School of Computer Science Algorithms & Programming. Fall Midterm Examination # 2 Wednesday, November 14, 2007.

Transcription

1 School of Computer Science Algorithms & Programming Fall 2007 Midterm Examination # 2 Wednesday, November 14, 2007 Marking Scheme Duration of examination: 75 minutes 1. Answer all questions on this examination paper in the space provided. 2. This is a closed-book examination no notes or books or electronic computing or storage or communications devices may be used. 3. Do not copy from other students or communicate in any way. All questions will be answered only by the attending proctors. 4. Each student must sign an exit attendance sheet before leaving the examination. 5. The examination must be surrendered immediately when the instructor announces the end of the test period. Failure to do so will result in a mark penalty of 5 marks. Total mark obtained: Maximum mark: 38 Page 1

2 Question 1. [ 8 marks ] This question deals with tracing. For the C program below, show what output would be produced. Your answers must be stated precisely, accounting for proper formatting. It is highly recommended that you show your tracing work deriving the answers, neatly, within the space provided beside the program codes. #include <stdio.h> #include <string.h> int main ( ) /* SHOW TRACE BELOW */ int L = 3, M = 4, N = 5 ; float X= ; char S[] = Spring flowers ; char *P, *Q ; P = &S[2]; Q = strstr( S, low ) ; printf( A1 = %s ** %5s\n, Q, low ) ; printf( A2 = %.1f ** %5.2f\n, X, X ) ; printf( A3 = %d ** %d\n, Q-P, strlen(s) ) ; printf( %d\*%dn, strchr(s, f )-&S[1], M<N&&L>N-M ) ; return 0 ; Neatly, print the output, if any, that would be produced by the program above. Proper formatting is very important so a grid is provided use one character per box. More than enough boxes are provided and the first row is labeled for easier referencing of column spacing A 1 = l o w e r s * * l o w A 2 = 2. 3 * * A 3 = 6 * * * 1 Carefully type in the program using a standard editor and then run the program to check the results. I have had some cases where students said they ran the program (as justification for their appeal) and obtained an output that was identical to what they placed on their exams. However, I have found in all cases that the programs they ran were NOT identical to this program, and therefore their outputs were not proper. Page 2

3 Question 2. [ 20 marks ] This question deals with programming of string handling functions. For each part you are given a specification for a function, including a prototype, description of input and output requirements and assumptions. For each part you are to write a complete function definition for each stated prototype. You must not use any char or string handling functions from the C libraries. However, you may reference any function that may be required from an earlier Part of this Question 2, regardless of whether you have fully or correctly defined the function logic. PART 2A. [ 3 marks ] Prototype: int StringLength ( char * S ) ; Input: S is a pointer to a collection of characters (char). Output: Returns the number (int) of characters (char) in S not equal to \0. Assumptions: All char values in S are valid ASCII. It is possible that \0 might not occur in this case, when the number of char values is greater than MAXLEN a return value of -1 is executed immediately. MAXLEN is defined by a #define statement located elsewhere (and previously). /* SOLUTION Q.2.A */ int StringLength ( char * S ) int L = 0 ; while( *S!= \0 ) L++ ; S++ ; if( L > MAXLEN ) return -1 ; return L ; Page 3

