How to use TypeName in C + + template (template) __c++

How to use TypeName in Template (template)

http://blog.csdn.net/caroline_wendy/article/details/23910709

when declaring template parameters, the prefix keyword class and typename can be interchanged;

use keyword typename to identify nested subordinate type names, but not in base class lists and member initialization lists.

subordinate name (dependent names): The name that appears within the template (template), dependent on a template (template) parameter, such as T t;

nested subordinate names (nested dependent names): subordinate names are nested within class, such as T::const_iterator CI;

Non-subordinate name (non-dependent names): does not rely on the name of any template parameter, such as int value;

If you do not specifically point to TypeName, nested subordinate names may result in parsing (parse) ambiguity.

Any time in the template (template) to the middle finger of a nested subordinate type name, you need to add a keyword in the previous position TypeName;

Otherwise the error (GCC): error:need ' TypeName ' before ' t::xxx ' because ' is a dependent scope

Code:

* * * BInsertSort.cpp * *  Created on:2014.4.17.
 *      author:spike * *

#include <iostream>
#include <string>
#include <vector>

using namespace std;

Template<typename t>
void print2nd (const t& container) {typename
	t::const_iterator iter ( Container.begin ()); Not add typename, error
	++iter;
	int value = *iter;
	Std::cout << value;
}

int main () {
	vector<int> VI = {1,2,3,4,5};
	print2nd (vi);

	return 0;
}

Output:

2

Exceptions:The nested subordinate type name, if it is List of base classes (base class list)And member initial Value column (members initialization list)In Do not useTypeName

Code:

* * * * BInsertSort.cpp * *
 Created on:2014.4.17 * author:spike * *

#include < iostream>
#include <vector>

using namespace std;

struct Number {number
	(int x) {
		std::cout << "number =" << x << std::endl;
	}
};

template<typename t>
struct base{
	typedef number Nested;

Template<typename t>
class derived:public base<t>::nested {//No typename public
:
	explicit Derived (int x): base<t>::nested (x) {//No typename
		TypeName base<t>::nested Temp (7);//must use c27/>}
};

int main () {
	derived<int> D (5);

	return 0;
}

Output:

Number = 5 number
= 7

When using Attribute Class (traits Class), you must use TypeName, such as

Code:

* * * * BInsertSort.cpp * *
 Created on:2014.4.17 * author:spike * *

#include < array>
#include <iostream>

using namespace std;

Template<typename t>
void Workwithiter (T iter) {
	typedef typename STD::ITERATOR_TRAITS<T>:: Value_type Value_type; Use typename
	value_type temp (*iter);
	Std::cout << "temp =" << temp << std::endl;

}

int main () {
	std::array<int, 5> ai = {1,2,3,4,5};
	Std::array<int, 5>::iterator aiiter = Ai.begin ();
	Workwithiter (aiiter);
	return 0;
}

Output:

temp = 1