实现C ++后缀增量运算符
我编译了以下示例：
#include <iostream>
#include <iterator>
using namespace std;
class myiterator : public iterator<input_iterator_tag, int>
{
int* p;
public:
myiterator(int* x) :p(x) {}
myiterator(const myiterator& mit) : p(mit.p) {}
myiterator& operator++() {++p;return *this;}
myiterator& operator++(int) {myiterator tmp(*this); operator++(); return tmp;}
bool operator==(const myiterator& rhs) {return p==rhs.p;}
bool operator!=(const myiterator& rhs) {return p!=rhs.p;}
int& operator*() {return *p;}
};
int main () {
int numbers[]={10,20,30,40,50};
myiterator beginning(numbers);
myiterator end(numbers+5);
for (myiterator it=beginning; it!=end; it++)
cout << *it << " ";
cout << endl;
return 0;
}
iterator.cpp: In member function 'myiterator& myiterator::operator++(int)':
iterator.cpp:13: warning: reference to local variable 'tmp' returned
myiterator operator++(int)
没有找到相关结果
已邀请:
4 个回复
嗓瑰
的东西”，其中是迭代器。在实践中，这意味着如果适用于您（它没有）或按值返回，您可以通过引用返回。 见24.1.3 / 1委婪绷冗诉
myiterator operator++(int) {myiterator tmp(*this); operator++(); return tmp;}才脊烽馈低
myiterator& operator++(int) {myiterator tmp(*this); operator++(); return tmp;}myiterator operator++(int) {myiterator tmp(*this); operator++(); return tmp;} // ^^^^ Not return by reference. // Don't worry the cost is practically nothing for your class // And will probably be optimized into copying the pointer back.myiterator(const myiterator& mit) : p(mit.p) {}bool operator==(const myiterator& rhs) const {return p==rhs.p;} bool operator!=(const myiterator& rhs) const {return !(*this == rhs);} // ^^^^^^^ Added constint& operator*() {return *p;} int const& operator*() const {return *p;}桑娠贯涤