CS007A Fall 12 Second Midterm

Transcription

1 CS007A Fall 12 Second Midterm 1. A programmer writes the following code to create a 3x2 matrix and print it out. a. Are there syntax errors? List all that you can find. b. Suppose the syntax errors are fixed. Are there run-time errors? List all that you can find. #include <iostream> using namespace std; int main() int coordinates[2][2] = 0,2,3,1,4,5; output(coordinates[][2]); void output(int coordinates) for(i = 0; i < 3; ++i) for(j = 0; j < 3; --j) cout << coordinates[j][i] << " "; cout << endl; 2. Consider following complete program: #include <iostream> using namespace std; const int arraysize = 4; void show2darray(int [][arraysize],int); int main() int a[arraysize][arraysize] = 0; for(int i=0;i<arraysize;++i) a[i][i] = i*i; show2darray(a,arraysize); return 0; void show2darray(int b[][arraysize],int n) for(int i=0;i<n;++i) for(int j=0;j<n;++j) cout << b[i][j] << " "; cout << endl; a. What is the function prototype here and what is it for? b. How many bytes are allocated in memory for array a? Hint: use the sizeof operator. c. In the function, show2darray(), which variable is passed by reference and which variable is passed by value? d. How would the program be altered to produce the output below?

2 3. Consider the following complete program. #include <iostream> using namespace std; const int dim = 3; void outerprod(double [], double [], double [][dim], int); int main() double prod[dim][dim] = 0; double fact01[] = 1.,2.,3.; double fact02[] = 4.,5.,6.; outerprod(fact01, fact02, prod, dim); outerprod(fact02, fact01, prod, dim); return 0; void outerprod(double f1[], double f2[], double p[][dim], int dim) for(int i = 0; i < dim; ++i) for(int j = 0; j < dim; ++j) p[i][j] = f1[i]*f2[j]; a. How many bytes are allocated in memory for prod? b. Why does dim have to be a const? c. Which variables are passed by reference in outerprod()? d. Fill in a table of values for i, j and p[i][j] corresponding to each of the function calls to outerprod(). That is, thread through the program, step by step, showing the values for i, j and p[i][j] at each step. 4. Write a program to simulate 128 rolls of two 6-sided dice and store the results in a 2x128 double-indexed array named Dice. Then write a function called Eights which takes this array and returns the number times the sum of the dice is 8. In the main() function, use cout << Eights(Dice) to print this result to the console. Use the const qualifier where appropriate. The program should be complete and run without bugs when typed in and compiled. 5. Suppose that p = 0x0012FF50, &a = 0x0012FF44, &p = 0x0012FF38. What will be output by the following code? int main() int a[] = 22, 33, 44, 55, 66, 77, 88, 99 ; int* p = &a[3]; // p points to a[3] cout << "p = " << p << ", *p = " << *p; cout << "\n&a = " << &a; // ok: a is an lvalue cout << "\n&p = " << &p; // ok: p is an lvalue cout << "\n&(a[5]) = " << &(a[5]); // ok: a[5] is an lvalue cout << "\n&(*(a+5)) = " << &(*(a+5));// ok: *(a+5) is an lvalue cout << "\n&(*(p+2)) = " << &(*(p+2));// ok: *(p+2) is an lvalue cout << "\n&(p+2) = " << p+2; //&(p+2); // ERROR: p+2 is not an lvalue cout << endl;

