Reputation: 15501
I am trying to assign a custom type as a key for std::map
. Here is the type which I am using as key:
struct Foo
{
Foo(std::string s) : foo_value(s){}
bool operator<(const Foo& foo1) { return foo_value < foo1.foo_value; }
bool operator>(const Foo& foo1) { return foo_value > foo1.foo_value; }
std::string foo_value;
};
When used with std::map
, I am getting the following error:
error C2678: binary '<' : no operator found which takes a left-hand operand of type 'const Foo' (or there is no acceptable conversion) c:\program files\microsoft visual studio 8\vc\include\functional 143
If I change the struct
to the one below, everything works:
struct Foo
{
Foo(std::string s) : foo_value(s) {}
friend bool operator<(const Foo& foo,const Foo& foo1) { return foo.foo_value < foo1.foo_value; }
friend bool operator>(const Foo& foo,const Foo& foo1) { return foo.foo_value > foo1.foo_value; }
std::string foo_value;
};
Nothing changed, except that the operator is overloaded as friend. Why does my first code not work?
Upvotes: 24
Views: 24011
Reputation: 9743
The other answers already solve your problem, but I'd like to offer an alternative solution. Since C++11 you can use a lambda expression instead of defining operator<
for your struct
. (operator>
is not needed for your map to work.) Providing a lambda expression to the constructor of a map has certain advantages:
struct
that you want to store in your map.struct
as key.operator<
differently and use it for a different purpose.As a result, you can keep your struct
as short as follows:
struct Foo {
Foo(std::string s) : foo_value(s) {}
std::string foo_value;
};
And your map can then be defined in the following way:
int main() {
auto comp = [](const Foo& f1, const Foo& f2) { return f1.foo_value < f2.foo_value; };
std::map<Foo, int, decltype(comp)> m({ {Foo("b"), 2}, {Foo("a"), 1} }, comp);
// Note: To create an empty map, use the next line instead of the previous one.
// std::map<Foo, int, decltype(comp)> m(comp);
for (auto const &kv : m)
std::cout << kv.first.foo_value << ": " << kv.second << std::endl;
return 0;
}
Output:
a: 1
b: 2
Upvotes: 0
Reputation: 399823
I suspect you need
bool operator<(const Foo& foo1) const;
Note the const
after the arguments, this is to make "your" (the left-hand side in the comparison) object constant.
The reason only a single operator is needed is that it is enough to implement the required ordering. To answer the abstract question "does a have to come before b?" it is enough to know whether a is less than b.
Upvotes: 33
Reputation: 79790
It's probably looking for const member operators (whatever the correct name is). This works (note const):
bool operator<(const Foo& foo1) const { return foo_value < foo1.foo_value;}
EDIT: deleted operator>
from my answer as it was not needed (copy/paste from question) but it was attracting comments :)
Note: I'm 100% sure that you need that const because I compiled the example.
Upvotes: 3