下面的代码编译得很好。但是当去连接,为什么下面的代码编译得很好,但使用静态链接时显示错误
它显示了以下错误
Undefined symbols for architecture x86_64:
"derived::counter", referenced from:
derived::getAddressCounter() in main.cpp.o
ld: symbol(s) not found for architecture x86_64
collect2: ld returned 1 exit status
我怀疑有什么错误与静态。但不知道为什么。因为一旦我拿出静态,代码链接很好。但是静态在这个代码中扮演什么角色?
#include <iostream>
#include <string>
struct base_result { };
struct result : public base_result {
int a;
std::string b;
};
struct base {
static base_result counter;
};
struct derived: public base {
static result counter;
result * getAddressCounter(){
counter.a = 10;
counter.b = "haha";
return &counter;
}
};
int main(){
derived d;
result * ptr;
ptr = d.getAddressCounter();
ptr->a = 20;
ptr->b = "baba";
std::cout << ptr->a << std::endl;
std::cout << ptr->b << std::endl;
return 0;
}
太好了,谢谢。 – 2012-03-09 07:54:13
因此,一个实例变量的正常定义将变成声明,如果它是静态变量? – 2012-03-09 08:07:27
没关系我知道了 – 2012-03-09 08:11:13