Using constexpr wrong from the c++11 standard or have I caught some compiler bug?

Question

I'm playing with the constexpr keyword and coded the following simple program: ``` #include using namespace std; template constexpr bool is_equal(T const* array1, T const* array2, size_t len) { return array1 == array2 || (len ? array1[len - 1] == array2[len - 1] && is_equal(array1, array2, len - 1) : true); } template constexpr bool is_equal(T const (&array1)[N1], T const (&array2)[N2]) { return N1 == N2 && is_equal(array1, array2, N1); } template constexpr size_t arraylength(T const (&array)[N]) { return N; } constexpr size_t stringlength(char const* str, size_t len=0) { return str ? (*str ? stringlength(str + 1, len + 1) : len) : 0; } constexpr size_t min(size_t one, size_t another) { return one < another ? one : another; } constexpr size_t min_length(char const* str1, char const* str2) { return min(stringlength(str1), stringlength(str2)); } template constexpr size_…

Accepted Answer

Update:

GCC chokes on your function `stringlength`. Try this:

```
constexpr size_t stringlength(char const* str, size_t i=0)
{
    return str ? (str[i] ? 1 + stringlength(str, i+1) : 0) : 0;
}
```

To GCC 4.6.3 this seems dubious:

```
static constexpr int const* c = &a[0];
static constexpr int const* d = &b[0];
```

Try this:

```
static constexpr int* c = &a[0];
static constexpr int* d = &b[0];
```

Old answer:

You code seems to be correct and with GCC 4.6.3 it also compiles.

```
int main()
{
    const char* a = "hallo";
    const char* b = "qallo";
    std::cout << is_equal(a,b,5) << std::endl;
    constexpr const char* c = "hallo";
    std::cout << is_equal(a,c,5) << std::endl;
}
```

Be careful, that the strings you are giving to your functions are constants.

See here what is allowed with `constexpr`.

Using constexpr wrong from the c++11 standard or have I caught some compiler bug?

Solution

Problem