Accessing Data in Memory

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1 Accessing Data in Memory You will implement a simple C function that parses a tangled list of binary records in memory, processing them nonsequentially, and produces a simple text report. The function must conform to the following interface specification: / Untangle() parses a chain of records stored in the memory region pointed to by pbuffer and writes a report to the supplied file stream. Formatting details are provided in the specification for the assignment. Pre: pbuffer points to a region of memory formatted as specified. Log points to an opened text file. Post: The target of pbuffer has been parsed and a report written as specified. Returns: the number of tokens in the supplied quotation. / uint8_t Untangle(const uint8_t pbuffer, FILE Log); The first two bytes of the memory region contain the offset at which you will begin processing records; the offset is stored as a uint16_t value. This offset is followed by a collection of word records, each consisting of an integer value, a sequence of characters, and another integer value: Length of word Characters in word Offset of next word uint8_t chars uint16_t The first value in each record specifies the number of characters in the sequence for that record. Since words are never very long, this value will be stored as a uint8_t, which has a range of This is followed by a sequence of ASCII codes for the characters that make up the word. (The term "word" is used a bit loosely here.) The last character is followed immediately (no string terminator!) by a uint16_t value specifying the offset of the next word record in the list. Note that the length of the record depends upon the number of characters in the word, and so these records are variable-length. While traversing the memory region, your program will write to a text file reporting the results of the traversal. Aside from a header, each line of the file will specify the hexadecimal offset at which a word record was found, and then the word from that record. Memory Layout: Your program will deal with a single chunk of memory, in the format described above. Here is a hex dump of a sample memory region: A 0B 0C 0D 0E 0F F _..species5..vir B 3A 00 0A 6D 6F F tue;:..moderatio E E E 7A 00 nu..vice.d..inz F is@..of#..but D 2D F 6D a...--I..Thomas E C 00 0A R..Paine...isl D 6F F 6E 2B C 77 Moderation+..alw E D ays...in...tempe C 5A 00 0A E C 65 2C 30 r,z..principle, C 00..a A 0B 0C 0D 0E 0F 1

2 Since I'm using x86 hardware, integer values are stored in memory in little-endian order; that is, the low-order byte is stored first (at the smallest address) and the high-order byte is stored last (at the largest address). So the bytes of a multibyte integer value appear to be reversed. As a programmer, you usually do not need to take the byte-ordering into account since the compiler will generate machine language compatible with your hardware, and that will make use of the bytes in the appropriate manner. But, when you're reading memory displays, you must take the byte-ordering into account. So, looking at the first two bytes of the memory block, we see that the word record we will process first occurs at relative offset 0x005F from the beginning of the memory block. The first word record is: 0A 4D 6F F 6E 2B 00 The length of the word is 0x0A or 10 in base-10. The ASCII codes are 4D 6F F 6E. The next record is at a relative offset of 0x002B. Here is the log file, named "Log.txt", corresponding to the given input files: 005F: Moderation 002B: in 007A: temper, 005A: is 006C: always 0091: a 000C: virtue; 003A: but 0016: moderation 0075: in 0084: principle, 0030: is 0040: a 0002: species 0035: of 0023: vice. 0044: : Thomas 0052: Paine Number of words reported: 19 To aid in your testing, a simple driver (driver.c) will be posted on the website. This driver will read a binary file and build a suitable in-memory data region, and then call your implementation of Untangle(). Some sample binary files will also be posted to use with this driver. In addition, a binary file generator (tangle) will be posted as a 32-bit Linux executable. You may use this program to generate additional binary test files for the aforementioned simple driver. Type tangle --help for more information. Submitting Your Program: You will submit this assignment to the Curator System (read the Student Guide), and it will be graded automatically. Instructions for submitting, and a description of how the grading is done, are contained in the Student Guide. For this project you should develop your implementation, using separate compilation, and then submit a single.c file containing your function implementation and any necessary include directives. You may write helper functions that are called by your code; if so, be sure to put any necessary declarations for those in your.c file, not in a header file. Your solution should not write to standard output. Submit your source file to the Curator system under the heading HW10. 2

3 You will be allowed multiple submissions for this assignment. Use them wisely. Test your program thoroughly before submitting it. If you do not get a perfect score, analyze the problem carefully and test your fix before submitting again. The highest score you achieve will be counted. The Student Guide and link to the submission client can be found at: Evaluation: Your submitted code will be compiled with a driver program, which will allocate and initialize a memory region in the format described above, open a FILE pointer on an empty log file, and call your implementation of the function Untangle(). An appropriate header file will be provided, containing a declaration for your function implementation and any include directives needed for the prototype. The supplied header file will be named Untangle.h, so if you need to include that header into your.c file, make sure you use that name for it. Your submitted program will be assigned a score based upon the runtime testing performed by the Curator System. After that, your program may be given a brief evaluation by a TA, who will consider whether you have followed the requirements in the specification above, and how well you have documented your implementation. See the Programming Standards page on the course website for specific requirements that should be observed in this course. This evaluation will produce a deduction (ideally zero, of course) that will be applied to your runtime testing score to produce your final score for the project. Pledge: Each of your program submissions must be pledged to conform to the Honor Code requirements for this course. Specifically, you must include the pledge statement provided with the earlier project specifications in the header comment for your source code file. 3

