Strings Investigating Memory Allocation Pointers Fixed-Point Arithmetic. Memory Matters. Embedded Systems Interfacing.

Transcription

1 22 September 2011

2 Strings Single character char ch; char ch = a ; char ch = 0x41; Array of characters char str[5]={ H, e, l, l, o }; Null-terminated string char str[ ]= Hello

3 String Runtime Library #include <string.h> Memory copy, compare, move and initialize (set) String copy, concatenate, compare (collate), scan, tokenize, length and error message strings Some functions have safe versions available Specify string length Does not just depend on \0

4 Memory Space Allocation char s[ ]= Hello Zippy! ; C-30 linker reserves contiguous memory in RAM (data space) which is part of ndata (near data section). C-30 linker stores initialized value in long table (in program memory) which is part of init code section. C-30 compiler creates routine to copy from init to ndata section which is part of c0 code. This uses twice the memory we would expect.

5 Memory Space Allocation (cont d) const char s[ ]= Hello Zippy! ; C-30 linker stores initialized value in long table (in program memory) which is part of init code section. Can not copy other string over top of this string because of const. This uses exactly the amount of memory we would expect.

6 Program Space Visibility

7 Program Space Visibility (cont d)

8 Program Space Visibility (cont d) PSV is used to manage constant arrays (numeric of string) so that a single type of pointer to the data memory bus Can be used uniformly for constants and variables Table access mechanism is used to perform variable initialization, limited to c0- segments, for maximum compactness and efficiency

9 #include <string.h> const char a[]= Zippy loves the PIC24. ; char b[100]; int main(){ strncpy(b, MPLAB C-30,sizeof(b)); }

10 Watch Window Before Execution After Execution

11 Dissassembly Listing

12 Map File Strings

13 .const in Program Memory

14 and Arrays: One-Dimensional const int Enrollment1K[ ]= {3383,1299,463,179,44,14}; int enrollment1ktotal,size,group; size=sizeof(enrollment1k)/sizeof(int); for(group=0;group<size;group++) { enrollment1ktotal+=*(enrollment1k+group); }

15 and Arrays: One-Dimensional const int Enrollment1K[ ]= {3383,1299,463,179,44,14}; int enrollment1ktotal,size,group; size=sizeof(enrollment1k)/sizeof(int); for(group=0;group<size;group++) { enrollment1ktotal+=*(enrollment1k+group); } Dereference Base Address Subscript

16 Strings int main(){ int *pi; // pointer to integer int i; // index or counter int a[10]; // array of ten integers for(i=0;i<sizeof(a)/sizeof(int);i++) a[i]=i; // access array with index } pi=a; // initialize array pointer for(i=0;i<sizeof(a)/sizeof(int);i++,pi++) *pi=i; //access array with pointer

17 The Heap Strings Heap is open unused RAM Allocate via command line argument for linker Allocate large area so malloc( ) and calloc( ) can make most efficient use of heap

18 C-30 Memory Model

19 More Information String.h contains addition function for memory block manipulations Ctype.h functions are used to determine if character is upper-case, lower-case, and to convert between Null (ASCII-Z) strings allow for compact code (see text) For a safety check of a pointer to see if it is null, compare it with NULL declared in stdlib.h

20 Introduction Numerical Operations The Need for No hardware support for floating-point Floating-point in software time consuming PIC C-18 supports 24-bit and 32-bit floating-point in software Microchip has floating-point libraries available

21 Introduction Numerical Operations A fixed point number places the decimal point between the whole and fractional parts of a number at a fixed location, providing f bits of fractional precision Example: an 8.24 fixed-point number has an eight-bit integer portion and twenty-four bits of fractional precision. Since the split between the whole and fractional portion is fixed, we know exactly what our range and precision will be.

22 16.16 Fixed-Point Format Introduction Numerical Operations This format fits perfectly into a typical 32-bit character in C We have 0xIIIIFFFF, where IIII is the integer portion and FFFF is the fraction portion. The integer part allows us 2 16 values from 32, 768 to +32, 767 as signed integers The fraction portion allows 2 16 discrete values to represent numbers between 0 and 1 This range is from 0 to 65,535 65, The resolution is 1 65,

23 Example Strings Introduction Numerical Operations 0x00104ACF approximately represents The high sixteen bits are 0x0010, which is 16. The low sixteen bits are 0x4ACF, so 0x4ACF / We can convert from fixed point to decimal by dividing the number by 2 f 0x00104ACF

24 Example (cont d) Strings Introduction Numerical Operations Note that the numbers are stored in two s comp xFFEFB531. The fraction portion of is 0xB531 0x4ACF. We cannot just mask the fraction portion to get the fraction.

25 Aside: Two s Comp Introduction Numerical Operations For n bits, 2 n numbers are represented, 2 n 1 positive and 2 n 1 negative. Positive numbers are represented as expected. Negative numbers are represented modulo-2 n 1. Practically, this is encoded by inverting each bit and then adding one Decoding the two s comp is accomplished by inverting each bit and adding one

26 Aside: Two s Comp (cont d) Introduction Numerical Operations Advantages of Two s Comp Zero has a single representation. Addition and subtraction using positive and negative numbers works. Disadvantages of Two s Comp Decoding negative numbers takes thought. The magnitude of a negative number is inversely proportional to the magnitude of its binary representation. For fixed-point representations, the fraction portions of a positve and negative number of the same magnitude are different.

27 Conversion Strings Introduction Numerical Operations Converting to and from fixed-point formats is fairly simple From integer the conversion is accomplished by shifting left f bits: F = i << f. From floating point, the conversion is accomplished by multiplying by 2 f : F = 2 f r Converting from fixed-point to floating point is trivial r = 2 f F Converting from fixed-point to integer is more difficult because of rounding questions Round-towards-zero Round-towards-positive-infinity Round-towards-negative-infinity Round-to-nearest

28 Addition Strings Introduction Numerical Operations It just works; A2 f + B2 f = (a + b) 2 f If you have two representations with f 1 and f 2 you have to align the decimal points You can convert both to f 1. You can convert both to f 2. You can convert both to some other format.

29 Multiplication Strings Introduction Numerical Operations This is a bit more complicated A larger size intermediate memory variable may be needed. Suppose A2 m and B2 n are being multiplied, and the answer outght to have f bits of precision. A2 m B2 n = AB2 m+n The product is over-scaled, so we need to get it back to the desirable range by multiplying by 2 f m n Thus, C2 f = AB2 m+n 2 f m n

30 Division Strings Introduction Numerical Operations This is similar to multiplication Suppose A2 m is to be divided by B2 n, and the answer ought to have f bits of precision. A2 m B2 n = A B 2m n C2 f = A2m 2 f m+n B2 n