CODEWITHSUNDEEP
c++
Kphysics
Kphysics

Reputation: 453

No Viable Overloaded +=, why?

addition for a user defined type was defined by operator+, but += does not resolve to it, why?

typedef std::array<std::uint64_t,4> myvec_t;

inline myvec_t operator+(myvec_t foo, myvec_t bar){return mod_add(foo,bar);}

myvec_t a,b; 
a+=b; // g++ says: error: no viable overloaded '+='

EDIT: several people sufficiently answered the narrow question, I will accept an answer which explains why did the standard allow people to overload '+='(a,b) to do something different from a=a+b ?

Upvotes: 2

Views: 1267

Answers (2)

Moia
Moia

Reputation: 2354

Because you didn't proved an overload for operator+=, you provided an overaload for operator+

operator+= overaload is done with

inline myvec_t& operator+=(myvec_t& foo, const myvec_t& bar)
{
   foo = mod_add(foo,bar);
   return foo;
}

Note that your operator+ implementation is very inefficent because it copies both operand and return a copy of the object while should make 1 only copy in total (the latter).

Upvotes: 0

Paul Evans
Paul Evans

Reputation: 27577

You declare/define an overload for operator+ and pass and return copies:

inline myvec_t operator+(myvec_t foo, myvec_t bar){return mod_add(foo,bar);}

you want to overload operator+= and pass and return references (assuming mod_add returns its result):

inline myvec_t& operator+=(myvec_t& foo, const myvec_t& bar)
{
    foo = mod_add(foo, bar);
    return foo;
}

Upvotes: 1

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