课程名称︰高等程式设计 课程性质︰选修 课程教师：刘邦锋 开课学院：电资学院 开课系所︰资工所、网媒所 考试日期（年月日）︰2013.04.16 考试时限（分钟）： 试题 : Advanced Computer Programming 2013 Midterm Examination 04/16/2013 The first part of the test is a written test. Please answer the following problem with sufficient details. (50%) 1. Explain the functionalities of gcc compiler options -c, -o , -E, -g, and -O. Also please describe in what situation you need to use them. (10%) Ans: Each option for [2%] ○ -c: Compile or assemble the source file but not link, so the output will be an object file. ○ -o: Specify the name of output. ○ -E: Stop after preprocessing stage. ○ -g: Produce debug information in the operating system's native format. ○ -O: Optimize. 2. What is the output of the following program? Please explain in the answer in details. (5%) #include <iostream> using namespace std; int main() { int i, j, *temp; int a[5] = {3, 5, 1, 2, 4}; int *ptr[5] = {&(a[0]), &(a[1]), &(a[2]), &(a[3]), &(a[4])}; int ref0 = a[0]; int ref1 = a[1]; int ref2 = a[2]; int ref3 = a[3]; int ref4 = a[4]; for (i = 0; i < 4; i++) for (j = 0; j < 4; j++) if (*(ptr[j]) > *(ptr[j + 1])) { temp = ptr[j]; ptr[j] = ptr[j + 1]; ptr[j + 1] = temp; } for (i = 0; i < 5; i++) cout << a[i] << ' ' << *(ptr[i]) << endl; cout << ref0 << endl; cout << ref1 << endl; cout << ref2 << endl; cout << ref3 << endl; cout << ref4 << endl; } Ans: ○ Each kind of output (a[i], ptr[i], refs) for [1%]* 3 1 5 2 1 3 2 4 4 5 3 5 1 2 4 ○ Explanation. [2%] Sort the pointers by its dereferenced value. The value in array a is not changed, so as the references. 3. In 1968 Professor Edsger W. Dijkstra wrote the following. Please describe his argument why he is against the use of goto. (10%) My second remark is that our intellectual powers are rather geared to master static relations and that our powers to visualize processes evolving in time are relatively poorly developed. For that reason we should do (as wise programmers aware of our limitations) our utmost to shorten the conceptual gap between the static program and the dynamic process, to make the correspondence between the program (spread out in text space) and the process (spread out in time) as trivial as possible. Ans: We can tread a block structure as a unit of code, making programmer easier to maintain. Using goto will break such block structure. 4. Describe the reason why we need to write the short statement in the then part. (5%) Ans: See section 3.4.5 in "programming-style.pdf" [5%] // Ahbong's comment I think that we should also keep "else" part short for the same reason. There are two acceptable style: ○ Use guard condition, then there's no "else" part. ○ "if" part should be the normal case and "else" part should be the abnormal case. For switch statement the most frequent condition should be at the top, and so on for other conditions. 5. Describe the caching behavior and performance difference in the following code when compiled with and without XY. What kind of tool can you use to confirm the caching behavior? (10%) #include <stdio.h> #define DIMX 1000 #define DIMY 100000 int array[DIMX][DIMY]; int main() { int x, y; #ifdef XY for (x = 0; x < DIMX; x++) for (y = 0; y < DIMY; y++) array[x][y] = x + y; #else for (y = 0; y < DIMY; y++) for (x = 0; x < DIMX; x++) array[x][y] = x + y; #endif return 0; } Ans: ○ State that array in C/C++ is row-major order [4%]* ○ Code compiled with defining XY has less cache miss and thus has better performance. [4%]* ○ valgrind [2%]* 6. What is the output of the following program if we run it in a UNIX environment? Please explain the answer in details. (10%) #include <stdio.h> #include <assert.h> void dump_file(char *filename, char *mode) { int c; int count = 0; FILE *fp = fopen(filename, mode); assert(fp != NULL); while ((c = fgetc(fp)) != EOF) { printf("%02x ", c); count++; if (count % 8 == 0) putchar('\n'); } fclose(fp); printf("\nThere are %d bytes in file %s\n", count, filename); return; } int main(void) { FILE *fp; char c; fp = fopen("binary", "wb"); assert(fp != NULL); fputs("hello\n", fp); fclose(fp); fp = fopen("text", "wt"); assert(fp != NULL); fputs("hello\n", fp); fclose(fp); dump_file("binary", "rb"); dump_file("text", "rb"); dump_file("text", "rt"); return 0; } Ans: ○ "printf("%02x", c)" prints the ASCII code in hex format of each character in file. [2%]* ○ Each of the output ("binary" in "rb", "text" in "rb", "text" in "rt") for [2%]* All of the three dump_file say that "There are 6 bytes in file ***" ○ Explanation [2%]* The treatment of binary and text file are the same in UNIX environment. The second part of the test is the programming part. (50%) 1. Please download the problem description and the buggy code. Then fix the buggy code and upload to judgegirl. (15%) 2. Please enhance the BasicLinkedList class with a new method called insert. The insert function will insert a key i into the list before the first key that is no less than it. For example, if the list is now 1, 2, and 3. Then after inserting 3 then list will become 1, 2, 3, and 3. Note that if we keep inserting using insert, the list will be sorted all the time. The prototype of insert is as follows. void insert(const int i); You should use public inheritance to build the class. (15%) 3. Please build a class Date, The class must support the following member functions. (20%) a. Date::Date(); A default constructor that initializes the date to 01/01/1970. b. Date::Date(const int y, const int m, const d); A constructor to initialize it to year, month, and date. You may assume that all input are valid. c. void Date::add(const int n); Add n days to the current date. For example, if date is 02/28/2000, then after adding one day, i.e. date.add(1), date becomes 02/29/2000. Note that n could be negative, but it is between -1000000 and 1000000. Also note that the year will never become non-positive even when the n is negative. We will use the standard rule to determine whether a year is a leap year. d. int between(const Date &d) const; Return the number of days between two Date, For example, it should return 366 between 01/01/2000 and 01/01/2001. e. void Date::setmode(DateMode m); This is a class method that controls the printing format. If the mode has been set using Date::setmode(dayfirst), the date will be printed as day first, like 29/02/2000. If the mode has been set using Date::setmode(monthfirst) then the date will be printed as 02/29/2000. If the mode has been set as Date::setmode(text), the output should be February 29, 2013. The default mode is monthfirst. monthfirst, dayfirst, and text should be defined as an enum type like this. enum DateMode {dayfirst, monthfirst, text}; f. void print() const; The following is an example on how to control the output. #include <iostream> #include "date.h" using namespace std; int main() { Date date(2000, 2, 29); date.print(); Date::setmode(dayfirst); date.print(); Date::setmode(text); date.print(); } The output is as follows. 2/29/2000 29/2/2000 February 29, 2000 --

※ 发信站: 批踢踢实业坊(ptt.cc), 来自: 61.231.132.45 ※ 文章网址: https://webptt.com/cn.aspx?n=bbs/NTU-Exam/M.1461408321.A.B99.html