3 6. Consider the code below. a. How many syntax errors can you find? b. Are there any logic errors? c. Are there any parameters that should be qualified as const? d. Fix the program so it works and write a couple of sentences describing what the program is intended to do. #include <iostream> #include <iomanip> using namespace std; void mean(int [], int ); void median( int [], int ); void mode( int [], int [], int ); void bubblesort( int[], int ); void printarray(int[], int ); int main() const int responsesize = 99; // size of array responses int frequency[ 10 ] = 1 ; // initialize array frequency // initialize array responses int response[ ] = 6, 7, 8, 9, 8, 7, 8, 9, 8, 9, 7, 8, 9, 5, 9, 8, 7, 8, 7, 8, 6, 7, 8, 9, 3, 9, 8, 7, 8, 7, 7, 8, 9, 8, 9, 8, 9, 7, 8, 9, 6, 7, 8, 7, 8, 7, 9, 8, 9, 2, 7, 8, 9, 8, 9, 8, 9, 7, 5, 3, 5, 6, 7, 2, 5, 3, 9, 4, 6, 4, 7, 8, 9, 6, 8, 7, 8, 9, 7, 8, 7, 4, 4, 2, 5, 3, 8, 7, 5, 6, 4, 5, 6, 1, 6, 5, 7, 8, 7 ; // process responses mean( response, responsesize ); median( response, responsesize ); mode( frequency, response, responsesize ); return 0; // indicates successful termination // end main // calculate average of all response values void mean(int answer[], int arraysize ) int total = 0; cout << "********\n Mean\n********\n"; // total response values for ( int i = 0; i < arraysize; i++ ) total += answer[ i ]; // format and output results cout << fixed << setprecision( 4 ); cout << "The mean is the average value of the data\n" << "items. The mean is equal to the total of\n" << "all the data items divided by the number\n" << "of data items (" << arraysize << "). The mean value for\nthis run is: " << total << " / " << arraysize << " = " << static_cast< double >( total ) / arraysize << "\n\n"; // end function mean

4 // sort array and determine median element's value void median( int answer[], int size ) cout << "\n********\n Median\n********\n" << "The unsorted array of responses is"; printarray( answer, size ); // output unsorted array bubblesort( answer, size ); // sort array cout << "\n\nthe sorted array is"; printarray( answer, size ); // output sorted array // display median element cout << "\n\nthe median is element " << size / 2 << " of\nthe sorted " << size << " element array.\nfor this run the median is " << answer[ size / 2 ] << "\n\n"; // end function median // determine most frequent response void mode( int freq[], int answer[], int size ) int largest = 0; // represents largest frequency int modevalue = 0; // represents most frequent response cout << "\n********\n Mode\n********\n"; // initialize frequencies to 0 for ( int i = 1; i <= 9; i++ ) freq[ i ] = 0; // summarize frequencies for ( int j = 0; j < size; j++ ) ++freq[ answer[ j ] ]; // output headers for result columns cout << "Response" << setw( 11 ) << "Frequency" << setw( 19 ) << "Histogram\n\n" << setw( 55 ) << " \n" << setw( 56 ) << " \n\n"; // output results for ( int rating = 1; rating <= 9; rating++ ) cout << setw( 8 ) << rating << setw( 11 ) << freq[ rating ] << " "; // keep track of mode value and largest fequency value if ( freq[ rating ] > largest ) largest = freq[ rating ]; modevalue = rating; // end if // output histogram bar representing frequency value for ( int k = 1; k <= freq[ rating ]; k++ ) cout << '*'; cout << '\n'; // begin new line of output // end outer for // display the mode value cout << "The mode is the most frequent value.\n" << "For this run the mode is " << modevalue << " which occurred " << largest << " times." << endl; // end function mode // function that sorts an array with bubble sort algorithm void bubblesort( int a[], int size ) int hold; // temporary location used to swap elements // loop to control number of passes for ( int pass = 1; pass < size; pass++ ) // loop to control number of comparisons per pass for ( int j = 0; j < size ; j++ ) // swap elements if out of order if ( a[ j ] > a[ j + 1 ] ) hold = a[ j ]; a[ j ] = a[ j + 1 ]; a[ j + 1 ] = hold;

5 // end if // end function bubblesort // output array contents (20 values per row) void printarray( const int a[], int size ) for ( i = 0; i < size; i++ ) if ( i % 20 == 0 ) // begin new line every 20 values cout << endl; cout << setw( 2 ) << a[ i ]; // end for // end function printarray 7. Consider the following code for manipulating polynomials (below). a. Get the program working in your compiler and use it to analyze the roots for i ii. 1 Describe your results. b. Make a flow chart to describe the operation of the i. findupperbound() ii. findzero(). iii. descartesrulepos() iv. findallzeros() // G Hagopian // Polynomial function handler #include < iostream> using namespace std; const int N = 100; // maximim size of the polynomial coeff array void getcoeff(double [], const int, int & degree); void clearcoeff(double c[]); // zero out coefficients double dodivision(double dividend[], double quotient[], int degree, double input, bool show); //double f(double [], int, double); void printpoly(double [], int degree); double findupperbound(double [], int degree); double findlowerbound(double [], int degree); int descartesrulepos(double [], int degree); int descartesruleneg(double [], int degree); double findzero(double [], double & left, double right, int degree,bool up); void findallzeros(double [], double [], double left, double right, int degree); void getmenu(double coeff[],double quotient[], double roots[], int & degree); double abs(double x) return x>0? x : -x; int main() int degree = 4; double coeff[n] = 1,-10,35,-50,24; // The coefficient of x^n is coeff[n] double quotient[n] = 0; double roots[n] = 0;// An array to hold the roots while(1) getmenu(coeff,quotient,roots,degree); return 0; void getmenu(double coeff[],double quotient[], double roots[], int & degree)

