C++ non-type template parameter const char*

Question

Let's say we have

template <const char*>
struct A{};

// static storage
const char a[] = "asd";
const char* p = "asd";

This instantiation

A<a>{};

is okay for compiler. And this is understandable - array a decays to pointer to first element. But if we instantiate A with p like this

A<p>{};

compiler gives an error:

error: non-type template argument of type 'char *' is not a constant expression

Why doesn't Standard allow to specify named variable of type const char* or just string literal "asd", which is btw lvalue itself, as a template argument?

http://stackoverflow.com/questions/3773378/passing-const-char-as-template-argument — David G, Jul 19 '15 at 18:24
I don't believe it's about lvalues as much as `p` doesn't have a precise memory location, but `a` does. — kirbyfan64sos, Jul 19 '15 at 18:27
This is no different from `template struct X{}; int p = 5; X
x;` — Kerrek SB, Jul 19 '15 at 19:08

Some programmer dude · Accepted Answer · 2015-07-19T18:38:07.500

The array a is a constant array of character, it's fully initialized at the time of compilation and also gets a know memory address by the compiler which is why it can decay to a pointer in a template.

But p is a pointer to a constant array of characters, but the pointer itself is not a compile-time constant, and can be changed to point to other strings, and it's is not initialized at time of compilation but either on time of linking (which happens after compilation) or when the program is loaded into memory. The address of p is known at time of compilation, but not the address of the string literal p points to.

To expand on the reason the address of the string literal is not known at time of compilation, it's because it's put in a special read-only segment by the compilers code-generator, and that read-only segment is then combined with the read-only segments from the other translation units when linking. This is why the final address of the string literal can't be known until the time of linking (at earliest).

thanks, makes sense. But why it doesn't work even if I use constexpr const char* p = "something"; ? — Bikineev, Jul 19 '15 at 21:11
@Bikineev Same issue really, the location of string literals can't be calculated at the time of compilation, when it's needed for templates. — Some programmer dude, Jul 20 '15 at 00:20

score 0 · Answer 2 · answered Jul 26 '17 at 23:10

0

It may help to understand more why A<p>{}; is not allowed if you imagine p is initialized by some function

const char* p = getP();

where getP() determines its result by some runtime data, for example, by reading a config file.

answered Jul 26 '17 at 23:10

bjhend

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C++ non-type template parameter const char*

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