Question

// std:: iterator sample
#include <iostream>  // std::cout
#include <iterator>  // std::iterator, std::input_iterator_tag

class MyIterator:public std::iterator<std::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 from(numbers);
MyIterator until(numbers+5);
for (MyIterator it=from; it!=until; it++)
std::cout << *it << '';
std::cout << '\n';

return 0;
};

当我试图更好地理解“迭代器”是什么时。我将这些代码复制到我的编译器（codeBlock）。有一个错误：“预期';'在'！'之前令牌”。这有什么关系？

Answer 1

operator!=中有拼写错误：

p!rhs.p

应该阅读

p != rhs.p

或者，更一般地说，

!(*this == rhs)

此行中还有一个无效的空字符常量：

std::cout << *it << '';
                    ^^  // Get rid of that, or change it to something sensible

Answer 2

看起来就是这一行：

bool operator!=(const MyIterator& rhs){return p!rhs.p;}

将其更改为：

bool operator!=(const MyIterator& rhs){return p != rhs.p;}

Answer 3

我建议定义！= as

bool operator==(const MyIterator& rhs){return p == rhs.p;}
bool operator!=(const MyIterator& rhs){return !(*this == rhs);}

因此，如果==变得更复杂，您不必在！=

中复制代码

同样，您可以定义＆lt; ＆GT; ＆lt; =＆gt; =仅仅＆lt;和==，尽量减少代码重复

bool operator == (const MyIterator& rhs) const {return p == rhs.p;}
bool operator <  (const MyIterator& rhs) const {return p < rhs.p;}

bool operator <= (const MyIterator& rhs) const {return (*this == rhs) || (*this < rhs);}
bool operator != (const MyIterator& rhs) const {return !(*this == rhs);}
bool operator >= (const MyIterator& rhs) const {return !(*this < rhs);}
bool operator >  (const MyIterator& rhs) const {return !(*this <= rhs);}

Answer 4

我相信!=运营商的过载。该行应该是：

bool operator!=(const MyIterator& rhs){return p!=rhs.p;}

错误：预期';'在'！'之前代币

4 个答案: