Homework 37

Part 1 of 2: Given the following information complete the table below:

1)  If the value of a variable is an address, simply write ADDR for the value (but still give the type)
2)  If the value of a variable is a struct, simply write STRUCT for the value (but still give the type)

Hint: Do this by hand, not by running in Visual Studio. You don't need to run/simulate the whole program above to be able to fill out the matrix, just look at the variable and figure out what its value would be.  Once a variable is initialized its value is not modified.

#include <iostream>

using namespace std;

struct myStruct {

double data;

myStruct *next;

};

int main () {

double y = 3.7;

int a = 5;

int b = 11;

double z;

myStruct *list;

myStruct firstNode;

firstNode.data = 5.3;

firstNode.next = 0;

list = &firstNode;

//complete the table below as if all of the

//instructions occurred at this point

//Assume each line of the table is the

//only line of code (ie assignments made

//based on questions in the table do not

//effect any other row of the table)

return 0;

}

Expression	Type	Value
&firstNode	myStruct*	ADDR
list->next	myStruct*	ADDR
list->data	Double	5.3
z = b / a	Double	2
z = b / double (a)	Double	2.2
(a < 10 && b < 10)	Bool	False
(a < 10 | | b < 10)	Bool	True
a = 6	Int	6
b == 6	Bool	False

Modify the following code so the function printList prints the data fields of the linked list from back to front.

class Node
{
public:
 int data;
 Node* next;
};

void printList(Node* p) {
    if (p!=0){
cout << p ->data << endl;
printList(p->next);
    }
}

Solution

void printList(Node* p)

{

if (p!=0)

{

printList(p->next); // just swap order

cout << p ->data << endl;

}

} // base case is when p==0, do nothing!