Fall 2001 CSCI 333 Homework 8 Solution

These are solutions to the problems of Homework 8 that were not assigned from the textbook.

Problem 1

The problem is to use the qsort routine to sort 3000 records of the following type:


  struct city {
    char Name[80] ;
    char State[2] ;
  }

First, you need a routine to compare the records. The problem requires us to sort first by city and then by state. The arguments to this routine are of type void * as required in the specification of qsort. Consequently, we must cast the arguments to type struct city * before accessing the fields of the record.


int cityCmp(const void *R1, const void *R2) {
  struct city *P1, *P2 ;
  int tempCmpRes ;
  P1 = (struct city *)R1 ;
  P2 = (struct city *)R2 ;
  /* Compare the states.  If unequal, return the result of the compare */
  if (tempCmpRes = strncmp(P1->State, P2->State, 2))
    return tempCmpRes ;
  else
    return strncmp(P1->Name, P2->Name, 80) ;
}

Assume Cities is our array of 3000 struct city records. Now we only need to call qsort. Of course, qsort expects the array to be passed as a void *.


  qsort((void *)Cities, 3000, sizeof(struct city), cityCmp) ;

Problem 3

The ASCII values for the letters 'U', 'N', 'C', and 'A' are, in hex, 0x55, 0x4e, 0x43, and 0x41, respectively.

C statement	`*key`	`h`	`g`
initally	`0x55` or `'U'`	`0x00000000`
`h = (h << 4) + *key++ ;`	`0x4e` or `'N'`	`0x00000055`
`unsigned long g = h & 0xF0000000L ;`	`0x4e` or `'N'`	`0x00000055`	`0x00000000`
`if (g) h ^= g >> 24 ;`	`0x4e` or `'N'`	`0x00000055`	`0x00000000`
`h &= ~g ;`	`0x4e` or `'N'`	`0x00000055`	`0x00000000`
`h = (h << 4) + *key++ ;`	`0x43` or `'C'`	`0x0000059e`
`unsigned long g = h & 0xF0000000L ;`	`0x43` or `'C'`	`0x0000059e`	`0x00000000`
`if (g) h ^= g >> 24 ;`	`0x43` or `'C'`	`0x0000059e`	`0x00000000`
`h &= ~g ;`	`0x43` or `'C'`	`0x0000059e`	`0x00000000`
`h = (h << 4) + *key++ ;`	`0x41` or `'A'`	`0x00005a23`
`unsigned long g = h & 0xF0000000L ;`	`0x41` or `'A'`	`0x00005a23`	`0x00000000`
`if (g) h ^= g >> 24 ;`	`0x41` or `'A'`	`0x00005a23`	`0x00000000`
`h &= ~g ;`	`0x41` or `'A'`	`0x00005a23`	`0x00000000`
`h = (h << 4) + *key++ ;`	`0x00`	`0x0005a271`
`unsigned long g = h & 0xF0000000L ;`	`0x00`	`0x0005a271`	`0x00000000`
`if (g) h ^= g >> 24 ;`	`0x00`	`0x0005a271`	`0x00000000`
`h &= ~g ;`	`0x00`	`0x0005a271`	`0x00000000`

So, the answer is 0x0005a271. The problem would have been more interesting had the string had more than eight characters.