复制构造函数在下面的代码中失败。我已经刮到了代码清晰模板的复制构造函数
#include <iostream>
#include <stdio.h>
#include <assert.h>
namespace my {
template <class T>
class Sptr {
private:
//some kind of pointer
//one to current obj
T* obj;
.
.
.
public:
.
.
.
Sptr(const Sptr &);
template <typename U>
Sptr(const Sptr<U> &);
.
.
.
};//Class ends
template <typename T>
template <typename U>
Sptr<T>::Sptr(U* u) {
//do something
}
template <typename T>
Sptr<T>::Sptr(const Sptr<T> ©Obj) {
//do copy constructor stuff
}
.
.
.
}
using namespace std;
using namespace my;
/* Basic Tests 1 ================================================================================ */
size_t AllocatedSpace;
class Base1 {
protected:
Base1() : derived_destructor_called(false) {
printf("Base1::Base1()\n");
}
private:
Base1(const Base1 &); // Disallow.
Base1 &operator=(const Base1 &); // Disallow.
public:
virtual ~Base1() {
printf("Base1::~Base1()\n");
assert(derived_destructor_called);
}
protected:
bool derived_destructor_called;
};
class Derived : public Base1 {
friend void basic_tests_1();
private:
Derived() {}
Derived(const Derived &); // Disallow.
Derived &operator=(const Derived &); // Disallow.
public:
~Derived() {
printf("Derived::~Derived()\n");
derived_destructor_called = true;
}
int value;
};
这个测试代码行产生错误Sptr<Base1> sp3(sp);
void basic_tests_1() {
//size_t base = AllocatedSpace;
// Test deleting through original class.
{
// Base1 assigned from Sptr<Derived>.
{
Sptr<Base1> sp2;
{
Sptr<Derived> sp(new Derived);
// Test template copy constructor.
Sptr<Base1> sp3(sp);
sp2 = sp;
sp2 = sp2;
}
}
}
}
错误:
/tmp/ccKrn1xG.o: In function `basic_tests_1()':
Sptr.cpp:(.text+0x81): undefined reference to `my::Sptr<Base1>::Sptr<Derived>(my::Sptr<Derived> const&)'
collect2: error: ld returned 1 exit status
这里有一件事你为什么在Sptr上有templte T :: Sptr?它已经在类声明中。它会影响它。 –
2013-03-31 19:06:18
不要“为了清晰而剪切代码”,因为现在它甚至没有接近有效的代码,它全是'.'字符,'SPTR'的类定义没有关闭'}',并且您已经定义了一个'SPTR (U *')构造函数没有声明。而是**将代码减少到仍然类似于有效代码的东西。请参阅http://sscce.org/并解决问题。 –