CS 2073 -- Review for Exam 2

CS 2073 Engineering Programming
Topics and Review Questions for
Exam 2, Fall 2005
Answers

Sample Questions: I will pick simple questions about many of the topics from the web pages and from class. Often it will involve giving you some code and asking you to explain, modify, or complete the code. Here are sample questions.

Arrays: This is the most important topic since the first exam.

Use of arrays as parameters: Write a main function that will:

declare an integer array a of size 40.
pass a to a function squares that will initialize each element a[i] with i*i.
pass this initialized array a to a function printarray that will print the elements of the array.
pass this initialized array a to another function addup that will add all the elements of the array and return the sum as an integer.
within main print the value of the sum as returned from the function addup.

#include <stdio.h>
#define LIM 40
void squares(int a[LIM]);
void printarray(int a[LIM]);
int addup(int a[LIM]);

int main() {
   int a[LIM];
   int sum;
   squares(a);
   printarray(a);
   sum = addup(a);
   printf("Sum of squares through %i = %i\n", LIM-1, sum);
   printf("Formula: LIM*(LIM-1)*(2*LIM-1)/6 = %i\n",
      LIM*(LIM-1)*(2*LIM-1)/6);
}

void squares(int a[LIM]) {
   int i;
   for (i = 0; i < LIM; i++)
         a[i] = i*i;
}

void printarray(int a[LIM]) {
   int i;
   for (i = 0; i < LIM; i++) {
      printf("%4i ", a[i]);
      if (i%10 == 9) printf("\n");
   }
   printf("\n");
}

int addup(int a[LIM]) {
   int i;
   int sum = 0;
   for (i = 0; i < LIM; i++)
      sum = sum + a[i];
   return sum;
}

Here is the output of a run:
   0    1    4    9   16   25   36   49   64   81
 100  121  144  169  196  225  256  289  324  361
 400  441  484  529  576  625  676  729  784  841
 900  961 1024 1089 1156 1225 1296 1369 1444 1521

Sum of squares through 39 = 20540
Formula: LIM*(LIM-1)*(2*LIM-1)/6 = 20540

2-dimensional arrays: Write a program that will use 2-dimensional arrays to print a 16-by-16 multiplication table. Use an array m, that is declared int m[17][17];. Use a separate function to initialize m[i][i]; to i*j. Then use another function to print the array, skipping the entries for i = 0 and for j = 0.

#include <stdio.h>
#define LIM 17
void initarray(int m[LIM][LIM]);
void printarray(int m[LIM][LIM]);
int main() {
   int m[LIM][LIM];
   initarray(m);
   printarray(m);
}

void initarray(int m[LIM][LIM]) {
   int i, j;
   for (i = 0; i < LIM; i++)
      for (j = 0; j < LIM; j++)
         m[i][j] = i*j;
}

void printarray(int m[LIM][LIM]) {
   int i, j;
   printf("   |");
   for (j = 1; j < LIM; j++) printf("%3i ", j);
   printf("\n---+");
   for (j = 1; j < LIM; j++) printf("----", j);
   printf("\n");

   for (i = 1; i < LIM; i++) {
      printf("%2i |", i);
      for (j = 1; j < LIM; j++)
         printf("%3i ", m[i][j]);
      printf("\n");
   }
}

Here is the output of a run:
   |  1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16
---+----------------------------------------------------------------
 1 |  1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16
 2 |  2   4   6   8  10  12  14  16  18  20  22  24  26  28  30  32
 3 |  3   6   9  12  15  18  21  24  27  30  33  36  39  42  45  48
 4 |  4   8  12  16  20  24  28  32  36  40  44  48  52  56  60  64
 5 |  5  10  15  20  25  30  35  40  45  50  55  60  65  70  75  80
 6 |  6  12  18  24  30  36  42  48  54  60  66  72  78  84  90  96
 7 |  7  14  21  28  35  42  49  56  63  70  77  84  91  98 105 112
 8 |  8  16  24  32  40  48  56  64  72  80  88  96 104 112 120 128
 9 |  9  18  27  36  45  54  63  72  81  90  99 108 117 126 135 144
10 | 10  20  30  40  50  60  70  80  90 100 110 120 130 140 150 160
11 | 11  22  33  44  55  66  77  88  99 110 121 132 143 154 165 176
12 | 12  24  36  48  60  72  84  96 108 120 132 144 156 168 180 192
13 | 13  26  39  52  65  78  91 104 117 130 143 156 169 182 195 208
14 | 14  28  42  56  70  84  98 112 126 140 154 168 182 196 210 224
15 | 15  30  45  60  75  90 105 120 135 150 165 180 195 210 225 240
16 | 16  32  48  64  80  96 112 128 144 160 176 192 208 224 240 256

Pointers in C:

Start with an array declared in main by:

int a[] = {12, 34, 21, 47, 52, 7, 41};
int asize = 7;

Pass this array and its size to a function maxmin that will calculate both the maximum and the minimum value in the array. The function should have two formal parameters max and min, each of type int *, that is, a pointer to an integer, or the address of an integer. Use these parameters to get the maximum and minimum values back to the function main, where they should be printed. (You must pass the addresses of two ordinary integers as the actual parameters.)

#include <stdio.h>
void maxmin(int a[], int asize, int *max, int *min);

int main() {
   int a[] = {12, 34, 21, 47, 52, 7, 41};
   int asize = 7;
   int max, min;
   maxmin(a, asize, &max, &min);
   printf("max: %i, min: %i\n", max, min);
}

void maxmin(int a[], int asize, int *max, int *min) {
   int i;
   *max = a[0];
   *min = a[0];
   if (asize == 1) return;
   for (i = 1; i < asize; i++) {
      if (a[i] > *max) *max = a[i];
      if (a[i] < *min) *min = a[i];
   }
}

Here is the output of a run:
max: 52, min: 7

macros in C:

Write a macro STARS that will take i and will print i stars with no newline. (Go ahead and put a for loop and a printf or a putchar inside the macro. In standard C you'll have to declare the loop index separately, but not in C++.)

#include <stdio.h>
#define STARS(i) for(j=0;j<i;j++)printf("*")   /* preprocessor macro */

int main() {
   int j;
   STARS(10);
   printf("\n");
   STARS(21);
   printf("\n");
   STARS(1);
   printf("\n");
   STARS(0);
   printf("\n");
}

Here is the result after preprocessing
(normally you pay no attention to this):

int main() {
   int j;
    for ( j = 0 ; j < 10 ; j ++ ) printf ( "*" );
   printf("\n");
    for ( j = 0 ; j < 21 ; j ++ ) printf ( "*" );
   printf("\n");
    for ( j = 0 ; j < 1 ; j ++ ) printf ( "*" );
   printf("\n");
    for ( j = 0 ; j < 0 ; j ++ ) printf ( "*" );
   printf("\n");
}

Here is the output of a run:
**********
*********************
*

CS 2073 Engineering Programming Topics and Review Questions forExam 2, Fall 2005 Answers

CS 2073 Engineering Programming
Topics and Review Questions for
Exam 2, Fall 2005
Answers