c++ - 遍历字符串集时没有可行的转换

Question

给定集合:

{“警报”、“警报”、“问题”

我想要输出:

alarm
alert
issue

我想向后或向前打印集合中的每个元素。以下是我尝试过的代码:

set<string> str_set = {"hi", "bye", "there"}; string cur_word;

for(long iter = str_set.begin(); iter!= set.end(); iter++) {
    cout << iter << endl;
}

我收到错误:

No viable conversion from 'std::__1::set<std::__1::basic_string<char>,
   std::__1::less<std::__1::basic_string<char> >, 
std::__1::allocator<std::__1::basic_string<char> > >::iterator' (aka 
'__tree_const_iterator<std::__1::basic_string<char>, 
std::__1::__tree_node<std::__1::basic_string<char>, void *> *, long>') to 
'long'

Answer 1

您的代码的问题在于 std::set<std::string>::begin() 返回的迭代器正在转换为 long时间。您应该定义 iter 的类型作为 std::set<std::string>::iterator或 auto .

---

std::set<std::string> str_set = {"hi", "bye", "there"};
for(auto iter = str_set.begin(); iter != str_set.end(); iter++) {
    std::cout << *iter << '\n';
}

与 auto相当于std::set<std::string>::iterator .

---

或

std::set<std::string> str_set = {"hi", "bye", "there"};
for(auto &value : str_set) {
    std::cout << value << '\n';
}

与 auto相当于std::string& .