4 Brief tutorial on using pointer arithmetic and pointer typecasts in C: This assignment does not require any esoteric portions of the C language, but it does require an understanding of pointer manipulations. First of all, you must understand the effect and uses of a pointer typecast. Consider the following snippet: uint16_t x = malloc(sizeof(short int)); // 1 x = 684; // 2 Statement 1 causes the allocation of a two-byte region of memory, whose address is stored in the pointer x. Statement 2 stores the value 684 (0x2AC or in binary) into that two-byte region. So the current situation in memory would be: x the Stack???? low byte (at low address) high byte (at next address) the Heap (The bytes are stored in little-endian order, just as they would be on any Intel-compatible system.) If you dereference the pointer x, you'll obtain the contents of the two bytes beginning at the address stored in x. That's because x was declared as a pointer to something of type uint16_t, and sizeof(uint16_t) is 2 (bytes). Now consider: uint8_t y = NULL; // 3 y = (uint8_t) x; // 4 uint8_t z = y; // 5 uint16_t w = x; // 6 The effect of statement 4 is that y takes on the same value as x; pointer variables are 32 bits wide, regardless of the type of target they may take, and so the value of x will fit neatly into y. So, why the typecast? Simply that C is somewhat picky about assigning a pointer of one type to a pointer of another type, and the typecast formalizes the logic so that the compiler will accept it. If you dereference y, you'll obtain the contents of the single byte at the address stored in y, since sizeof(uint8_t) is 1. Hence, statement 5 will assign the value 0xAC or 172 to z, but statement 6 will assign the two-byte value 0x02AC or 684 to the variable w. Here's a brief summary: 4

5 Note how we can use the type of a pointer to determine how many bytes of memory we obtain when we dereference that pointer, as well as how those bytes are interpreted. This can be really useful. The second thing to understand is how pointer arithmetic works. Here is a simple summary: T p, q; // Take T to represent a generic type.... // Assume p gets assigned a target in here. q = p + K; // Let K be an expression that evaluates to an integer. Then the value of q will be: p + K sizeof(t). Note well: this is very dangerous unless you understand how to make use of it. In some respects, this is really quite simple; maybe too simple. The essential thing you must always remember is that if you want to move a pointer by a specific number of bytes, it's simplest if the pointer is a char or uint8_t, since the arithmetic will then provide you with byte-level control of the pointer's logical position in memory. The following loop would walk the pointer y through the bytes of the target of x (the uint16_t seen earlier): uint8_t y = (uint8_t) x; uint32_t bytesread = 0; while ( bytesread < sizeof(uint16_t) ) { printf("%"prix8"\n", y); // print value of current byte in hex; // 'x' causes output in hex; // 'X' capitalizes the hex } ++y; ++bytesread; // step to next byte of target of x You might want to experiment with this a bit Now for copying values from memory into your variables (which are also in memory also, of course)... The simplest approach is use appropriate variable types and pointer typecasts. Suppose that the uint8_t pointer p points to some location in memory and you want to obtain the next four bytes and interpret them as an int32_t value; then you could try these: int32_t N = p; // NO. This takes only 1 byte! int32_t q = (int32_t) p; // Slap an int32_t onto the location; int32_t N = q; // so this takes 4 bytes as desired. int32_t N = ((int32_t) p); // Or do it all in one fell swoop. The last form is the most idiomatic in the C language; it creates a temporary, nameless int32_t from p and then dereferences it to obtain the desired value. Note that this doesn't move p. So, if you wanted to copy the next few bytes you'd need to apply pointer arithmetic to move p past the bytes you just copied: p = p + sizeof(int32_t); // Since p is a uint8_t, this will move p forward // by exactly the size of an int32_t. One final C tip may be of use. The C Standard Library includes a function that will copy a specified number of bytes from one region of memory into another region of memory: memcpy(). You can find a description of how to use the function in any decent C language reference, including the C tutorial linked from the Resources page of the course website. 5

6 The C Standard, section says: When an expression that has integer type is added to or subtracted from a pointer, the result has the type of the pointer operand. If the pointer operand points to an element of an array object, and the array is large enough, the result points to an element offset from the original element such that the difference of the subscripts of the resulting and original array elements equals the integer expression. In other words, if the expression P points to the i-th element of an array object, the expressions (P)+N (equivalently, N+(P)) and (P)-N (where N has the value n) point to, respectively, the i+n-th and i n-th elements of the array object, provided they exist. Moreover, if the expression P points to the last element of an array object, the expression (P)+1 points one past the last element of the array object, and if the expression Q points one past the last element of an array object, the expression (Q)-1 points to the last element of the array object. If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow; otherwise, the behavior is undefined. If the result points one past the last element of the array object, it shall not be used as the operand of a unary operator that is evaluated. A useful Linux utility: This assignment requires processing in-memory data. The supplied test driver makes use of an input file that stores the exact contents of a sample memory block of the type that your function must process. Because the file contains exactly the same bytes as would be stored in memory, there is no point in trying to view the contents of the file in a typical editor, or in cat'ing it to the console window. However, there is a handy Linux utility: hexdump. The example below uses the "canonical" switch to display the contents of a binary file as formatted text. The left side shows the offset of the beginning of each "line". The middle portion shows the bytes contained in the file, formatted in hexadecimal and chopped into 16-byte chunks. The right side shows the corresponding ASCII characters (for bytes whose values are valid ASCII codes). MSI-Ubuntu HW > hexdump -C Data01.bin c f 75 3c the...you<..D e f 6f c 1c ay...choose,...d c b f f ay,...yourg..cho e 0e e 64 4d f ac ice...andm..do e is...way...wh e at...integritya f c your,..characte a f e 6b 2c r...youy..think, f d d 2d 9a your b e f a y...yourt..herac a0 6c d0 litus...is`..is b ea f 75 be what...you...b c f 6d 65 2e c f0 00 ecome...light d f b8 00 0c e 79 2e 2e.who...destiny e0 2e e f it...is...you f f that...guides f d f.yourF..The...co e e f ntent...of%..wha ts

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