CODEWITHSUNDEEP
c++operator-overloadingtemplate-argument-deduction
Ranoiaetep
Ranoiaetep

Reputation: 6637

Specifying template parameters on overloaded operator()

struct foo{
    template<typename T>
    auto operator()(T arg) { return T{}; }
}

To use the operator(), I would call it like: foo()(1), which T would be deduced to int.

However, if I want to specify T as something else, like long, the only way that seems to work is, which kind of defeats the reason of using operator():

foo().operator()<long>(1);

Is there a better option?

I'm thinking about letting foo take a template parameter U and defaults U to void; T would defaults to U if U is not void. However I'd prefer to only change the operator() function. Maybe using a lambda instead?

Upvotes: 0

Views: 110

Answers (1)

Remy Lebeau
Remy Lebeau

Reputation: 595782

You can use the L or l suffix to specify that the integer literal 1 should be treated as a long instead of an int, eg:

foo()(1L)

Upvotes: 3

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