6 double r, start = 0, max = 1000, min; int choice, k; // i for choice 9. cout << "Choose what you want to do: " << endl << "1: Print the current polynomial." << endl << "2: Enter a new polynomial." << endl << "3: Do synthetic division by x - c." << endl << "4: Evaluate the polynomial at a point." << endl << "5: Find lower and upper bounds on the zeros." << endl << "6: List the possible number(s) of positive and negative roots using D's rule of signs." << endl << "7: Find a positive zero above a given number, if there is one." << endl << "8: Find a negative zero below a given number, if there is one." << endl << "9: Find all real zeros of a polynomial." << endl; do // get choice //cin.clear(); cin.ignore(); cout << "\nchoice: "; cin >> choice; while(choice < 1 choice > 9); switch (choice) case 1 : cout << "\nhere's your polynomial: "; printpoly(coeff,degree); cout << endl << endl; cin.ignore(); cin.get(); break; case 2 : getcoeff(coeff, N, degree); cout << "\nyou entered\n"; printpoly(coeff, degree); cout << endl << endl; cin.ignore(); cin.get(); break; case 3 : cout << "\nenter a value of c, and we'll divide f(x) by x-c: "; cin >> r; cout << "\nthe value of f(" << r << ") is " << dodivision(coeff,quotient,degree,r,true) << endl << endl; cin.ignore(); cin.get(); break; case 4 : cout << "\nenter a value of c, and we'll evaluate f(c): " ; cin >> r; cout << "\nthe value of f(" << r << ") is " << dodivision(coeff,quotient,degree,r,false) << endl << endl; cin.ignore(); cin.get(); break; case 5 : cout << "\nthe zeros of the polynomial are between x = "; cout << findlowerbound(coeff, degree); cout << " and x = "; cout << findupperbound(coeff, degree) << endl << endl; cin.ignore(); cin.get(); break; case 6 : cout << "\nthe possible numbers of positive zeros is "; for(int i = descartesrulepos(coeff, degree); i >= 0; i -= 2) cout << i << " "; cout << "\nthe possible numbers of negative zeros is ";

7 for(int i = descartesruleneg(coeff, degree); i >= 0; i -= 2) cout << i << " "; cin.ignore(); cin.get(); break; case 7 : max = findupperbound(coeff,degree); do cout << "\nenter the left endpoint of an interval where you seek a positive zero: "; cin >> start; while (start > max); cout << "\nthere is a postive zero near where x = "; cout << findzero(coeff,start,max,degree,true); cout << endl << endl; cin.ignore(); cin.get(); break; case 8 : max = findlowerbound(coeff,degree); do cout << "\nenter the right endpoint of an interval where you seek a negative zero: "; cin >> start; while (start < max); cout << "\nthere is a negative zero near where x = "; cout << findzero(coeff,start,max,degree,false); cout << endl << endl; cin.ignore(); cin.get(); break; case 9 : for(int j = 0; j < N; j++) roots[j]=0; // clear roots array min = findlowerbound(coeff, degree); max = findupperbound(coeff, degree); findallzeros(coeff,roots,min,max,degree); cout << "\nthe roots are "; k = 0; while(roots[k]!=0 && roots[k++]<10e99) cout << roots[k-1] << " "; cin.ignore(); cin.get(); break; default : break; void clearcoeff(double c[]) for(int i = 0; i < N; ++i) c[i]=0; void getcoeff(double c[], const int N, int & degree) do cout << "\nenter the degree of the polynomial, " << "\na number less than " << N << " : "; cin >> degree; while (degree >= N); for(int i = 0; i <= degree; ++i) cout << "\nenter the coefficient of x^" << i << ": "; cin >> c[i];

