字符串文字vs const char *函数重载

Question

I have a function that I want to work for const char* 's but it only works for string literals because they are given a special rule to be allowed to initialize arrays. The second overload, foo(const char*) will be preferred for both string literals and const char* s, but my template overload will not work for const char* s.

// Errors for const char*.
template<typename T, size_t n>
void foo(const T (&s)[n])
{
}

// Selected both times if both overloads are present.
void foo(const char*)
{
}

int main()
{
    foo("hello");
    const char* s = "hello";
    foo(s); // Error if foo(const char*) is absent.
}

Is there a way to allow const char* s to initialize arrays?

---

I've tried this:

template<typename T, size_t n>
void _foo(const T (&s)[n])
{
    std::cout << __PRETTY_FUNCTION__;
}

#define _expand(s) #s
#define expand(s) _expand(s)

void foo(const char* s)
{
    _foo(expand(s));
}

Answer 1

I think that

const char* s = "hello";

is a pointer to the string literal somewhere in read only memory, and compiler cannot deduce the array size, so it chooses second overload.

You can use

 const char s[] = "hello";

Answer 2

No, you can't initialize an array with a pointer. The string literal syntax is a special shorthand for a const char s[] and your working code roughly equivalent to

static const char s[]{'h','e','l','l','o','\0'};
foo(s);

So, you can pass arrays to your template function, including string literals, but you cannot pass pointers, including pointers to string literals.

On the other hand, arrays can decay to pointers, which is why both arrays and pointers can be passed to your second overload.