Inspection and Procedural Abstraction

Course theme: problem decomposition

• Achieve control of a complex system by
  1. Dividing it into relatively independent parts.
  2. Documenting the interdependencies that remain.

• Steps 1 and 2 are both important.

• The result of a well-designed and carefully documented decomposition is a system that is easier to understand, inspect, test, and modify.

Procedural abstraction

• Abstraction: intentionally ignoring certain aspects of a problem to simplify analysis and to focus attention on the remaining aspects.

• In procedural abstraction the procedure implementation is ignored to focus attention on the service provided by the procedure.

• Successful procedural abstraction requires a precise and complete specification of the service offered by the procedure.

Procedural abstraction in inspection

• Suppose that you are given specifications and implementations for the C functions F and G, and that F calls G.

• To inspect G: show that G's implementation behaves as required by its specification.

• To inspect F: show that F's implementation behaves as required by its specification. When reasoning about the call to G in F's implementation, refer to G's specification and not its implementation.

Example

• Use procedural abstraction to inspect the findLongestPlateau and removeLongestPlateau functions shown below.

• First show that findLongestPlateau is correct or produce a list of the faults found.

• Then show that removeLongestPlateau is correct or produce a list of the faults found. When reasoning about the call to findLongestPlateau, refer to its specification not its implementation.

• Note: A plateau in a sequence of numbers is a subsequence of one or more consecutive numbers of the same value. In the sequence

  	S = <1,2,2,3,2,0,0,0>
  

  there are many plateaus including

  	<2,2>
  	<3>
  	<0>
  	<0,0>
  	<0,0,0>
  

  In S, the longest plateau is <0,0,0>. If a sequence has two or more "longest plateaus" then longest plateau refers to the leftmost one.

/* Assign to *pStart and *pLen the starting position and length
 * of the longest plateau in a[0..aLen-1].
 * 
 * Assumed: a has at least aLen elements and aLen > 0
 */
void findLongestPlateau(int a[],int aLen,int* pStart,int *pLen)
{
	int tmpStart,tmpLen,i;

	*pStart = 0;
	*pLen = 1;
	tmpStart = 0;
	tmpLen = 1;
	for (i = 0; i < aLen-1; i++) {
		if (a[i] == a[i+1]) {
			tmpLen++;
		} else {
			if (tmpLen >= *pLen) {
				*pStart = tmpStart;
				*pLen = tmpLen;
			}
			tmpStart = i+1;
			tmpLen = 1;
		}
	}
}

/* Remove P, the longest plateau in a, shifting left all the elements
 * to the right of P and decreasing *aLen appropriately.
 * 
 * Assumed: a has at least *aLen elements and *aLen > 0
 */
void removeLongestPlateau(int a[],int *aLen)
{
	int i,pStart,pLen;

	findLongestPlateau(a,*aLen,&pStart,&pLen);
	for (i = 0; i < *aLen-pStart-pLen; i++) {
		a[i+pStart] = a[i+pStart+pLen];
	}
}

int main()
{
	int i,xLen,x[100];

	/* longest plateau at end */
	xLen = 4; x[0] = 1; x[1] = 2; x[2] = 3; x[3] = 3;
	removeLongestPlateau(x,&xLen);
	for (i = 0; i < xLen; i++)
		printf("%d\n",x[i]);
	printf("\n");

	/* two longest plateaus */
	xLen = 6; x[0] = 1; x[1] = 2; x[2] = 2; x[3] = 3; x[4] = 3; x[5] = 1;
	removeLongestPlateau(x,&xLen);
	for (i = 0; i < xLen; i++)
		printf("%d\n",x[i]);
}