8 double dodivision(double dividend[], double quotient[], int degree, double r, bool show) clearcoeff(quotient); quotient[degree-1] = dividend[degree]; double value = dividend[degree]; for(int i = degree; i > 1; i--) value *= r; value += dividend[i-1]; quotient[i-2] = value; value *= r; value += dividend[0]; if(show) cout << "\nthe result of Polynomial division is \n\n"; printpoly(dividend,degree); if(r>=0) cout << endl << " = (x - " << r << ")*("; else cout << endl << " = (x + " << -r << ")*("; printpoly(quotient,degree-1); cout << ")"; if(value!=0) if(value>0) cout << " + " << value; else cout << " - " << -value; cout << endl; return value; void printpoly(double c[], int degree) if (degree <= 1) if(degree == 1) // it's a linear polynomial if(c[1]==1) cout << "x"; else cout << c[1] << "*x"; c[0] > 0? cout << " + " << c[0] : cout << " - " << -c[0]; else cout << c[0]; // it's a constant else // degree>1 // First print the leading term if(c[degree] == 1) cout << "x^" << degree; // treat the cases where leading coeff is 1 separately else if(c[degree] == -1) cout << "-x^" << degree; else // leading coeff is not 1 or -1 cout << c[degree]<< "x^" << degree; // Print out the rest of the terms for(int i = degree-1; i>1; i--) if(c[i]>0) if(c[i]==1) cout << " + x^" << i; //don't print coeff 1 else cout << " + " << c[i] << "*x^" << i; if(c[i]<0) if(c[i]==-1) cout << " - x^" << i; // don't print coeff -1 else cout << " - " << -c[i]<< "*x^" << i;

9 if((degree-i+1)%10 == 0) cout << endl; if(c[1] > 0) cout << " + " << c[1] << "*x"; if(c[1] < 0) cout << " - " << -c[1]<< "*x"; if(c[0] > 0) cout << " + " << c[0]; if(c[0] < 0) cout << " - " << -c[0]; double findupperbound(double c[], int degree) bool found = false; double quot[n] = 0; double incrmnt = 1/(2*c[degree]), remainder; double x = 0., sign; int count = degree-1; //, parity=0; while(!found) // && diff>bound/1000) remainder = dodivision(c, quot, degree, x,false); sign = quot[count]>0? 1. : -1.; count = degree-1; while(sign*quot[count-1]>=0 && count > 0) --count; if(count==0 && sign*remainder>=0) return x; else x += incrmnt; //cout << "\nx = " << x; return 0; cout << endl << endl; double findlowerbound(double c[], int degree) bool found = false; double sign; double quot[n] = 0; //This is an attempt to scale the increment proportional to the //leading coefficient double decrmnt = 1/(2*c[degree]), remainder; double x = 0.; int count = degree-1;//, parity=0; while(!found) // this conditional can likely be refined remainder = dodivision(c, quot, degree, x, false); //evaluate but don't show results sign = quot[count] < 0? -1. : 1.; // since count is initialized to 1 less than degree, it's the // degree of quot if(degree > 2) while(quot[count-1]*sign<=0 && quot[count]*quot[count-2]>=0 && count > 1) --count; if(count==1 && remainder*quot[0]<=0) return x; else x -= decrmnt; else if(quot[1]*remainder >=0 && quot[1]*quot[0] <=0)

