模板参数中的 const

这个模板中const关键字的作用是什么?

模板<class T, int const ROWNUM, int const COLNUM>类矩阵

这是否意味着这个模板只接受一个 const 作为参数?如果是这样，有没有办法将变量作为 COLNUM 和 ROWNUM 传递?

(当我尝试将变量作为模板的 COLNUM 传递时，它会给出错误:IntelliSense:表达式必须具有常量值")

解决方案

忽略:

<块引用>

[C++11: 14.1/4]: 非类型模板参数应具有以下之一(可选cv-qualified) 类型:

• 整数或枚举类型，
• 指向对象的指针或指向函数的指针，
• 对对象的左值引用或对函数的左值引用，
• 指向成员的指针，
• std::nullptr_t.

[C++11: 14.1/5]: [ 注意: 其他类型在下面明确地或被管理 形式的规则隐含地禁止模板参数 (14.3).―结束注释 ] 在确定其类型时，模板参数上的顶级cv-qualifiers被忽略.

相同的措辞出现在 C++03 中的相同位置.

这部分是因为模板参数必须在编译时知道.因此，无论您是否有 const，您可能不会传递一些变量值:

模板<int N>无效 f(){N = 42;}模板<int const N>无效 g(){N = 42;}主函数(){f<0>();g<0>();静态常量 int h = 1;f<h>();g<h>();}

<块引用>

prog.cpp:在函数'void f() [with int N = 0]'中:
prog.cpp:15:从这里实例化
prog.cpp:4: 错误:需要左值作为赋值的左操作数
prog.cpp:在函数'void g() [with int N = 0]'中:
prog.cpp:16:从这里实例化
prog.cpp:10: 错误:需要左值作为赋值的左操作数
prog.cpp:在函数'void f() [with int N = 1]'中:
prog.cpp:19:从这里实例化
prog.cpp:4: 错误:需要左值作为赋值的左操作数
prog.cpp:在函数'void g() [with int N = 1]'中:
prog.cpp:20:从这里实例化
prog.cpp:10: 错误:需要左值作为赋值的左操作数

What is the effect of the const keyword in this template?

template <class T, int const ROWNUM, int const COLNUM> 
class Matrix

Does it mean that this template only accept a const as parameter? If so, is there a way to pass a variable as the COLNUM and ROWNUM?

(when I try to pass a variable as the COLNUM for the template, it gives an error: "IntelliSense: expression must have a constant value")

解决方案

It's ignored:

[C++11: 14.1/4]: A non-type template-parameter shall have one of the following (optionally cv-qualified) types:

• integral or enumeration type,
• pointer to object or pointer to function,
• lvalue reference to object or lvalue reference to function,
• pointer to member,
• std::nullptr_t.

[C++11: 14.1/5]: [ Note: Other types are disallowed either explicitly below or implicitly by the rules governing the form of template-arguments (14.3). ―end note ] The top-level cv-qualifiers on the template-parameter are ignored when determining its type.

The same wording is present at the same location in C++03.

This is partially because template arguments must be known at compile-time anyway. So, whether you have the const there or not, you may not pass some variable value:

template <int N>
void f()
{
    N = 42;
}

template <int const N>
void g()
{
    N = 42;
}

int main()
{
    f<0>();
    g<0>();

    static const int h = 1;
    f<h>();
    g<h>();
}

prog.cpp: In function ‘void f() [with int N = 0]’:
prog.cpp:15: instantiated from here
prog.cpp:4: error: lvalue required as left operand of assignment
prog.cpp: In function ‘void g() [with int N = 0]’:
prog.cpp:16: instantiated from here
prog.cpp:10: error: lvalue required as left operand of assignment
prog.cpp: In function ‘void f() [with int N = 1]’:
prog.cpp:19: instantiated from here
prog.cpp:4: error: lvalue required as left operand of assignment
prog.cpp: In function ‘void g() [with int N = 1]’:
prog.cpp:20: instantiated from here
prog.cpp:10: error: lvalue required as left operand of assignment