我上这堂课
namespace baseUtils {
template<typename AT>
class growVector {
int size;
AT **arr;
AT* defaultVal;
public:
growVector(int size, AT* defaultVal ); //Expects number of elements (5) and default value (NULL)
AT*& operator[](unsigned pos);
int length();
void reset(int pos); //Resets an element to default value
void reset(); //Resets all elements to default value
~growVector();
};
}
这是operator []的实现
template<typename AT>
AT*& growVector<AT>::operator [](unsigned pos){
if (pos >= size){
int newSize = size*2;
AT** newArr = new AT*[newSize];
memcpy(newArr, arr, sizeof(AT)*size);
for (int i = size; i<newSize; i++)
newArr[i] = defaultVal;
size = newSize;
delete arr;
arr = newArr;
}
return arr[pos];
}
（是的，我的确意识到我不检查size * 2> = pos ...但这不是重点）
如果我在类似的代码中使用它：
int main() {
growVector<char> gv();
char* x = NULL;
for (int i = 0; i< 50; i++){
gv[i] = x;
}
gv.reset();
return 0;
}
编译器说
../src/base.cpp:98: warning: pointer to a function used in arithmetic
../src/base.cpp:98: error: assignment of read-only location ‘*(gv + ((unsigned int)i))’
../src/base.cpp:98: error: cannot convert ‘char*’ to ‘baseUtils::growVector<char>()’ in assignment
参照线gv [i] = x; （似乎没有看到[]的重新定义）
为什么？？？？我想念什么？
更正构造函数问题后，我有了链接器sayng：
/home/dario/workspace/base/Debug/../src/base.cpp:95: undefined reference to `baseUtils::growVector<char>::growVector(int, char*)'
/home/dario/workspace/base/Debug/../src/base.cpp:98: undefined reference to `baseUtils::growVector<char>::operator[](unsigned int)'
/home/dario/workspace/base/Debug/../src/base.cpp:100: undefined reference to `baseUtils::growVector<char>::reset()'
/home/dario/workspace/base/Debug/../src/base.cpp:101: undefined reference to `baseUtils::growVector<char>::~growVector()'
/home/dario/workspace/base/Debug/../src/base.cpp:101: undefined reference to `baseUtils::growVector<char>::~growVector()'
像它无法链接...为什么？ ：O
最佳答案
问题是你的声明
growVector<char> gv();
编译器将此解释为声明一个名为
gv的函数，该函数返回一个growVector<char>，而不是您所意图的对象。由于没有默认的构造函数，因此无论如何都不会编译。更改为：growVector<char> gv(0,0);