10 return x; else x -= decrmnt; return 0; cout << endl << endl; int descartesrulepos(double c[], int degree) int count = 0; for(int i = 0; i < degree; ++i) if(c[i]*c[i+1] <= 0) ++count; return count; int descartesruleneg(double c[], int degree) int count = 0; for(int i = 0; i < degree; ++i) if(c[i]*c[i+1] >= 0) ++count; return count; // findzero takes a polynomial (coefficient array), an initial x value, a // bound, and the degree // bool up is used to tell whether to search up the x-axis or back // to the right. double findzero(double c[], double & x, double max, int degree, bool up) if(max == 0) return 10e100; // max is an upper/lower bound on the roots. If max = 0, none. double q[n]=0, step; // if(up) step = 1/(10*c[degree]); // for searching through positives else step = -1/(10*c[degree]); // for searching through negatives double x_next = x+step, y = dodivision(c,q,degree,x,false); double y_next = dodivision(c,q,degree,x_next,false); double midx, midy, toler=1.e-10; // First walk through some smallish x increments to find a change of // signs: while (y*y_next>0 && abs(x_next) < abs(max)) // y has not changed signs & not beyond bound y = y_next; x_next += step; // if not up, incr is negative y_next = dodivision(c,q,degree,x_next,false); if(y_next == 0) x = x_next; // set reference variable return x; // Got lucky! if(abs(x_next) > abs(max)) x = x_next; return 10e100; x = x_next;

11 x_next -= step; y = y_next; y_next = dodivision(c,q,degree,x_next,false); midy = y_next; while(abs(midy)>1e-10) midx = (x+x_next)/2; midy = dodivision(c,q,degree,midx,false); if(midy*y<0) // root is on the left side x_next = midx; else x = midx; if(up && x_next<=max) x = x_next; return x_next; else if(!up && x_next>=max) x = x_next; return x_next; else return 10e100; void findallzeros(double c[], double roots[], double negmax, double posmax, int degree) double step = 1/(2*c[degree]); // Fairly large step size scaled to leading coefficient double x = step; // start looking in positive direction int i=0; while(x<posmax) // posmax is determined by findupperbound() roots[i++] = findzero(c,x,posmax,degree,true); x += step; x = -step; // start looking in negative direction while(x>negmax) //negmax is determined by findlowerbound() roots[i++] = findzero(c,x,negmax,degree,false); cout << "\nroots[" << i-1 << "] = " << roots[i-1]; x -= step;

12 CS007A Second Midterm Solutions 1. A programmer writes the following code to create a 3x2 matrix and print it out. a. Are there syntax errors? List all that you can find. b. Suppose the syntax errors are fixed. Are there run-time errors? List all that you can find. #include <iostream> using namespace std; int main() int coordinates[2][2] = 0,2,3,1,4,5; output(coordinates[][2]); void output(int coordinates) for(i = 0; i < 3; ++i) for(j = 0; j < 3; --j) cout << coordinates[j][i] << " "; cout << endl; SOLN: There are these syntax errors. (1) Either a prototype for output() should be given, or the definition should precede main(). void output(int [][2]); would be a good prototype. (2) int coordinates[2][2] = 0,2,3,1,4,5; Here coordinates should be initialized with dimensions, [3][2]. (3) The call to output() in main() should contain only the name of the array, output(coordinates), not the dimensions. (4) The definition of output() should include the dimensions of the array at least the second size should be specified: (5) There are a bunch of run-time errors and syntax errors in the body of the definition of output(). The integer counters for the for loops are not declared, and the column index should only count two, instead of three. Here s one way to fix this function definition: void output(int [][2]); int main() int coordinates[3][2] = 0,2,3,1,4,5; output(coordinates); void output(int coordinates[][2]) for(int i = 0; i < 3; ++i) for(int j = 0; j < 2; ++j) cout << coordinates[i][j] << " "; cout << endl;

13 2. Consider following complete program: #include <iostream> using namespace std; const int arraysize = 4; void show2darray(int [][arraysize],int); int main() int a[arraysize][arraysize] = 0; for(int i=0;i<arraysize;++i) a[i][i] = i*i; show2darray(a,arraysize); return 0; void show2darray(int b[][arraysize],int n) for(int i=0;i<n;++i) for(int j=0;j<n;++j) cout << b[i][j] << " "; cout << endl; a. What is the function prototype here and what is it for? SOLN: The prototype is void show2darray(int [][arraysize],int); and is used to declare there is a function by that name and to specify its formal parameter list and what sort of return value it has (none, in this case.) b. How many bytes are allocated in memory for array a? Hint: use the sizeof operator. SOLN: The array a requires 4 bytes for each of its 16 integers for a total of 64 bytes. c. In the function, show2darray(), which variable is passed by reference and which variable is passed by value? SOLN: Array a (or b, as it is renamed in the body of the definition) is passed by reference (implicitly, since it s the name of an array) and arraysize (renamed n by the function definition) is passed by value. d. How would the program be altered to produce the output below? SOLN: You d change arraysize from 4 to 5. Then add a nested for loop in the main() function like so: const int arraysize = 5; int main() int a[arraysize][arraysize] = 0; for(int i=0;i<arraysize;++i) for(int j = 0; j < arraysize; ++j) a[i][j] = (i+1)*(j+1); show2darray(a,arraysize); cout << endl << sizeof(a); return 0;

