CODEWITHSUNDEEP
c++templatesc++14non-type
BeeOnRope
BeeOnRope

Reputation: 64895

Deducing the type of a pointer non-type template parameter

I have a function which accepts as a non-type template parameter a pointer O to an arbitrary object (with linkage), like so:

Foo foo;

...

template <typename T, T* O>
ftype* make_function() {
    // do something with O
}

To pass the pointer to an foo object of type Foo, I need to call it like this:

make_function<Foo, &foo>()

However, I'd like to simply call it like make_function<&foo>(), i.e., have the type Foo of the pointer deduced, as it could be in similar patterns (e.g., if I was passing foo as an argument).

Here's a more fleshed out example on Godbolt.

Upvotes: 1

Views: 52

Answers (1)

Jarod42
Jarod42

Reputation: 217135

C++17 allows

template <auto* O>
ftype* make_function() {
    // do something with O
}

Before that, you need indeed

template <typename T, T* p>

Macro can help to reduce verbosity at usage:

#define AUTO(p) decltype(p), p

// ...

make_function<AUTO(&foo)>();

Upvotes: 4

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