Cx51 User's GuideCx51 Introduction Compiling Programs Language Extensions Preprocessor Advanced Programming Customization Files Startup Code Variable Initialization Code XBANKING.A51 Basic I/O Memory Allocation Optimizer Optimizer Options General Optimizations 8051-Specific Optimizations Global Register Coloring Segment Naming Conventions Data Objects Program Objects Parameters and Local Variables Interfacing C to Assembler Global Variables Function Parameters Passing in Memory Passing in Registers Function Return Values Using the SRC Directive Register Usage Overlaying Segments Small Model Example Compact Model Example Large Model Example Interfacing C programs to PL/M-51 Data Storage Formats Bit Variables 1-Byte Scalars 2-Byte Scalars 4-Byte Scalars Generic and Far Pointers Floating-point Numbers Floating-point Errors Absolute Memory Locations Absolute Access Macros Linker Location Controls The _at_ Keyword Debugging 8051 Derivatives Error Messages Library Reference Appendix
Scalars of type float are stored using four bytes (32-bits). The format used follows the IEEE-754 standard.
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 byte boundaries in the following format:
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:
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:
From this illustration, you can determine the following:
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:
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:
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.
of your data.