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Practice 7: Adopt life cycle configuration management The manager includes in the process the use of a configuration management tool for all documents and includes process steps to enter all documents and changes into the configuration management tool Practice 8: Manage and trace requirements The manager includes process steps to acquire the requirements from the user and steps to trace each requirement to the current phase of development Practice 9: Use system-based software design The manager includes steps to ensure a system-based design Practice 10: Ensure data and database interoperability The manager includes steps to check for interoperability between the data and the database Practice 11: Define and control interfaces The manager includes steps to define and baseline the interfaces Practice 12: Design twice, code once The manager includes design review steps Practice 13: Assess reuse risks and costs.

P(5,2) =

5! A-=-= 120=2()

(5-2)! 3! 6

So there are 20 different permutations of 2 items taken from a set of 5. The code below implements this formula for the permutation function:

The manager includes steps to identify areas of potential reuse and steps to assess costs and risks Practice 14: Inspect requirements and design The manager includes inspection steps in both the requirements and design phases Practice 15: Manage testing as a continuous process The manager includes testing steps in all phases Practice 16: Compile and smoke-test frequently The manager includes frequent testing steps in the implementation phase..

// Returns P(n;k), the number of permutations of k from n: int perm(int n, int k) -t if (n < 0 II k < 0 II k > n) return 0; return factorial(n)/factorial(n-k); >

Notice that the condition (n < 0 11k < 0 I I k > n) is used to handle the cases where either parameter is out of range. In these cases the function returns an impossible value, 0, to indicate that its input was erroneous. That value would then be recognized by the calling program as an error flag. Here is a test driver for the perm ( ) function:

int perm(int,int);

main0 1 for (int i = -1; i < 8; i++) { for (int j = -1; j <= i+l; j++) tout << " il << perm(i,j); tout << endl; > >

Label the decisions in the control flow graph of Example 5.6 with lowercase letters. As shown in Fig. 5-3, from node a, there are three arcs, so there must be two decisions, labeled a and b. From nodes c and f, there are two arcs and so one decision each. The other nodes have at most one exit and so no decisions. There are four decisions, so C 4 1 5.

4.7 void FUNCTIONS A function need not return a value. In other programming languages, such a function is called a procedure or a subroutine. In C++, such a function is identified simply by placing the keyword void where the function s return type would be. A type specifies a set of values. For example, the type short specifies the set of integers from -32,768 to 32,767. The void type specifies the empty set. Consequently, no variable can be declared with void type. A void function is simply one that returns no value.

This function prints the date in literal form, given its month, day, and year in numeric form:

void main0 int month, day, year; tin >> month >> day >> year; printDate(month,day,year); } while (month > 0); printDate(int, int, int);

m, int d, int y)

Fig. 5-3.

if (m < 1 II m > 12 II d < 1 II d > 31 II y < 0) { tout C-C 'Error: parameter out of range.\n"; return; switch (m) { case 1: case 2: case 3: case 4: case 5: . case 6: case 7: case 8: case 9: case 10: case 11: case 12: tout << d <<

"January "; "February '; "March "; "April "; " May " ; "June "; "July "; "August "; "September "; "October '; "November "; "December ';

break; break; break; break; break; break; break; break; break; break; break; break;

C-C y << endl;