4 Question 2. ( Continued ) PART 2B. [ 10 marks ] Prototype: int StringCompareN ( char * S, char * F, unsigned int N ) ; Input: S and F are pointers to different, non-overlapping, collections of characters (char). N is an unsigned integer value (>= 0). Output: Returns an integer value (int) indicating: 1 if S > F; 0 if S == F; -1 if S < F. The comparisons indicated represent lexical ordering of the strings S and F. Assumptions: All char values in S and F are valid ASCII strings, delimited by \0. The string comparison is lexical (ie. dictionary). It is possible that either S or F or both are null strings (with length 0). Strings of length zero have minimum lexical value so all non- NULL strings are defined as greater than 0 length strings. It is possible that N is zero (0), in which case a value of 0 is returned. It is possible that N is less than the minimum length of S or F, in which case only up to the first N characters of string F are compared with string S. You may use the StringLength() function defined in Part 2A. /* SOLUTION Q.2.B */ int StringCompareN ( char * S, char * F, unsigned int N ) int SL, FL, k, L ; if( N == 0 ) return 0 ; /* Zero length comparison*/ SL = StringLength(S) ; FL = StringLength(F) ; if( SL == 0 && FL == 0 ) return 0 ; /* Zero length strings */ if( SL == 0 ) return -1 ; /* S is zero length, S<F */ if( FL == 0 ) return 1 ; /* F is zero length, S>F */ if( SL > FL ) L = FL ; /* Determine the */ else L = SL ; /* minimum of N, SL, FL */ if( N < L ) L = N ; /* and initialize L */ for( k=0 ; k<l ; k++, S++, F++ ) /* Compare each char */ if( *S > *F ) return 1 ; /* S > F case */ if( *S < *F ) return -1 ; /* S < F case */ return 0 ; /* S == F case */ Page 4

5 Question 2. ( Continued ) PART 2C. [ 7 marks ] Prototype: char * StringFind ( char * S, char * F ) ; Input: Output: S and F are pointers to different, non-overlapping, collections of characters (char). Returns a pointer to the location in S where F is found. If F is not found in S, a value of NULL is returned. Assumptions: All char values in S and F are valid ASCII strings, delimited by \0. It is possible that either S or F or both are null strings. It is possible that the length of S is less than the length of F, in which case a NULL value can be returned immediately. You may use the StringLength() function defined in Part 2A. You may use the StringCompareN() function defined in Part 2B. /* SOLUTION Q.2.C */ char * StringFind ( char * S, char * F ) int L, SL, FL, j, k ; SL = StringLength(S) ; FL = StringLength(F) ; if( SL < FL ) return NULL ; /* S is smaller than F */ /* Determine maximum number of comparison substrings in S */ L = SL FL + 1 ; for( j=0 ; j < L ; j++ ) if( StringCompareN( S+j, F, FL ) == 0 ) return (S+j) ; /* F is found in S, return location */ return NULL ; /* F is not found in S, return NULL */ Page 5

6 Question 3. [ 10 marks ] Define the function whose prototype is given below. You may not use any functions from the string.h library, but you may use any other functions defined in other C-libraries. Prototype: int StringCatExtra ( char * S, char * F ) ; Input: Output: S and F are pointers to different collections of characters (char). If the length of F is less than or equal to the length of S, then S is modified by concatenating the character representation of the string length to S, and the new length of S is returned. If the length of F is greater than the length of S then only the characters in F occurring after the length of S are concatenated to S, along with the character representation of the new string length to S, creating a new valid string, and the length of the new string S is returned. Assumptions: All char values in S and F are valid ASCII strings, delimited by \0. It is possible that either S or F or both are null strings. You may use the StringLength() function defined in Question 2A, StringCompareN() function defined in Question 2B, and/or StringFind() function defined in Question 2C. The following are examples that illustrate how the function should work. Input S Input F Output S StringCatExtra output 0 1 (two null strings are inputted) William Mary William7 8 Mary William Maryiam7 8 Sister Brother-in-law Sisterr-in-law14 16 /* SOLUTION TO Q3 */ int StringCatExtra ( char * S, char * F ) int k, LS, LF ; LS = StringLength( S ); FS = StringLength( F ); k = LS; if( LS < FS ) /* catenate F chars onto S */ for(; k<fs; k++ ) *(S+k) = *(F+k) ; sprintf( S+k, %d, FS ); *(S+FS) = \0 ; return StringLength( S ); NOTE: In the original exam the first example Output S value should have been 0, not 1. This was a mistake in the exam and students should not have had marks deducted for this specific mistake. Page 6