The objects in the std::set are stored as const StudentT. So when you try to call getId() with the const object the compiler detects a problem, mainly you're calling a non-const member function on const object which is not allowed because non-const member functions make NO PROMISE not to modify the object; so the compiler is going to make a safe assumption that getId() might attempt to modify the object but at the same time, it also notices that the object is const; so any attempt to modify the const object should be an error. Hence compiler generates an error message.

解决方案很简单:将函数设为const as:

int getId() const {
    return id;
}
string getName() const {
    return name;
}

这是必要的，因为现在你可以在const对象上调用getId()和getName():

void f(const StudentT & s)
{
     cout << s.getId();   //now okay, but error with your versions
     cout << s.getName(); //now okay, but error with your versions
}

作为旁注，你应该实现操作符< As:

inline bool operator< (const StudentT & s1, const StudentT & s2)
{
    return  s1.getId() < s2.getId();
}

注形参现在是const引用。

The objects in the std::set are stored as const StudentT. So when you try to call getId() with the const object the compiler detects a problem, mainly you're calling a non-const member function on const object which is not allowed because non-const member functions make NO PROMISE not to modify the object; so the compiler is going to make a safe assumption that getId() might attempt to modify the object but at the same time, it also notices that the object is const; so any attempt to modify the const object should be an error. Hence compiler generates an error message.

解决方案很简单:将函数设为const as:

int getId() const {
    return id;
}
string getName() const {
    return name;
}

这是必要的，因为现在你可以在const对象上调用getId()和getName():

void f(const StudentT & s)
{
     cout << s.getId();   //now okay, but error with your versions
     cout << s.getName(); //now okay, but error with your versions
}

作为旁注，你应该实现操作符< As:

inline bool operator< (const StudentT & s1, const StudentT & s2)
{
    return  s1.getId() < s2.getId();
}

注形参现在是const引用。

实际上，c++标准(即c++ 0x草案)说(tnx @Xeo和@Ben Voigt指出了这一点):

23.2.4 Associative containers 5 For set and multiset the value type is the same as the key type. For map and multimap it is equal to pair. Keys in an associative container are immutable. 6 iterator of an associative container is of the bidirectional iterator category. For associative containers where the value type is the same as the key type, both iterator and const_iterator are constant iterators. It is unspecified whether or not iterator and const_iterator are the same type.

所以vc++ 2008 Dinkumware实现是错误的。

旧的回答:

你得到这个错误是因为在std lib的某些实现中，set::iterator与set::const_iterator相同。

例如，libstdc++(随g++一起发布)就有它(查看这里的整个源代码):

typedef typename _Rep_type::const_iterator            iterator;
typedef typename _Rep_type::const_iterator            const_iterator;

在SGI的文档中，它写道:

iterator       Container  Iterator used to iterate through a set.
const_iterator Container  Const iterator used to iterate through a set. (Iterator and const_iterator are the same type.)

另一方面，vc++ 2008 Express编译你的代码时不会抱怨你在set::迭代器上调用了非const方法。

让我举一个更详细的例子。对于下面的结构:

struct Count{
    uint32_t c;

    Count(uint32_t i=0):c(i){}

    uint32_t getCount(){
        return c;
    }

    uint32_t add(const Count& count){
        uint32_t total = c + count.getCount();
        return total;
    }
};

如你所见，IDE(CLion)将给出提示，非const函数“getCount”在const对象上被调用。在该方法中，add count被声明为const对象，但getCount方法不是const方法，因此count.getCount()可能会改变count中的成员。

编译错误如下(核心消息在我的编译器):

error: passing 'const xy_stl::Count' as 'this' argument discards qualifiers [-fpermissive]

要解决上述问题，您可以:

改变uint32_t getCount()方法{…} to uint32_t getCount() const{…}。因此count. getcount()不会改变count中的成员。

or

修改uint32_t add(const count & count){…} to uint32_t add(count & count){…}。所以count并不关心其中成员的变化。

至于你的问题，std::set中的对象被存储为const StudentT，但方法getId和getName不是const，所以你给出了上述错误。

你也可以看到这个问题“const”在类的函数声明中的最后含义?更多细节。

不修改类实例的成员函数应该声明为const:

int getId() const {
    return id;
}
string getName() const {
    return name;
}

当你看到"discards qualifiers"时，它指的是const或volatile。