3 BYTE 4 ... BYTE n real*4 XXXX XXXX XMMM MMMM MMMM MMMM MMMM MMMM     real*8 SXXX XXXX XXXX MMMM MMMM MMMM MMMM MMMM ...  MMMM MMMM real*10 SXXX XXXX XXXX XXXX 1MMM MMMM MMMM MMMM ...  MMMM MMMM

注：研究的是4字节浮点数，即float型数。

S represents the sign bit, the X's are the exponent bits, and the M's are the mantissa bits. Note that the leftmost bit is assumed in real*4 and real*8 formats, but is present as "1" in BYTE 3 of the real*10 format.

To shift the binary point properly, you first unbias the exponent and then move the binary point to the right or left the appropriate number of bits.

Examples

The following are some examples in real*4 format:

In the following example, the sign bit is zero, and the stored exponent is 128, or 100 0000 0 in binary, which is 127 plus 1. The stored mantissa is (1.) 000 0000 ... 0000 0000, which has an implied leading 1 and binary point, so the actual mantissa is one.

SXXX XXXX XMMM MMMM ... MMMM MMMM 2 = 1 * 2**1 = 0100 0000 0000 0000 ... 0000 0000 = 4000 0000 Same as +2 except that the sign bit is set. This is true for all IEEE format floating-point numbers.

-2 = -1 * 2**1 = 1100 0000 0000 0000 ... 0000 0000 = C000 0000 Same mantissa, exponent increases by one (biased value is 129, or 100 0000 1 in binary.

4 = 1 * 2**2 = 0100 0000 1000 0000 ... 0000 0000 = 4080 0000 Same exponent, mantissa is larger by half — it's (1.) 100 0000 ...0000 0000, which, since this is a binary fraction, is 1 1/2 (the values of the fractional digits are 1/2, 1/4, 1/8, and so forth).

6 = 1.5 * 2**2 = 0100 0000 1100 0000 ... 0000 0000 = 40C0 0000 Same exponent as other powers of two, mantissa is one less than two at 127, or 011 1111 1 in binary.

1 = 1 * 2**0 = 0011 1111 1000 0000 ... 0000 0000 = 3F80 0000 The biased exponent is 126, 011 1111 0 in binary, and the mantissa is (1.) 100 0000 ... 0000 0000, which is 1 1/2.

.75 = 1.5 * 2**-1 = 0011 1111 0100 0000 ... 0000 0000 = 3F40 0000 Exactly the same as two except that the bit that represents 1/4 is set in the mantissa.

2.5 = 1.25 * 2**1 = 0100 0000 0010 0000 ... 0000 0000 = 4020 0000 1/10 is a repeating fraction in binary. The mantissa is just shy of 1.6, and the biased exponent says that 1.6 is to be divided by 16 (it is 011 1101 1 in binary, which is 123 in decimal). The true exponent is 123 – 127 = –4, which means that the factor by which to multiply is 2**–4 = 1/16. Note that the stored mantissa is rounded up in the last bit — an attempt to represent the unrepresentable number as accurately as possible. (The reason that 1/10 and 1/100 are not exactly representable in binary is similar to the reason that 1/3 is not exactly representable in decimal.)

0.1 = 1.6 * 2**-4 = 0011 1101 1100 1100 ... 1100 1101 = 3DCC CCCD 0 = 1.0 * 2**-128 = all zeros--a special case.

本文地址：http://com.8s8s.com/it/it24135.htm