EXERCISES ON ARRAYS

READING 
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Notes for lecture 2.
Sedgewick, p. 83-84
Kernighan and Ritchie, 1.6
King, Chapter 8 (Arrays)

SUPPLEMENTAL READING
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Deitel and Deitel, 6.1-6.3


EXERCISES
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 1. What happens if you declare an array with 
        #define N 1000
        int a[N];

 2. What happens if you declare an array with 
        #define N 1000
        int a[N*N];

 3. What happens if you declare an array with 
        #define N 1000
        int a[N*N*N];

 4. Explain what the following program does:

          #include <stdio.h>
          main()
            {
              int num, i, maxi;
              int a[100];
              for (i = 0; i < 99; i++)
                a[i] = 0;
              while (scanf("%2d", &num) != EOF) 
                  a[num]++;
              maxi = 0;
              for (i = 1; i < 99; i++)
                if (a[i] > a[maxi]) maxi = i;
              printf("%d\n", maxi);
            }

 5. Write a program that reads a sequence of integers between 0 and 99
    from standard input and prints the median value.

          % echo 11 9 99 34 12 18 1 | a.out
          12
          % echo 1 1 1 1 1 2 3 3 3 | a.out
          1
          % 

 6. What is in a[8] after the following code is executed?

          for (i = 0; i < 10; i++) a[i] = 9-i;
          for (i = 0; i < 10; i++) a[i] = a[a[i]];



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		    ANSWERS TO EXERCISES ON ARRAYS

 1. Equivalent to a[1000]. This is a convenient way to name the size of
    the array at *compile* time, so we can use the name in for loops and
    elsewhere in the program.  If we want to change the size, we only have
    to change it in the define line. The size is constant at *run* time:
    N doesn't change when the program runs.

 2. Equivalent to a[1000000]. Simple expressions involving defined variables
    can be evaluated at *compile* time. In ancient times (10 years ago), not 
    many computers could hold such a big array; nowadays most can.

 3. Equivalent to a[1000000000]. Might work; but your computer probably
    can't hold such a big array.  Exact behavior depends on the compiler.

          % more bigarray.c
          #include <stdio.h>
          #define N 1000
          main()
            { int i;
              int a[N*N*N];
              for (i = 0; i < N*N*N; i++)
                a[i] = 0;
            }
          % cc bigarray.c
          % a.out
          Segmentation Fault (core dumped)
          % lcc bigarray.c
          bigarray.c:5: size of `array of int' exceeds 2147483647 bytes
          bigarray.c:8: warning: missing return value
          % 

 4. Computes the value that occurs most often in an integer sequence
    of values between 0 and 99 (this is called the mode, if there is only
    one such value). The "%2d" in the scanf format specification protects 
    against bad input (only reads two digits), but not totally: what about 
    negative numbers, or character strings?   

 5. If N is odd, this program prints the exact median.  If N is even,
    the number of inputs that are <= the printed value is 1 greater than 
    the number of inputs that are >= the printed value.
 
          #include <stdio.h>
          main()
            {
              int num, N = 0, i, cnt = 0;
              int a[100];
              for (i = 0; i < 99; i++)
                a[i] = 0;
              while (scanf("%2d", &num) != EOF) 
                { N++; a[num]++; }
              for (i = 0; i < 100; i++, cnt += a[i])
                if (cnt > N/2) break;
              printf("%d\n", i);
            }

 6. 1
                          i:  0  1  2  3  4  5  6  7  8  9
        a[i] after 1st loop:  9  8  7  6  5  4  3  2  1  0
        a[i] after 2nd loop:  0  1  2  3  4  4  3  2  1  0