C ++程序不会编译 [英] C++ program will not complie

问题描述

以下简单程序无法编译：

The following simple program will not compile:

#include <map>
     2
     3 using namespace std;
     4 ^M
     5 template<class T, class A>
     6 void ShowMap(const map<T, A>& v)
     7 {
     8    map<T, A>::const_iterator ci = v.begin();
     9    return;
    10
    11 }

编译器产生以下错误信息;



g ++ program.cpp -o program

2 program.cpp：在函数'void ShowMap（const std :: map< t，>&）'中：

3 program.cpp：8：4：错误：在'std :: map< t，> :: const_iterator'之前需要'typename'因为'std :: map< t，>'是依赖范围

4 program.cpp：8：30：错误：预期';'在'ci'之前

5 make：*** [程序]错误1

~

有谁可以向我解释这个问题？



非常感谢



TCNM

The compiler produces the following error message;

g++ program.cpp -o Program
2 program.cpp: In function ‘void ShowMap(const std::map<t,>&)’:
3 program.cpp:8:4: error: need ‘typename’ before ‘std::map<t,>::const_iterator’ because ‘std::map<t,>’ is a dependent scope
4 program.cpp:8:30: error: expected ‘;’ before ‘ci’
5 make: *** [Program] Error 1
~
Can anyone explain the problem to me?

Thank you very much

TCNM

推荐答案

因为const_iterator是由std :: map类中的typedef创建的，依赖于尚未确定的类型T您的代码被解析为新的C ++ 11标准，您需要将其称为

Because of the way const_iterator is created by a typedef inside the std::map class dependent on the as yet undecided type T when your code is parsed the new C++11 standard says you need to refer to it as

typename map< T, A >::const_iterator ci = v.begin();

换句话说''通过在std :: map<中查找const_iterator来命名的类型; T，A>

模板很有趣，只要你不赶时间。

in other words ''the type that is named by looking up const_iterator in std::map< T, A >
templates are fun as long as you''re not in a hurry.

试试

...
auto ci = v.begin();
...

干杯

Andi

Cheers
Andi

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