MISCELLANEOUS EXERCISES

 1. The file badnews.c contains 

          #include badnews.c

    What happens when it is compiled?

 2. What does the following program do?

          #include <stdio.h>
          int f()
            { int i = 3;
              while (i > 1)
                { int i = 3;
                  printf("%d ", i--);
                  if (i > 1) break;
                }
              return i--;
            }
          main()
            { int i = 3;
              printf("%d %d %d\n", f(), f(), f());
            }

 3. Replace "int" by "static int" in the first line of f in the program
    above.  Now what does it do? (See page 83 of K and R.)

 4. Suppose that a disk has 2^24 blocks, 2^12 bytes per block.  All 
    files are defined by a block with 2^6 bytes of descriptive info, 
    followed by pointers to data blocks. How big a file can be accommodated?
     
 5. The same disk is configured with 2^16 blocks, 2^20 bytes per block.  
    How big a file can be accommodated?

 6. Give an input string of length 10 for which the tree of Programming
    Assignment 7 produces a tree with all null left links.


























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		  ANSWERS TO MISCELLANEOUS EXERCISES


 1. Puts the preprocessor in a loop.  Different compilers react differently.
        % lcc badnews.c
        badnews.c:1: badnews.c: Too many open files
        % cc badnews.c
        cpp: error ./badnews.c:2: Unreasonable include nesting
        badnews.c:1: warning: empty input file
        % gcc badnews.c
        In file included from badnews.c:1, from badnews.c:1, from badnews.c:1,
        from badnews.c:1, from badnews.c:1, from badnews.c:1, from badnews.c:1,
         ...
        badnews.c:1: macro or `#include' recursion too deep

 2. 3 3 3 3 3 3

 3. 3 3 3 2 1

 4. Pointers must be 3 bytes to accommodate 24-bit addresses. Thus, 1344
    pointers ((2^12 - 2^6) / 3) can fit in the defining block, so the file size
    is 1344 * 2^12 = 5505024.

 5. (2^20 - 2^6 / 2) * 2^20 = 1099444518912.  (Use bc). 

 6. aaaaaaaaaa