今晚中秋之夜,苦逼的程序员只能呆在寝室。
正无聊的时候,想起最近学的数据结构,想完善一下代码,于是想给线性表重载一下<<运算符。
Out()
template<class T> void LineList<T>::Out(ostream& os) { for(int p = 0; p < length; p++) os<<array[p]<<endl; }
friend ostream& operator<<(ostream& out,const LineList<T>&line) { line.Out(out); return out; }
但是问题出现了。。
error:passing ‘const LineList<int>’ as ‘this’ argument of ‘void LineList<T>::Out(ostream&) [with T = int]’ discards qualifiers百思不得其解,于是百度,google吧。。
发现Stackoverflow上也有人有相同的问题
下面是他的问题:
For my compsci class, I am implementing a Stack template class, but have run into an odd error:
Stack.h: In member function ‘
const T Stack<T>::top() const
[with T = int]’:Stack.cpp:10: error: passing ‘
const Stack<int>
’ as ‘this
’ argument of ‘void Stack<T>::checkElements()
[with T = int]’ discards qualifiers
Stack<T>::top()
looks like this:
const T top() const { checkElements(); return (const T)(first_->data); }
Stack<T>::checkElements()
looks like this:
void checkElements() { if (first_==NULL || size_==0) throw range_error("There are no elements in the stack."); }
The stack uses linked nodes for storage, so first_
is a pointer to the first node.
Why am I getting this error?
正确答案是
Your checkElements()
function is not marked as const
so you can't call it on const
qualified objects.
top()
, however is const qualified so in top()
, this
is a pointer to a const Stack
(even if theStack
instance on which top()
was called happens to be non-const
), so you can't callcheckElements()
which always requires a non-const
instance.
意思是说 在一个加了const限定符的成员函数中,不能够调用 非const成员函数。
因为如果调用了非const成员函数,就违反了const成员函数不改变对象的规定。
而error:...discards qualifiers 的意思就是缺少限定符
因此,他的解决办法很简单了 ,只要将checkElements()函数申明改为 checkElements()const就行了
因为重载<<中有一个参数为 const LineList<T>&line
注意:const对象只能调用const成员函数;
所以调用line.Out(cout)的时候,Out()函数必须是const成员函数,因而出现了错误discards qualifiers
解决办法就是将Out(ostream& os)改为 Out(ostream& os)const 。