c++ - Why is const required for 'operator>' but not for 'operator

Monday, 9 October 2017

c++ - Why is const required for 'operator>' but not for 'operator

style="font-weight: bold;">

Answer

style="font-weight: bold;">

Answer

Consider this piece of
code:

#include
            

#include 
#include
            
#include 
using namespace
            std;

struct MyStruct
{
 int key;

            std::string stringValue;


 MyStruct(int k, const
            std::string& s) : key(k), stringValue(s) {}

 bool operator <
            (const MyStruct& other) {
 return (key < other.key);

            }
};

int main() {
 std::vector < MyStruct
            > vec;


 vec.push_back(MyStruct(2, "is"));

            vec.push_back(MyStruct(1, "this"));
 vec.push_back(MyStruct(4,
            "test"));
 vec.push_back(MyStruct(3, "a"));


            std::sort(vec.begin(), vec.end());

 for (const MyStruct& a :
            vec) {
 cout << a.key << ": " << a.stringValue <<
            endl;

            }

}

It
compiles fine and gives the output one would expect. But if I try to sort the structures
in descending order:

#include
            
#include 
#include
            
#include 
using namespace
            std;


struct MyStruct
{
 int
            key;
 std::string stringValue;

 MyStruct(int k, const
            std::string& s) : key(k), stringValue(s) {}

 bool operator >
            (const MyStruct& other) {
 return (key >
            other.key);

 }
};


int
            main() {
 std::vector < MyStruct > vec;


            vec.push_back(MyStruct(2, "is"));
 vec.push_back(MyStruct(1,
            "this"));
 vec.push_back(MyStruct(4, "test"));


            vec.push_back(MyStruct(3, "a"));

 std::sort(vec.begin(), vec.end(),
            greater());

 for (const MyStruct& a : vec)
            {
 cout << a.key << ": " << a.stringValue <<
            endl;

            }
}

This
gives me an error. rel="noreferrer">Here is the full
message:

/usr/include/c++/7.2.0/bits/stl_function.h: In instantiation of 'constexpr bool
std::greater<_Tp>::operator()(const _Tp&, const _Tp&) const [with _Tp =
MyStruct]':
/usr/include/c++/7.2.0/bits/stl_function.h:376:20: error: no
match for 'operator>' (operand types are 'const MyStruct' and 'const MyStruct') />{ return __x > __y;
}

It seems to be
because this function right here doesn't have a const
qualifier:

bool operator >
            (const MyStruct& other) {
 return (key >
            other.key);

}

If
I add it,

bool operator >
            (const MyStruct& other) const {
 return (key >
            other.key);
}

Then
everything is fine again. Why is this so? I'm not too familiar with operator
overloading, so I've just put it in memory that we need to add the
const but it's still weird why it works for
operator< without the
const.

class="post-text" itemprop="text">

class="normal">Answer

You get
different behaviors because you are in fact calling two different (overloaded) href="http://en.cppreference.com/w/cpp/algorithm/sort"
rel="noreferrer">sort functions.

In the first case you call the two parameter
std::sort, which uses operator<
directly. Since the iterators to your vector elements produce non-const references, it
can apply operator< just
fine.

In the second case, you are using the
three parameter version of std::sort. The one that accepts a
functor. You pass std::greater. And that functor has an
operator() declared as
follows:

constexpr bool
            operator()( const T& lhs, const T& rhs )
            const;

Note
the const references. It binds the elements it needs to compare to const references. So
your own operator> must be const correct as
well.

If you were to call
std::sort with href="http://en.cppreference.com/w/cpp/utility/functional/less"
rel="noreferrer">std::less, your
operator< will produce the same error, because it's not
const-correct.

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Monday, 9 October 2017

c++ - Why is const required for 'operator>' but not for 'operator

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