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Asked: 2026-05-18T09:34:50+00:00

I recently wrote a function template which takes a reference to a C-array: template

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I recently wrote a function template which takes a reference to a C-array:

template <class T, size_t N>
void foo(T(&c_array)[N]);

Assuming T is a char, the length of the C-string is N - 1 due to the null-terminator. I realized I should probably handle the edge-case where N == 0, because then N - 1 would be std::numeric_limits<std::size_t>::max().

So in order to avoid the chaos that might ensue in the rare case that someone passes a zero-length array to this function, I placed a check for N == 0.

However, to my surprise, it seems that a zero-length array is actually not even an array type – or at least, that’s what GCC seems to believe. In fact, a zero-length array doesn’t even bind to the above function signature, if a function with a pointer-type signature is available as a candidate.

Consider the following code:

template <class T, size_t N>
void foo(T(&array)[N])
{
    std::cout << "Array" << std::endl;
}

void foo(const void* p)
{
    std::cout << "Pointer" << std::endl;
}

int main(int argc, char** argv)
{
    char array1[10] = { };
    const char* pointer = 0;
    char array2[0] = { };

    foo(array1);
    foo(pointer);
    foo(array2);
}

With GCC 4.3.2, this outputs:

Array
Pointer
Pointer

Oddly, the zero-length array prefers to bind to the function that takes a pointer type. So, is this a bug in GCC, or is there some obscure reason mandated by the C++ standard why this behavior is necessary?

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1 Answer

  1. Editorial Team

    As arrays must have greater than zero length, if your compiler erroneously accepts a definition of a zero-sized array then you’re “safely” outside of the scope of the language standard. There’s no need for you to handle the edge case of N == 0.

    This is true in C++: 8.3.5 [dcl.array]: If the constant-expression (5.19) is present, it shall be an integral constant expression and its value shall be greater than zero.

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