仿函数经验总结

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transform

transform的行为

template <class InputIterator, class OutputIterator, class UnaryOperator>  OutputIterator transform (InputIterator first1, InputIterator last1,                            OutputIterator result, UnaryOperator op){  while (first1 != last1) {    *result = op(*first1);  // or: result=binary_op(*first1,*first2++);    ++result; ++first1;  }  return result;}

从第6行可以看到传入的仿函数必须有返回值
比如预先定义的仿函数negate：求相反数的

template <class T> struct negate {  T operator() (const T& x) const {return -x;}  typedef T argument_type;  typedef T result_type;};

下面是一个实例：把list里面的元素平方后输出

#include <iostream>#include <list>#include <algorithm>#include <iterator>using namespace std;template <class T1, class T2>inline void PRINTCOLL(const T1 &coll, const char * optcstr=" "){    std::cout << optcstr;    copy(coll.begin(), coll.end(),        ostream_iterator<T2>(cout, " "));    std::cout <<"\n";}template <class T>class MySquare{    public:        MySquare(){};        T operator()(T & elem)const{            return elem * elem;        };};template <class T>class MyMutiply{    public:        MyMutiply(T val){};        MyMutiply(){};        T operator()(T & elem1, T& elem2)const{            return elem1 * elem2;        };};int main(){    list<int> coll;    list<int> coll2;    for(int i = 0; i < 10 ; ++i){        coll.push_back(i);    }    transform(coll.begin(), coll.end(),\        coll.begin(), MySquare<int>());    PRINTCOLL<list<int>, int>(coll);    transform(coll.begin(), coll.end(),\        coll.begin(), coll.begin(), MyMutiply<int>());    PRINTCOLL<list<int>, int>(coll);//  transform(coll.begin(), coll.end(),\        back_inserter(coll2), \        bind2nd(MyMutiply<int>(), 10));    transform(coll.begin(), coll.end(),\        back_inserter(coll2), \        bind2nd(multiplies<int>(), 10));    PRINTCOLL<list<int>, int>(coll2);}

第50行报错，暂时没找到解决办法，等看了后面的章节再回过头来调试

for_each

for_each的行为

template<class InputIterator, class Function>  Function for_each(InputIterator first, InputIterator last, Function fn){  while (first!=last) {    fn (*first);    ++first;  }  return fn;      // or, since C++11: return move(fn);}

这里不要求传入的函数必须有返回值

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