声明一个C ++集合迭代器 [英] declaring a C++ set iterator

问题描述

可能重复：



c ++模板typename迭代器

以下代码将无法编译：

  #include< iostream> 
 #include< set> 
 using namespace std; 
 
 template< class T> 
 void printSet（set< T> s）{
 set< T> :: iterator it; 
} 
 
 int main（int argc，char ** argv）{
 set< int> s; 
 printSet< int>（s）; 
 return 0; 
} 
  

我得到一个错误说：

  set.cpp：在函数'void printSet（std :: set< T，std :: less< _Key>，std :: allocator< _CharT> ：
 set.cpp：7：error：expected`;'before'it'
 set.cpp：在函数'void printSet（std :: set< T，std :: less< _Key& std :: allocator< _CharT>>）[with T = int]'：
 set.cpp：12：从这里实例化
 set.cpp：7：error：dependent-name'std :: set< T，std :: less< _Key>，std :: allocator< _CharT> > :: iterator'被解析为非类型，但实例化产生类型
 set.cpp：7：note：say'typename std :: set< T，std :: less< _Key>，std :: allocator< _CharT> > :: iterator'如果类型意味着
  

我做错了什么？我觉得我几乎没有写任何东西，并且已经C ++给我这个可怕的消息。

如果它是有帮助的，看起来像，如果我注释掉行用迭代器，没有错误。但是，到目前为止，我在线上看到的所有示例都是这样声明迭代器。我认为。

解决方案

在关键字 typename 前声明：

  typename set< T> :: iterator it; 
  

每当引用依赖于模板参数的类型时，您需要这样做。在这种情况下， iterator 取决于T.否则，编译器很难设法找出迭代器无论是类型，还是静态成员或其他任何东西。

请注意，如果你有一个混凝土类型，例如：

  set< int> :: iterator it; 
  

不需要 typename p>

Possible Duplicate:
Where and why do I have to put the “template” and “typename” keywords?
c++ template typename iterator

The following code will not compile:

#include <iostream>
#include <set>
using namespace std;

template<class T>
void printSet(set<T> s){
    set<T>::iterator it;
}

int main(int argc, char** argv){
    set<int> s;
    printSet<int>(s);
    return 0;
}

I get an error saying:

set.cpp: In function ‘void printSet(std::set<T, std::less<_Key>, std::allocator<_CharT> >)’:
set.cpp:7: error: expected `;' before ‘it’
set.cpp: In function ‘void printSet(std::set<T, std::less<_Key>, std::allocator<_CharT> >) [with T = int]’:
set.cpp:12:   instantiated from here
set.cpp:7: error: dependent-name ‘std::set<T,std::less<_Key>,std::allocator<_CharT> >::iterator’ is parsed as a non-type, but instantiation yields a type
set.cpp:7: note: say ‘typename std::set<T,std::less<_Key>,std::allocator<_CharT> >::iterator’ if a type is meant

What did I do wrong? I feel like I've hardly written anything, and already C++ gives me this scary message.

In case it is helpful, it looks like, if I comment out the line with the iterator, there are no errors. However, all the examples I've seen online so far seem to declare iterators this way. I think.

解决方案

Precede your declaration with the keyword typename, thusly:

typename set<T>::iterator it;

You need to do this whenever you refer to a type that is dependent upon a template argument. In this case, iterator is dependent on T. Otherwise, the compiler has difficulty trying to figure out what iterator is, whether it's a type, or a static member or something else. (well, technically, it doesn't try to figure it out, its decision is already made, it just happens to be wrong)

Note that if you have a concrete type, for example:

set<int>::iterator it;

The typename is not necessary.

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