14 3. Consider the following complete program. #include <iostream> using namespace std; const int dim = 3; void outerprod(double [], double [], double [][dim], int); int main() double prod[dim][dim] = 0; double fact01[] = 1.,2.,3.; double fact02[] = 4.,5.,6.; outerprod(fact01, fact02, prod, dim); outerprod(fact02, fact01, prod, dim); return 0; void outerprod(double f1[], double f2[], double p[][dim], int dim) for(int i = 0; i < dim; ++i) for(int j = 0; j < dim; ++j) p[i][j] = f1[i]*f2[j]; a. How many bytes are allocated in memory for prod? SOLN: Since const int dim = 3 and each int is 4 bytes, the 3*3=9 ints require 9*4 = 36 bytes. b. Why does dim have to be a const? SOLN: Because it s used to allocate memory for an array, which must be of constant size. c. Which variables are passed by reference in outerprod()? SOLN: Only dim is not passed by reference, double f1[], double f2[], double p[][dim] are arrays and, as such, are implicitly passed by reference. d. Fill in a table of values for i, j and p[i][j] corresponding to each of the function calls to outerprod(). That is, thread through the program, step by step, showing the values for i, j and p[i][j] at each step. i j p i j p

15 4. Write a program to simulate 128 rolls of two 6-sided dice and store the results in a 2x128 double-indexed array named Dice. Then write a function called Eights which takes this array and returns the number times the sum of the dice is 8. In the main() function, use cout << Eights(Dice) to print this result to the console. Use the const qualifier where appropriate. The program should be complete and run without bugs when typed in and compiled. SOLN: #include <iostream> #include <cstdlib> // rand() #include <ctime> //srand(time()) using namespace std; int Eights(int Dice[2][128]) int cntr = 0; for(int i = 0; i<128; ++i) if(dice[0][i]+dice[1][i]==8) ++cntr; return cntr; int main() srand(unsigned(time(0))); int Dice[2][128]; for(int i = 0; i<128; ++i) Dice[0][i] = 1+rand()%6; Dice[1][i] = 1+rand()%6; cout << Eights(Dice) << endl; 5. Suppose that p = 0x0012FF50, &a = 0x0012FF44, &p = 0x0012FF38. What will be output by the following code? int main() int a[] = 22, 33, 44, 55, 66, 77, 88, 99 ; int* p = &a[3]; // p points to a[3] cout << "p = " << p << ", *p = " << *p; cout << "\n&a = " << &a; // ok: a is an lvalue cout << "\n&p = " << &p; // ok: p is an lvalue cout << "\n&(a[5]) = " << &(a[5]); // ok: a[5] is an lvalue cout << "\n&(*(a+5)) = " << &(*(a+5));// ok: *(a+5) is an lvalue cout << "\n&(*(p+2)) = " << &(*(p+2));// ok: *(p+2) is an lvalue cout << "\n&(p+2) = " << p+2; //&(p+2); // ERROR: p+2 is not an lvalue cout << endl; SOLN: Clearly, the statement int* p = &a[3]; assigns the address of the 4 th element of the array a to the pointer, p. The dereferencing operator, when applied to point p, the produces the value stored at that address, which is the fourth element of the array a[3] = 55. So the first cout produces: p = 0x0012FF50, *p = 55

