在 C++ 中访问静态类变量？
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Accessing static class variables in C++?
提问by Paul Wicks
Duplicate:
C++: undefined reference to static class member
If I have a class/struct like this
如果我有这样的类/结构
// header file
class Foo
{
public:
static int bar;
int baz;
int adder();
};
// implementation
int Foo::adder()
{
return baz + bar;
}
This doesn't work. I get an "undefined reference to `Foo::bar'" error. How do I access static class variables in C++?
这不起作用。我收到“对‘Foo::bar’的未定义引用”错误。如何在 C++ 中访问静态类变量？
回答by Drakosha
You must add the following line in the implementationfile:
您必须在实现文件中添加以下行：
int Foo::bar = you_initial_value_here;
This is required so the compiler has a place for the static variable.
这是必需的，以便编译器为静态变量留出空间。
回答by Chris Jester-Young
It's the correct syntax, however, Foo::barmust be defined separately, outside of the header. In one of your .cppfiles, say this:
这是正确的语法，但是，Foo::bar必须在标头之外单独定义。在你的一个.cpp文件中，这样说：
int Foo::bar = 0; // or whatever value you want
回答by Artyom
You need add a line:
您需要添加一行：
int Foo::bar;
That would define you a storage. Definition of static in class is similar to "extern" -- it provides symbol but does not create it. ie
这将为您定义一个存储。类中静态的定义类似于“extern”——它提供符号但不创建它。IE
foo.h
foo.h
class Foo {
static int bar;
int adder();
};
foo.cpp
文件
int Foo::bar=0;
int Foo::adder() { ... }
回答by BattleTested
for use of static variable in class, in first you must give a value generaly (no localy) to your static variable (initialize) then you can accessing a static member in class :
要在类中使用静态变量，首先您必须为静态变量（初始化）提供一个通用值（无局部），然后您可以访问类中的静态成员：
class Foo
{
public:
static int bar;
int baz;
int adder();
};
int Foo::bar = 0;
// implementation
int Foo::adder()
{
return baz + bar;
}
