C++ 结构模板变量快捷方式定义不起作用

Question

template <class T>
Struct st
{
   ... 
} a<T>;

Doesnt work althogh non template shortcut works how ia tge right way to do the template shorycut?

Answer 1

The C++ grammar generally lets you do type name; to declare a variable, where type is a type and name is a name. The syntax struct foo { } is considered a type by the grammar, which makes struct foo { } bar; legal (declare a type named foo and a variable of type foo with the name bar ).

However, in your example, st is not actually a type but rather a type template.

Type templates aren't types in their own right; they're "recipes" for the compiler to automatically create types on demand. st<int> is a type, but st by itself isn't.

Because of this, the shortcut is not grammatically correct, and is nonsensical when you think about it. You can't declare a variable of something that isn't a type. The workaround would be to simply not use the shortcut and declare a variable separately.