discard qualifiers的錯誤

寫代碼時，沒有注意到或者說不清楚的問題，discard qualifiers的問題，網上有人碰到此類問題並附介紹：

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?

正確答案是

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.

意思是說  在一個加了const限定符的成員函數中，不能夠調用 非const成員函數。

因爲如果調用了非const成員函數，就違反了const成員函數不改變對象的規定。

而error：...discards qualifiers 的意思就是缺少限定符

因此，他的解決辦法很簡單了 ，只要將checkElements()函數申明改爲  checkElements()const就行了