C++ 类实例上的静态成员方法调用
这是一个小测试程序:
#include <iostream>
class Test
{
public:
static void DoCrash(){ std::cout<< "TEST IT!"<< std::endl; }
};
int main()
{
Test k;
k.DoCrash(); // calling a static method like a member method...
std::system("pause");
return 0;
}
在 VS2008 + SP1 (vc9) 上它编译得很好:控制台只显示“TEST IT!”。
据我所知,不应在实例化对象上调用静态成员方法。
- 我错了吗? 从标准的角度来看,这段代码正确吗?
- 如果这是正确的,为什么会这样呢? 我找不到为什么允许它,或者可能是为了帮助在模板中使用“静态或非静态”方法?
Here is a little test program:
#include <iostream>
class Test
{
public:
static void DoCrash(){ std::cout<< "TEST IT!"<< std::endl; }
};
int main()
{
Test k;
k.DoCrash(); // calling a static method like a member method...
std::system("pause");
return 0;
}
On VS2008 + SP1 (vc9) it compiles fine: the console just display "TEST IT!".
As far as I know, static member methods shouldn't be called on instanced object.
- Am I wrong? Is this code correct from the standard point of view?
- If it's correct, why is that? I can't find why it would be allowed, or maybe it's to help using "static or not" method in templates?
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该标准规定不需要通过实例调用该方法,但这并不意味着您不能这样做。 甚至有一个使用它的例子:
C++03, 9.4 static Members
The standard states that it is not necessary to call the method through an instance, that does not mean that you cannot do it. There is even an example where it is used:
C++03, 9.4 static members
静态函数不需要实例化对象来调用,因此
完全相同
其行为与使用范围解析运算符 (::) 来确定类内的静态函数
。 请注意,在这两种情况下,编译器都不会将
this
指针放入堆栈中,因为静态函数不需要它。Static functions doesn´t need an instanciated object for being called, so
behaves exactly the same as
using the scope resolution operator (::) to determine the static function inside the class.
Notice that in both case the compiler doesn´t put the
this
pointer in the stack since the static function doesn't need it.它在多种情况下可能很有用:
[您建议的模板中的“是否静态”方法:]当可以为模板指定许多类型,并且模板随后想要调用成员:提供静态函数的类型可以使用与成员函数相同的符号来调用 - 前者可能更有效(没有
this
指针来传递/绑定),而后者允许多态(虚拟
)成员数据的分派和使用最小化代码维护
如果一个函数从需要特定于实例的数据演变为不需要它 - 因此被设为
静态
以允许轻松的无实例使用并防止意外使用实例数据 - 所有现有客户端使用点无需费力更新如果类型发生更改,
var.f()
调用将继续使用var
类型的函数,而Type::f()< /code> 可能需要手动更正
当您有一个返回 a 的表达式或函数调用时值并且想要调用(可能或总是)
static
函数,.
表示法可能会阻止您使用decltype
或支持模板来访问该类型,这样有时变量名只是更短、更方便,或者以更自记录的方式命名
It's potentially useful in several scenarios:
[the '"static or not" method in templates' you suggest:] when many types could have been specified to a template, and the template then wants to invoke the member: the types providing a static function can be called using the same notation as a member function - the former may be more efficient (no
this
pointer to pass/bind), while the latter allows polymorphic (virtual
) dispatch and use of member dataminimising code maintenance
if a function evolves from needing instance-specific data to not needing it - and is therefore made
static
to allow easy instance-free use and prevent accidental use of instance data - all the points of existing client usage don't need to be labouriously updatedif the type's changed the
var.f()
invocation continues to use thevar
type's function, whereasType::f()
may need manual correctionwhen you have an expression or function call returning a value and want to invoke the (potentially or always)
static
function, the.
notation may prevent you needing to usedecltype
or a supporting template to get access to the type, just so you can use the::
notationsometimes the variable name is just much shorter, more convenient, or named in a more self-documenting way
静态方法也可以使用类的对象来调用,就像在 Java 中一样。 尽管如此,您不应该这样做。 使用作用域运算符,如
Test::DoCrash();
也许您会想到命名空间:它只能由
Test::DoCrash();
从该命名空间外部调用,如果该函数未使用using 指令/声明
显式导入到调用者的作用域中。static methods can be called also using an object of the class, just like it can be done in Java. Nevertheless, you shouldn't do this. Use the scope operator like
Test::DoCrash();
Maybe you think of namespaces:which can only be called by
Test::DoCrash();
from outside that namespace if the function is not imported explicitly using ausing directive/declaration
into the scope of the caller.