16 The second cout shows the address of a, so cout << "\n&a = " << &a; produces: &a = 0x0012FF44 The 3 rd cout tells us the address of the pointer itself, so cout << "\n&p = " << &p; produces: &p = 0x0012FF38 The 4 th cout tells the address of the 6 th element of the array, and since each element is a int requiring 4 bytes, this is 8 bytes beyond a[3], so cout << "\n&(a[5]) = " << &(a[5]); produces: &(a[5]) = 0x0012FF58 The 5 th cout prints the address of the contents at the address which is 5 ints beyond a (note the & and * essentially just cancel each other out) so that cout << "\n&(*(a+5)) = " << &(*(a+5)); produces 20 bytes (0x14) more than a, or the same thing as above: &(*(a+5)) = 0x0012FF58 The 6 th is two ints beyond where p started, which is, again, the address of a[5]. So we d get &(*(p+2)) = 0x0012FF58 Similarly, the address of a[5] can be obtained by p+2; So cout << "\n&(p+2) = " << p+2; also produces &(p+2) = 0x0012FF58 but the left side of this equation is misleading, since pointer arithmetic doesn t work with address of the pointer itself! Trying to print out &(p+2) will lead to a syntax error. 6. Consider the code [above #6]. a. How many syntax errors can you find? b. Are there any logic errors? c. Are there any parameters that should be qualified as const? d. Fix the program so it works and write a couple of sentences describing what the program is intended to do. SOLN: There is a syntax error in printarray(). The for loop index i has not been initialized. This is easily remedied like so: void printarray( const int a[], int size ) for (int i = 0; i < size; i++ ) //and so on. Attempting to compile the code after correcting this minor error leads to the cryptic linker error, error LNK2019: unresolved external symbol "void cdecl printarray(int * const,int)" (?printarray@@yaxqahh@z) referenced in function "void cdecl median(int * const,int)", which appears to reference some parameter list involving constant pointers to integers. A careful inspection of the code reveals the prototype: void printarray(int[], int ) which doesn t comport with this formal parameter list in the definition: printarray(const int a[], int size). Changing the prototype to match the definition, the code compiles, but there is a run-time error after

17 producing this much output: ******** Mean ******** The mean is the average value of the data items. The mean is equal to the total of all the data items divided by the number of data items (99). The mean value for this run is: 681 / 99 = ******** Median ******** The unsorted array of responses is The sorted array is The median is element 49 of the sorted 99 element array. For this run the median is 7 ******** Mode ******** It appears the sorted array is a bit out of whack. Analyzing the bubblesort() function we see that it has some serious logic errors, which we can rebuild like so: void bubblesort( int a[], int size ) int hold; // temporary location used to swap elements // loop to control number of passes for(int pass = 1; pass < size; pass++) // loop to control number of comparisons per pass for ( int j = 1; j <= size-pass ; j++ ) // swap elements if out of order if ( a[ j-1 ] > a[ j ] ) hold = a[ j-1 ]; a[ j-1 ] = a[ j ]; a[ j ] = hold; // end if // end function bubblesort

18 After running the program with bubblesort() fixed, we get this output, which appears to be correct: ******** Mean ******** The mean is the average value of the data items. The mean is equal to the total of all the data items divided by the number of data items (99). The mean value for this run is: 681 / 99 = ******** Median ******** The unsorted array of responses is The sorted array is The median is element 49 of the sorted 99 element array. For this run the median is 7 ******** Mode ******** Response Frequency Histogram * 2 3 *** 3 4 **** 4 5 ***** 5 8 ******** 6 9 ********* 7 23 *********************** 8 27 *************************** 9 19 ******************* The mode is the most frequent value.

19 For this run the mode is 8 which occurred 27 times. Press any key to continue Consider the following code for manipulating polynomials ([above]). a. Get the program working in your compiler and use it to analyze the roots for i ii. 1 Describe your results. b. Make a flow chart to describe the operation of the i. findupperbound() ii. findzero(). iii. descartesrulepos() iv. findallzeros() SOLN (a i) Running the program and choosing (2) from the menu to enter the polynomial and then (9) to find all zeros, we get the following: Choose what you want to do: 1: Print the current polynomial. 2: Enter a new polynomial. 3: Do synthetic division by x - c. 4: Evaluate the polynomial at a point. 5: Find lower and upper bounds on the zeros. 6: List the possible number(s) of positive and negative roots using D's rule of signs. 7: Find a positive zero above a given number, if there is one. 8: Find a negative zero below a given number, if there is one. 9: Find all real zeros of a polynomial. choice: 2 Enter the degree of the polynomial, a number less than 100 : 4 Enter the coefficient of x^0: 27 Enter the coefficient of x^1: -45 Enter the coefficient of x^2: -120 Enter the coefficient of x^3: 20 Enter the coefficient of x^4: 48 You entered 48x^4 + 20*x^3-120*x^2-45*x + 27 Choose what you want to do: 1: Print the current polynomial. 2: Enter a new polynomial. 3: Do synthetic division by x - c. 4: Evaluate the polynomial at a point. 5: Find lower and upper bounds on the zeros.

