Why const char * function is prefered over template const char (&arr)[]

Asked May 08 '18 at 10:37

Active May 08 '18 at 10:43

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I have a Variant class that has string constructors. And I am trying to introduce variant type for string literals - to keep the pointer to source literal without copying it. If I remove the const char * constructor, than my new constructor is called:

Variant v = "asdf";
....

template <int size>
Variant(const char (&c_str) [size])

But I need the const char * constructor too - not all my stings are literals. And when there is such constructor, my new constructor becomes unreachable code. Can somebody explain me why? Visual Studio 2017

asked May 08 '18 at 10:37

Simo Simov

Actually this doesn't help in my case, because currently it's used mainly with constructor syntax. It may work if I use template instead of const char *. – Simo Simov May 08 '18 at 11:35
This works: template ::value > > Variant(ConstCharPtr sz) – Simo Simov May 08 '18 at 12:23

Why const char * function is prefered over template const char (&arr)[]

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