Trying to brush up on my C++ and STL proficiency, running into a problem with std::map keyed by a structure I've defined. Relevant code:
typedef struct key_t {
int a;
int b;
bool operator==(const key_t& rhs)
{
return (a == rhs.a) && (b == rhs.b);
}
bool operator<(const key_t& rhs) //added the when I saw this error, didn't help
{
return a < rhs.a;
}
} key_t;
std::map<key_t, int> fooMap;
void func(void)
{
key_t key;
key.a = 1;
key.b = 2;
fooMap.insert(std::pair<key_t, int>(key, 100));
}
Error looks like this:
"/opt/[redacted]/include/functional", line 133: error: no operator "<" matches these operands
operand types are: const key_t < const key_t
detected during:
instantiation of "bool std::less<_Ty>::operator()(const _Ty &, const _Ty &) const [with _Ty=key_t]" at line 547 of "/opt/[redacted]/include/xtree"
instantiation of "std::_Tree<_Traits>::_Pairib std::_Tree<_Traits>::insert(const std::_Tree<_Traits>::value_type &) [with _Traits=std::_Tmap_traits<key_t, UI32, std::less<key_t>, std::allocator<std::pair<const key_t, UI32>>, false>]"
What am I doing wrong? Is it just flat-out awful/impossible to use structures as a map key? Or something else I'm overlooking?
This
bool operator<(const key_t& rhs)
needs to be a const method
bool operator<(const key_t& rhs) const
The two are different signatures, and std::less
looks for the latter. The latter, as a const method, implying that it won't modify the object. The former however without const may imply that a modification to this
may be performed.
In general its a good idea to have const
methods, even if you can cast in away, it implies a promise to the client that no modifications will take place.
For starters, the operators have to be const
. (And you don't need the ==
operator.)
And where did you learn to use a typedef for a struct
. There's no reason for it.
And finally, if you want both elements to participate as part of the key, you'll have to compare both of them:
struct Key
{
int a;
int b;
bool operator<( Key const& rhs ) const
{
return a < rhs.a
|| ( !(rhs.a < a) && b < rhs.b );
}
};
Otherwise, Key( 1, 2 )
and Key( 1, 3 )
will effectively be equal.
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