我有以下代码来测试智能指针作为std::map
的关键,我在Mac和Linux上运行代码,但我观察到不同的输出,这是一个错误还是我做错了什么?
#include <iostream>
#include <memory>
#include <string>
#include <map>
using namespace std;
class Dog {
public:
typedef shared_ptr<Dog> sptr;
Dog(const string &name) : name_(name) { }
friend bool operator<(const Dog &lhs, const Dog &rhs) {
cout << "Dog::operator< called" << endl;
return lhs.name_ < rhs.name_;
}
friend bool operator<(const sptr &lhs, const sptr &rhs) {
cout << "Dog::operator< sptr called" << endl;
return lhs->name_ < rhs->name_;
}
private:
string name_;
};
void test_raw_object_as_map_key() {
cout << "raw object as map key ============== " << endl;
map<Dog, int> m;
m[Dog("A")] = 1;
m[Dog("B")] = 2;
m[Dog("C")] = 3;
m[Dog("A")] = 4;
cout << "map size: " << m.size() << endl;
}
void test_smart_pointer_as_map_key() {
cout << "smart pointer as map key ============== " << endl;
map<Dog::sptr, int> m;
m[make_shared<Dog>("A")] = 1;
m[make_shared<Dog>("B")] = 2;
m[make_shared<Dog>("C")] = 3;
m[make_shared<Dog>("A")] = 4;
cout << "map size: " << m.size() << endl;
}
int main(int argc, const char *argv[]) {
test_raw_object_as_map_key();
test_smart_pointer_as_map_key();
return 0;
}
在Mac上:
neevek@MAC$ g++ --version
Configured with: --prefix=/Applications/Xcode.app/Contents/Developer/usr --with-gxx-include-dir=/usr/include/c++/4.2.1
Apple LLVM version 5.0 (clang-500.2.79) (based on LLVM 3.3svn)
Target: x86_64-apple-darwin13.1.0
Thread model: posix
neevek@MAC$ ./a.out
raw object as map key ==============
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
map size: 3
smart pointer as map key ==============
Dog::operator< sptr called
Dog::operator< sptr called
Dog::operator< sptr called
Dog::operator< sptr called
Dog::operator< sptr called
Dog::operator< sptr called
Dog::operator< sptr called
Dog::operator< sptr called
Dog::operator< sptr called
map size: 3
在Linux上:
neevek@LINUX$ g++ --version
g++ (Ubuntu 4.8.2-19ubuntu1) 4.8.2
Copyright (C) 2013 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
neevek@LINUX$ ./a.out
raw object as map key ==============
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
Dog::operator< called
map size: 3
smart pointer as map key ==============
map size: 4
答案 0 :(得分:8)
GCC是正确的(在Mac上,因为g++
实际上是铿锵声):std::map
使用std::less<T>
比较密钥。这反过来在参数上调用operator <
,但查找首先在namespace std
中执行,因此它找到shared_ptr
的默认实现,比较内部指针。要完成这项工作,您必须为std::less
专门设置shared_ptr<Dog>
:
namespace std {
template<>
struct less<shared_ptr<Dog>> {
bool operator() (const shared_ptr<Dog>& lhs, const shared_ptr<Dog>& rhs) {
return *lhs < *rhs;
}
};
}
答案 1 :(得分:4)
Compare
的默认std::map
对象为std::less<Key>
,在您的情况下为std::shared_ptr<Dog>
。所以它查找std::less< std::shared_ptr<Dog> >
实现,它只是比较指针地址。
要指定Compare
对象,您可以自己定义并在map
class MyCompare {
public:
bool operator() (const sptr& l, const sptr& r) {
return *l < *r; // Invokes your Dog's operator <
}
};
然后
map<sptr, int, MyCompare> m;
m[make_shared<Dog>("A")] = 1;
m[make_shared<Dog>("B")] = 2;
m[make_shared<Dog>("C")] = 3;
m[make_shared<Dog>("A")] = 4;
cout << "map size: " << m.size() << endl;
输出map size: 3