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发表于 2009-10-8 14:54:45
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float Scalars
Scalars of type float are stored using four bytes (32-bits). The format used follows the IEEE-754 standard and is stored little endian or LSB first.
A floating-point number is expressed as the product of two parts: the mantissa and a power of two. For example:
±mantissa × 2exponent
The mantissa represents the actual binary digits of the floating-point number.
The power of two is represented by the exponent. The stored form of the exponent is an 8-bit value from 0 to 255. The actual value of the exponent is calculated by subtracting 127 from the stored value (0 to 255) giving a range of –127 to +128.
The mantissa is a 24-bit value (representing about seven decimal digits) whose most significant bit (MSB) is always 1 and is, therefore, not stored. There is also a sign bit that indicates whether the floating-point number is positive or negative.
Floating-point numbers are stored on word boundaries in the following format:
Address+0 Address+1 Address+2 Address+3
Contents MMMM MMMM MMMM MMMM EMMM MMMM SEEE EEEE
Where
S represents the sign bit where 1 is negative and 0 is positive.
E is the exponent with an offset of 127.
M is the 24-bit mantissa (stored in 23 bits).
Zero is a special value denoted with an exponent field of 0 and a mantissa of 0.
Using the above format, the floating-point number -12.5 is stored as a hexadecimal value of 0xC1480000. In memory, this value appears as follows:
Address+0 Address+1 Address+2 Address+3
Contents 0x00 0x00 0x48 0xC1
It is fairly simple to convert floating-point numbers to and from their hexadecimal storage equivalents. The following example demonstrates how this is done for the value -12.5 shown above.
The floating-point storage representation is not an intuitive format. To convert this to a floating-point number, the bits must be separated as specified in the floating-point number storage format table shown above. For example:
Address+0 Address+1 Address+2 Address+3
Format MMMMMMMM MMMMMMMM EMMMMMMM SEEEEEEE
Binary 00000000 00000000 01001000 11000001
Hex 00 00 48 C1
From this illustration, you can determine the following:
The sign bit is 1, indicating a negative number.
The exponent value is 10000010 binary or 130 decimal. Subtracting 127 from 130 leaves 3, which is the actual exponent.
The mantissa appears as the following binary number:
10010000000000000000000
There is an understood binary point at the left of the mantissa that is always preceded by a 1. This digit is omitted from the stored form of the floating-point number. Adding 1 and the binary point to the beginning of the mantissa gives the following value:
1.10010000000000000000000
To adjust the mantissa for the exponent, move the decimal point to the left for negative exponent values or right for positive exponent values. Since the exponent is three, the mantissa is adjusted as follows:
1100.10000000000000000000
The result is a binary floating-point number. Binary digits to the left of the decimal point represent the power of two corresponding to their position. For example, 1100 represents (1 × 23) + (1 × 22) + (0 × 21) + (0 × 20), which is 12.
Binary digits to the right of the decimal point also represent the power of two corresponding to their position. However, the powers are negative. For example, .100... represents (1 × 2-1) + (0 × 2-2) + (0 × 2-3) + ... which equals .5.
The sum of these values is 12.5. Because the sign bit was set, this number should be negative.
So, the hexadecimal value 0xC1480000 is -12.5.
Copyright (c) Keil Software, Inc. and Keil Elektronik GmbH. All rights reserved.
看看你的keil中索引float，上面是完整的拷贝。现在的老师也够懒的，出的题居然从keil里找，呵呵。你自己翻译吧。 |
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