我有一个类模板,它将输出存储在数组中的对象列表。我收到以下错误,我很困惑,因为错误是在.obj和.exe文件中导致错误。
1个未解析的外部(proj08.exe第1行)
未解析的外部符号“class std :: basic_ostream>& __cdecl operator<<(class std :: basic_ostream>&,class MyVector)”(?? 6 @ YAAAV?$ basic_ostream @ DU?$ char_traits @ D @ std @@@ std @@ AAV01 @ V?$ MyVector @ N @@@ Z)在函数_main(porj08.obj第1行)中引用
proj08.cpp
#include "stdafx.h"
#include <string>
#include "MyVector.h"
const double FRACTION = 0.5;
int main()
{
cout << "\nCreating a vector of doubles named Sam\n";
MyVector<double> sam;
cout << "\nPush 12 values into the vector.";
for (int i = 0; i < 12; i++)
sam.push_back(i + FRACTION);
cout << "\nHere is sam: ";
cout << sam;
cout << "\n---------------\n";
cout << "\nCreating an empty vector named joe";
MyVector<double> joe;
// test assignment
joe = sam;
cout << "\nHere is joe after doing an assignment:\n ";
cout << joe;
cout << "\n---------------\n";
// test the copy constructor
MyVector<double> bill = sam;
cout << "\nHere is bill after creating it using the copy constructor:\n ";
cout << bill;
cout << "\n---------------\n";
cout << endl;
system("PAUSE");
return 0;
}
MyVector.h
#pragma once
#include <iostream>
#include "stdafx.h"
using namespace std;
template <class T>
class MyVector
{
private:
int vectorSize;
int vectorCapacity;
T *vectorArray;
public:
MyVector() {
vectorArray = new T[10];
}
T size();
T capacity();
void clear();
void push_back(T n);
T at(int n);
friend ostream& operator<<(ostream& os, MyVector<T> vt);
MyVector<T> operator=(MyVector<T>&);
};
/*
* TEMPLATE FUNCTIONS
*/
//Return array size
template<class T>
T MyVector<T>::size()
{
return vectorSize;
}
// Return array capacity
template<class T>
T MyVector<T>::capacity()
{
return vectorCapacity;
}
// clear array values
template<class T>
void MyVector<T>::clear()
{
for (int i = 0; i < vectorSize; i++)
{
vectorArray[i] = '\0';
}
vectorSize = 0;
vectorCapacity = 2;
}
// Add number to array and double array size if needed
template<class T>
void MyVector<T>::push_back(T n)
{
int test = 100;
if (vectorCapacity > vectorSize)
{
vectorArray[vectorSize] = n;
vectorSize++;
}
else {
if (vectorCapacity == 0) {
vectorArray = new T[4];
vectorArray[0] = n;
vectorCapacity = 4;
vectorSize++;
}
else {
int newCapacity = vectorCapacity * 2;
// Dynamically allocate a new array of integers what is somewhat larger than the existing array.An algorithm that is often used is to double the size of the array.
int *tempArray = new int[newCapacity];
// Change capacity to be the capacity of the new array.
vectorCapacity = newCapacity;
// Copy all of the numbers from the first array into the second, in sequence.
for (int i = 0; i < MyVector::size(); i++)
{
tempArray[i] = vectorArray[i];
}
delete[] vectorArray;
vectorArray = new T[newCapacity];
for (int i = 0; i < MyVector::size(); i++)
{
vectorArray[i] = tempArray[i];
}
delete[] tempArray;
// Add the new element at the next open slot in the new array.
vectorArray[vectorSize] = n;
// Increment the size;
vectorSize++;
}
}
}
// Return Value and given point in array
template<class T>
T MyVector<T>::at(int n)
{
return vectorArray[n];
}
// Set one vector to equil another
template<class T>
MyVector<T> MyVector<T>::operator=(MyVector<T>& right) {
if (vectorCapacity < right.vectorCapacity) {
if (vectorCapacity != 0)
delete[] vectorArray;
vectorArray = new T[right.vectorCapacity];
vectorCapacity = right.vectorCapacity;
}
vectorSize = right.size();
// Assign values from left to right
for (int i = 0; i < vectorSize; i++)
{
vectorArray[i] = right.at(i);
}
return *this;
}
// Cout Vector
template<class T>
ostream& operator << (ostream& os, MyVector<T> vt)
{
T size = vt.size();
for (T i = 0; i < size; i++) {
os << "index " << i << " is " << vt.at(i) << endl;
}
return os;
}
答案 0 :(得分:8)
如果要匹配您定义的模板,则必须将friend
声明为函数模板:
template <typename U> // use U, so it doesn't clash with T
friend ostream& operator<<(ostream& os, MyVector<U> vt);
如果为类模板声明了friend
函数,则不会使其成为函数模板。
答案 1 :(得分:6)
@LogicStuff's answer is perfectly valid.我想澄清代码中究竟发生了什么,以及如何使用替代方法来避免它,因为我相信大多数C ++程序员至少在这个问题上遇到过一次。基本上,当模板被实例化时,比如说
MyVector<int> something;
然后自动将friend
声明绑定到模板类型,在本例中为int
,因此编译器生成
friend ostream& operator<<(ostream& os, MyVector<int> vt);
但是,这只是一个声明。你的后一个定义
template<class T>
ostream& operator << (ostream& os, MyVector<T> vt)
与int
无关,因为前者是更好的匹配,无论何时尝试
cout << something;
编译器尝试调用int
版本(没有定义)。所以你得到一个链接器错误。
另一种广泛使用的方法是定义您的运算符内联类,如
friend ostream& operator << (ostream& os, MyVector<T> vt)
{
T size = vt.size();
for (T i = 0; i < size; i++) {
os << "index " << i << " is " << vt.at(i) << endl;
}
return os;
}
现在,MyVect
的每个实例化都会生成与相应模板类型相关的operator<<
的有效定义。请注意,运算符本身不是成员函数,并且只能通过Argument Dependent Lookup (ADL)在全局命名空间中显示。这个技巧被称为朋友名称注入,在the Barton–Nackman trick中广泛使用,你可以成功使用它,因为在像
cout << something;
电话转换为
operator<< (std::cout, something)
并且something
的类型为MyVector<int>
,operator<<
的定义可通过ADL找到。如果您operator<<
需要, int
作为第二个参数,它不会通过ADL找到,因为基本类型没有关联的命名空间。