gcc 返回嵌套类错误
我尝试使用嵌套类的完全限定名称，如下所示，但编译器犹豫不决！
template <class T> class Apple {
//constructors, members, whatevers, etc...
public:
class Banana {
public:
Banana() {
//etc...
}
//other constructors, members, etc...
};
};
template <class K> class Carrot{
public:
//etc...
void problemFunction()
{
Apple<int>::Banana freshBanana = someVar.returnsABanana(); //line 85
giveMonkey(freshBanana); //line 86
}
};
我的问题是，编译器说：
Carrot.h:85: error: expected ';' before 'freshBanana'
Carrot.h:86: error: 'freshBanana' was not declared in this scope
我曾认为使用完全限定名称允许我访问这个嵌套类？它可能会打我的脸，但我到底在这里没看到什么？
I am attempting to use the fully qualified name of my nested class as below, but the compiler is balking!
template <class T> class Apple {
//constructors, members, whatevers, etc...
public:
class Banana {
public:
Banana() {
//etc...
}
//other constructors, members, etc...
};
};
template <class K> class Carrot{
public:
//etc...
void problemFunction()
{
Apple<int>::Banana freshBanana = someVar.returnsABanana(); //line 85
giveMonkey(freshBanana); //line 86
}
};
My issue is, the compiler says:
Carrot.h:85: error: expected ';' before 'freshBanana'
Carrot.h:86: error: 'freshBanana' was not declared in this scope
I had thought that using the fully qualified name permitted me to access this nested class? It's probably going to smack me in the face, but what on earth am I not seeing here??
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评论（1）
这可能不是您在代码中所做的。错误消息看起来像你这样做
编译器在解析代码之前必须知道名称是否命名了类型。在这种情况下，当它解析时，它无法知道，因为
K是什么类型，尚不知道（您可能有一个Apple的专门化，但它不有那个嵌套类）。因此它假设Apple::Banana 不是一种类型。但是，它是一个表达式，后面需要一个运算符或分号。您可以通过插入
typename来修复它：断言名称是一种类型，然后编译器知道将其解析为声明。
That's probably not what you do in your code. The error message looks like you do this
The compiler has to know before it parses the code whether a name names a type or not. In this case, when it parses, it cannot know because what type
Kis, is not yet known (you could have a specialization forApple<int>that doesn't have that nested class). So it assumesApple<K>::Bananais not a type. But then, it is an expression and an operator is needed after it or a semicolon.You can fix it by inserting
typename:That asserts the name is a type, and the compiler then knows to parse this as a declaration.