C++ 明显调用的括号前的表达式必须具有（指向）函数类型
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expression preceding parentheses of apparent call must have (pointer-to-) function type
提问by S.yao
I am learning C++ template on vs2015 community.Here is my code, I want to define a template class and call the member function in the main()function.
我在vs2015社区学习C++模板。这里是我的代码，我想定义一个模板类并调用函数中的成员main()函数。
template <typename T>
class Arithmetic {
T _a;
T _b;
Arithmetic() {};
public
Arithmetic(T a, T b) :_a(a), _b(b) {};
T max const() { return _a + _b; };
T minus const() { return _a - _b; };
};
int main() {
Arithmetic<int> ar(5,6);
cout << ar.max() << endl;
}
When I build this program, I get error at the last line. It says:
当我构建这个程序时，我在最后一行出现错误。它说：
Expression preceding parentheses of apparent call must have (pointer-to-) function type
明显调用的括号前面的表达式必须具有（指向）函数类型
What should I do?
我该怎么办？
回答by srinivirt
The error indicates trying to call a function max() that is not defined as a function. Change parenthesis after const keyword to after the identifier max:
该错误表示尝试调用未定义为函数的函数 max()。将 const 关键字后的括号更改为标识符 max 后：
T max const()...
to
到
T max() const ...
回答by MikeCAT
- Add required header inclusion and
using - Add
:afterpublic - Move
constto proper position
- 添加所需的标头包含和
using - 添加
:后public - 移动
const到合适的位置
#include <iostream>
using std::cout;
using std::endl;
template <typename T>
class Arithmetic {
T _a;
T _b;
Arithmetic() {};
public:
Arithmetic(T a, T b) :_a(a), _b(b) {};
T max() const { return _a + _b; };
T minus() const { return _a - _b; };
};
int main() {
Arithmetic<int> ar(5,6);
cout << ar.max() << endl;
}
