template function issue

Hello everyone,


In the following code, how GetArrayLength(arr1) is matched to template function size_t GetArrayLength(const T(&arr)[size])? My confusion is how arr1 is matched to const T(&arr)[size]? I have tried to change const T(&arr)[size] to const T(arr)[size]) but it does not work.

Code:

---

template<size_t size, typename T>
size_t GetArrayLength(const T(&arr)[size])
{
        return size;
}

int main()
{
  char arr1[] = "Hello World";
  std::cout << GetArrayLength(arr1) << std::endl;

  return 0;
}

---

thanks in advance,
George

I would guess that if you pass a non-reference, the array decays to a pointer and the size argument is just ignored (for one dimension).

If you pass a reference to an array, its size needs to be known and can be deduced by the compiler.

Code:

---

#include <iostream>

void foo(const char c[10])
{
   
    std::cout << c << '\n';
}

void bar(const char (&c)[10])
{
    std::cout << c << '\n';
}

int main()
{
    char ten[] = "ABCDEFGHI";
    char twelve[] = "Hello world";
    foo(ten);
    foo(twelve); //number of characters doesn't matter
    bar(ten);
    //bar(twelve); //ERROR, expected char[10], got char[12]
}

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Edit: if your compiler has the __PRETTY_FUNCTION__ macro, you might see that the first signature is seen as void foo(const char* c)

The variable is just a placeholder which allows the important information -- the value of size -- to flow in. You could replace it with: T(&)[size] if you wanted:

Code:

---

template<size_t size, typename T>
size_t GetArrayLength(const T(&)[size])
{
        return size;
}

---

This ONLY WORKS for real arrays, i.e. where the length is actually known to the compiler.

Save yourself some trouble with this, though. You can use std::string and not have to worry about size.

Quote:

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Originally Posted by Elysia

Save yourself some trouble with this, though. You can use std::string and not have to worry about size.

---

This is applicable to far more than just strings.