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Note that like insert, this operation also leaves the current and previous pointers unchanged. The remove ( ) function deletes the current node: template<class T> void ListItercT>::remove() -l if (current == list.first) list.first = current->next; else previous->next = current->next; delete current; current = 0; > ' It leaves the previous pointer unchanged and the current pointer NULL.

After:

int main() { int a[] = {22,55,66,88,11,33,44,77,99}; print(a,9); inplace_merge(a,a+4,a+9); print(a,9); } n=9: {22,55,66,88,11,33,44,77,99} n=9: {11,22,33,44,55,66,77,88,99}

Summary

The remove operation can be visualized like this:

iter_swap(p,q); // swaps the elements *p and *q;

Afier:

Here is a test driver for the list iterator:

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int main() { int a[] = {11,22,33,44,55,66,77,88,99}; int b[] = {10,20,30,40,50,60,70,80,90}; print(a,9); print(b,9); iter_swap(a+4,b+7); print(a,9); print(b,9); } n=9: {11,22,33,44,55,66,77,88,99} n=9: {10,20,30,40,50,60,70,80,90} n=9: {11,22,33,44,80,66,77,88,99} n=9: {10,20,30,40,50,60,70,55,90}

#include #include #include main0 <i.ostream.h> "List1ter.h" "String.h"

lexicographical_compare(p,q,pp,qq); // compares the two segments [pp,qq[ and [p,q[ lexicographically; // returns true iff the first precedes the second; // invariant: [p,q[ and [pp,qq[ are left unchanged;

List<String> friends; ListIter<String> it(friends); it.insert("Bowen, Van"); it++;

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it.insert("Dixon, Tom"); it++; it.insert("Mason, Joe"); it++; it.insert('White, Ann"); it++; friends.print(); it.reset(); it++; it = "Davis, Jim"; it++; it.remove(); friends.print(); if (!it) it.preInsert("Morse, friends.print(); for (it.reset(); !it; it++) i t = 1 [ 1' + it() + "1"; friends.print();

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void test(char*,int,char*,int); int main() { char* s1="COMPUTER"; char* s2="COMPUTABLE"; char* s3="COMPUTE"; test(s1,3,s2,3); test(s1,8,s2,10); test(s1,8,s3,7); test(s2,10,s3,7); test(s1,7,s3,7); } char* sub(char*,int); void test(char* s1, int n1, char* s2, int n2) { bool lt=lexicographical_compare(s1,s1+n1,s2,s2+n2); bool gt=lexicographical_compare(s2,s2+n2,s1,s1+n1); if (lt) cout << sub(s1,n1) << " < " << sub(s2,n2) << "\n"; else if (gt) cout << sub(s1,n1) << " > " << sub(s2,n2) << "\n"; else cout << sub(s1,n1) << " == " << sub(s2,n2) << "\n"; } char* sub(char* s, int n) { char* buffer = new char(n+1); strncpy(buffer,s,n); buffer[n] = 0; return buffer; } COM == COM COMPUTER > COMPUTABLE COMPUTER > COMPUTE COMPUTABLE < COMPUTE COMPUTE == COMPUTE

sets current to first item sets current to second item replace with new name sets current to third item removes third item

lower_bound(p,q,x); // returns the position of the first occurrence of x in [p,q[; // precondition: the segment must be sorted; // invariant: [p,q[ is left unchanged;

The for loop changes each data value in the list by prepending a left bracket and appending a right bracket to each string. Note that the assignment it = I [ I + it ( ) + I ] I calls the operator ( > ( > class as well as the constructor and operator=0 functions of the ListIter&tring> String(const char*) and operator+= () functiondefinedinthe String class. To give List~ter objects the access to the protected members of List objects that they need to do their job, we need to declare the Li s t I t er class a friend of the L i s t class:

template<class T> class List { friend class ListItereT>; public: // other members protected: ListNodecT>* first; // other members

int main() { int a[] = {11,22,22,33,44,44,44,55,66}; int* p = lower_bound(a,a+9,44); cout << "*p=" << *p << '\n'; cout << "p-a=" << p-a << '\n'; } *p=44 p-a=4

iterators also need the access to the protected members of Lis tNode objects:

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