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### c17s99f

Course: CSE 17, Spring 2008
School: Lehigh
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Word Count: 951

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17 CSc Final Examination Wednesday 12 May 1999 4 PM Page 1 !!!!!!!!!!!!!!!SUGGESTED ANSWERS!!!!!!!!!!!!!!!!! 1. (10 pts) Assume class Staque implements an ADT which has the properties of both a stack and a queue. Further assume that Staque has the member functions push(int), enqueue(int), int pop(), int dequeue(), bool full(), and empty(), where the functions push and enqueue have exactly the same effect. For the...

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17 CSc Final Examination Wednesday 12 May 1999 4 PM Page 1 !!!!!!!!!!!!!!!SUGGESTED ANSWERS!!!!!!!!!!!!!!!!! 1. (10 pts) Assume class Staque implements an ADT which has the properties of both a stack and a queue. Further assume that Staque has the member functions push(int), enqueue(int), int pop(), int dequeue(), bool full(), and empty(), where the functions push and enqueue have exactly the same effect. For the code below, list the contents of aStaque after each push, pop, enqueue, or dequeue is called. List the entries from left to right, with the latest entry on the right. Staque aStaque(100); //create a Staque with capacity for 100 entries for(int j=5;j<7;j++){ aStaque.push(j); 5 aStaque.push(j+10); 5 15 aStaque.enqueue(aStaque.dequeue()); 15 5 } 15 5 6 while(!aStaque.empty()){ 15 5 6 16 aStaque.pop(); 5 6 16 15 aStaque.dequeue(); 5 6 16 } 6 16 6 <empty> 2. (25 pts) Assume we have the following data structure for building a binary tree: class TNode{ public: int data; TNode *child[2]; }; class Tree{ private: TNode *root; //possibly other stuff... public: //other stuff.... }; Write a member function for Tree called recPrintRange such that the call recPrintRange(root,low,high) displays on the screen all entries in the tree that are greater than the int low and less than the int high. void Tree::recPrintRange(TNode *rt,int low, int high){ if(rt!=NULL){ if(rt->data>low && rt->data<high) cout<<rt->data<<endl; recPrintRange(rt->child[0],low,high); recPrintRange(rt->child[1],low,high); } } 3. (20 pts) For class Tree in question 2, I wrote the following member function which assumes a Tree constructed in ascending order, but I forgot to provide documentation. Provide it. //Purpose: To return the largest entry in the tree. If the tree is // emtpy, return 0. //Pre-condition(s): The TNodes actually form a binary tree; each value // of data is valid; each root and each child are defined; There is only // one way to go from the root to any child. //Post-condition(s): The tree is unchanged. The largest value (or 0 if // the tree is empty) has been returned. int Tree::Puzzling(){ TNode *temp; if(root!=NULL){ temp=root; while(temp->child[1]!=NULL) temp=temp->child[1]; return temp->data; } else return 0; } 4. (20 pts) For the class TNode in question 2 overload the << operator so that for the declaration TNode t, the statement "cout<<t" displays t.data on the screen. Write both the prototype and the definition. friend ostream & operator <<(ostream &out,const TNode &t); ostream & operator <<(ostream &out,const TNode &t){ out<<t.data; return out; } 5. (20 pts) For the class TNode in question 2 write the function "output" so that for the declaration TNode t, the statement "t.output(cout)" displays t.data on the screen. Write both the prototype and the definition. void output(ostream &out); TNode::output(ostream&out){ void out<<data; } 6. (30 pts) Suppose we have the declaration: int x[100], first, last. Suppose the call sum(x,first,last) returns the sum of x[first], x[first+1], ..., x[last]. Write a version of sum that uses the following recursvie algorithm: To sum the numbers between first and last, inclusive, take the entry in the middle and add it to the sum of the entries to the left of the entry in the middle and the sum of the entries to the right of the entry in the middle. int sum(int x[],int first, int last){ if(first>last) return 0; if(first==last) return x[first]; int middle=(first+last)/2; return(sum(x,first,middle-1)+x[middle]+sum(x,middle+1,last)); } 7. (10 pts) Write one, and only one, definition of the function swap so that the following code compiles (even though the code may not make much sense). Swap is supposed to interchange the values of its two arguments. int a,b; TNode c,d; //TNode is defined in question 2 swap(a,b); swap(c,d); template <class T> void swap(T &u,T&v){ T temp; temp=u; u=v; v=temp; } 8. (5 pts) What is the output of the following code? char line[]="One fine day", *p; p=line; cout<<line[4]<<p[5]<<endl; fi cout<<p<<endl; One fine day cout<<(p+1)[4]<<endl; i cout<<(p+1)<<endl; ne fine day line[3]='\0'; cout<<line<<endl; One 9. (5 pts) Write a prototype for the function junk so that the following code will compile: int x[5], y[5][2], z[2][4][2], w[5][4][2]; junk(x,z,z[3]); junk(y[2],w,y); junk(w[3][3],w,y); void junk(int x[],int[][4][2],int [][2]); 10. (30 pts) Write a function "charCount" which counts the number of times a given character appears as the first character on a line in a file. In particular, suppose we have the declaration: istream in; The call charCount(in,'a') should return the number of times the character 'a' appears as the first character of a line in the file "in". If the file contained the lines of this paragraph, right justified, then the call charCount(in,'c') would return 4 (c appears on lines 6 thorugh 9), and the call charCount(in,'T') would return 1 (T appears on line 4). Your function can assume the file is opened and at the beginning. int charCount(istream &in,char target){ char ch; int count; count=0; in.get(ch); while(!in.eof()){ if(ch==target) count++; while(ch!='\n' && !in.eof()) in.get(ch); if(!in.eof()) in.get(ch); } return count; } 11. (25 pts) Write a function "sort" which sorts an array of doubles into ascending order. In particular, if x is an array of doubles, and n is an int which counts the number of entries in x, after the call sort(x,n) the first n entries in x should be in ascending order. void sort(double x[],int last){ bool sorted; do{ sorted=true; for(int j=1;j<last;j++) if(x[j-1]>x[j]){ swap(x[j-1],x[j]); //use the swap of question 7 sorted=false; } }while(!sorted); }
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