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remove_copy_if() copies all the nonmatching elements to the container specified by result The returned OutputIterator marks one past the last element copied The original container is unchanged

#include <algorithm> #include <vector> #include <iostreamh> /* generates: original element sequence: 0 1 1 2 3 5 8 13 21 34 sequence after applying remove_if < 10: 13 21 34 sequence after applying remove_copy_if even: 1 1 3 5 13 21

file:///F|/WinDDK/resources/CPPPrimer/c++primerhtm (1046 / 1065) [2001-3-29 11:32:16]

*/ class EvenValue { public: bool operator()( int value ) { return value % 2 false : true; } }; int main() { int ia[] = { 0, 1, 1, 2, 3, 5, 8, 13, 21, 34 }; vector< int, allocator >::iterator iter; vector< int, allocator > vec( ia, ia+10 ); ostream_iterator< int > ofile( cout, " " ); cout "original element sequence:\n"; copy( vecbegin(), vecend(), ofile ); cout '\n'; iter = remove_if( vecbegin(), vecend(), bind2nd(less<int>(),10) ); vecerase( iter, vecend() ); cout "sequence after applying remove_if < 10:\n"; copy( vecbegin(), vecend(), ofile ); cout '\n'; vector< int, allocator > vec_res( 10 ); iter = remove_copy_if( ia, ia+10, vec_resbegin(), EvenValue() ); cout "sequence after applying remove_copy_if even:\n"; copy( vec_resbegin(), iter, ofile ); cout '\n'; } replace() template< class ForwardIterator, class Type > void replace( ForwardIterator first, ForwardIterator last, const Type& old_value, const Type& new_value );

In addition, the pseudo-geometric routing problem on the ring can be solved (locally) with two anchors; furthermore, the chosen route between source and destination is a shortest path In a grid on n nodes, with one anchor alone at least n nodes will have the same coordinate in the grid On the other hand, as depicted in Figure 25, if we choose one anchor A in the upper left corner and another anchor A in the upper right corner of the grid, all nodes have different coordinates Therefore, if chosen properly, two anchors are suf cient in order to solve the naming problem in the grid However, if the anchors are chosen arbitrarily, then at least ( n 1)2 + 2 anchors are needed to solve the naming problem Concerning routing, if anchors A, A are placed as depicted in Figure 2.

a_by_2(n 1..0) b(n 1..0)

replace() replaces all instances of old_value with new_value within the range specified by the iterator pair [first, last)

replace_copy() template< class InputIterator, class OutputIterator, class Type > OutputIterator replace_copy( InputIterator first, InputIterator last, OutputIterator result, const Type& old_value, const Type& new_value );

replace_copy() exhibits the same behavior as replace() except that the new sequence is copied into the container beginning at result The returned OutputIterator points one past the last value copied The original sequence is unchanged

#include <algorithm> #include <vector> #include <iostreamh> /* generates:

5, a node can compute the position of each anchor, based on its coordinate and the coordinate of its neighbors It is easy to see that the pseudo-geometric routing problem on the grid can be solved with two anchors Furthermore, the chosen route between source and destination is a shortest path As shown in Wattenhofer et al [12], in the case of trees it is not possible to solve the naming problem locally If the anchors are chosen arbitrarily, up to n 1 nodes are required to solve the naming problem However, in a tree it is always suf cient to choose all leaves as anchors For unit disk graphs, at least (( n + 9)/12)2 anchors are.

file:///F|/WinDDK/resources/CPPPrimer/c++primerhtm (1047 / 1065) [2001-3-29 11:32:16]

original element sequence: Christopher Robin Mr Winnie the Pooh Piglet Tigger Eeyore sequence after applying replace(): Christopher Robin Pooh Piglet Tigger Eeyore sequence after applying replace_copy(): Christopher Robin Mr Winnie the Pooh Piglet Tigger Eeyore */ int main() { string oldval( "Mr Winnie the Pooh" ); string newval( "Pooh" ); ostream_iterator< string > ofile( cout, " " ); string sa[] = { "Christopher Robin", "Mr Winnie the Pooh", "Piglet", "Tigger", "Eeyore" }; vector< string, allocator > vec( sa, sa+5 ); cout "original element sequence:\n"; copy( vecbegin(), vecend(), ofile ); cout '\n'; replace( vecbegin(), vecend(), oldval, newval ); cout "sequence after applying replace():\n"; copy( vecbegin(), vecend(), ofile ); cout '\n'; vector< string, allocator > vec2; replace_copy( vecbegin(), vecend(), inserter( vec2, vec2begin() ), newval, oldval ); cout "sequence after applying replace_copy():\n"; copy( vec2begin(), vec2end(), ofile ); cout '\n'; } replace_if() template< class ForwardIterator, class Predicate, class Type > void replace_if( ForwardIterator first, ForwardIterator last, Predicate pred, const Type& new_value );

replace_if() replaces all elements within the range specified by [first, last) with new_value, for which pred evaluates as true

replace_copy_if() template< class ForwardIterator, class OutputIterator, class Predicate, class Type > OutputIterator replace_copy_if( ForwardIterator first, ForwardIterator last, OutputIterator result, Predicate pred, const Type& new_value );