20 6: List the possible number(s) of positive and negative roots using D's rule of signs. 7: Find a positive zero above a given number, if there is one. 8: Find a negative zero below a given number, if there is one. 9: Find all real zeros of a polynomial. choice: 9 roots[3] = roots[4] = -1.5 roots[5] = 1e+101 The roots are Hmmm.evidently, some debugging is in order. To make it a bit easier to analyze , hard-wire it as the default polynomial in main(): double coeff[n] = 27,-45,-120,20,48; Then, since puts the 1e+101 capstone at the end of the list of roots, insert --i; //back up a notch to account for findzero's capstone between the two while loops in findallzeros() to be sure that this value is overwritten between the positive and negative zeros. The code now works and we see that Choose what you want to do: 1: Print the current polynomial. 2: Enter a new polynomial. 3: Do synthetic division by x - c. 4: Evaluate the polynomial at a point. 5: Find lower and upper bounds on the zeros. 6: List the possible number(s) of positive and negative roots using D's rule of signs. 7: Find a positive zero above a given number, if there is one. 8: Find a negative zero below a given number, if there is one. 9: Find all real zeros of a polynomial. choice: 9 roots[0] = roots[1] = -1.5 roots[2] = 1e+101 roots[2] = roots[3] = 1.5 roots[4] = 1e+101 The roots are So For part (ii) first use a calculator to compute the floating point approximations for the coefficients: , , , 0.5 and so on, so we can hardwire the polynomial with double coeff[n] = 1,0,-0.5,0, e-2,0, e- 3,0, e-5; Now running the program causes it to hang. Why? We

21 must debug! And, when in doubt...cout! Inseting these lines into the while loop of findzero() we that the initial x values are way too far out! choice: 9 x = y = e+029 x = y = e+030 A little reflection helps us realize that if the leading coefficient (1/40320) is very small, then the value of x will be too large (in absolute value) so we go in and alter the code in findzero() to read if(up) step = 1e-1;//1/(10*c[degree]); // for searching through positives else step = -1e-1;//(10*c[degree]); // for searching through negatives Then we see these values: choice: 9 x_next = -0.1 y = 1 x_next = -0.2 y = x_next = -0.3 y = [.. and so on..] x_next = -1.4 y = x_next = -1.5 y = roots[0] = roots[1] = x_next = y = e-011 x_next = y = x_next = y = x_next = y = [.. and so on.. ] x_next = y = x_next = y = x_next = y = roots[2] = x_next = y = e-010 x_next = y = [.. and so on.. ] x_next = y = e+011 x_next = y = e+011 A couple of anomalies here: (1) For some reason, is assigned to be a root twice, and (2) it doesn t find an upper bound where it should this is confirmed by inserting the code cout << "\nmax = " << max; into the findzero() function and viewing the result: max = (which is actually a min..). Max and min bounds on the roots are computed in the switch 9 section: case 9 : for(int j = 0; j < N; j++) roots[j]=0; // clear roots array min = findlowerbound(coeff, degree); max = findupperbound(coeff, degree); findallzeros(coeff,roots,min,max,degree); cout << "\nthe roots are "; k = 0; while(roots[k]<1e99) //roots[k]!=0 && cout << roots[k] << " "; ++k; cin.ignore(); cin.get(); break;

22 Which are apparently not being correctly computed perhaps because of the zero coefficients in polynomial? Perhaps. It s a debugging opportunity for you! The actual roots are approximately (according to the TI-85) and as shown in the (Maple V) graph shown below: The flow charts for part (b) follow. Note that

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24 Here s findzero(). Could use some work too And descartesrulepos():

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