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#include <stdio.h> int main (int argc, const char * argv[]) { int i, num; long fac; num = 5; fac = 1; for ( i=1; i<=num; i++ ) fac *= i; printf( "%d factorial is %ld.", num, fac ); return 0; }

Here s the basic business of selecting a seller (it s not that difficult): sellers want to do business with you as long as the negotiated terms are good for them You want to do business with them as long as it s good for your program The evaluation criteria you determined as part of the seller responses are referenced here again to confirm the seller you re considering really does qualify

2 public class Salmon extends Thread { 3 public static long id; 4 public void run() { 5 for(int i = 0; i < 4; i++) { 6 // insert code here 7 new Thread(new Salmon())start(); 8 throw new Error(); 9 } 10 Systemoutprint(i + " "); 11 } } 12 public static void main(String[] args) { 13 Thread t1 = new Salmon(); 14 id = t1getId(); 15 t1start(); 16 } }

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If you are interested in trying this code, you ll find it in the Learn C Projects folder, under the subfolder named 11.01 - iterate.

And the two code fragments: I if(i == 2 && id == ThreadcurrentThread()getId()) { II if(i == 2) {

If you ran this program, you d see this line printed in the console window:

You ll also rely on your qualified sellers list to confirm that the sellers do qualify and that they can work for your organization If not, you give em the boot You ll also review the proposals from the sellers to determine if what they ve proposed is on target or off the mark Proposal evaluation is a tricky business You and the seller may bounce ideas back and forth to adjust the proposal to be more in alignment with what the program deliverables should be Proposal writing is sometimes difficult for the vendor, because it may mean rounds of analysis, writing, and work and then the vendor may not even get the gig that s being proposed That s a sad day for the vendor

When inserting either fragment, independently at line 6, which are true (Choose all that apply) A Both fragments produce the same output B Both fragments will end in about the same amount of time C Compilation fails, regardless of which fragment is inserted D Regardless of which fragment is inserted, output ends once the Error is thrown E Regardless of which fragment is inserted, output continues after the Error is thrown

As you can see from the source code, the algorithm steps through (iterates) the numbers 1 through 5, building the factorial with each successive multiplication.

You can use a recursive approach to solve the same problem. For starters, you ll need a function to act as a base for the recursion, a function that will call itself. You ll need two things to build into your recursive function. First, you ll need a mechanism to keep track of the depth of the recursion. In other words, you ll need a variable or parameter that changes depending on the number of times the recursive function calls itself.

Answer (for Objective 42): E is correct In either case, before the Error is thrown a new thread is created, and the new thread will execute independently of the Error A and B are incorrect because fragment II creates a sort of recursive, endless loop C and D are incorrect based on the above

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