Why is 0 == ("abcde"+1) not a constant expression? - c++

Question

Why doesn't the following code compile?
// source.cpp
int main()
{
constexpr bool result = (0 == ("abcde"+1));
}
The compile command:
$ g++ -std=c++14 -c source.cpp
The output:
source.cpp: In function ‘int main()’:
source.cpp:4:32: error: ‘((((const char*)"abcde") + 1u) == 0u)’ is not a constant expression
constexpr bool result = (0 == ("abcde"+1));
~~~^~~~~~~~~~~~~~~
I'm using gcc6.4.

Answer 1

The restrictions on what can be used in a constant expression are defined mostly as a list of negatives. There's a bunch of things you're not allowed to evaluate ([expr.const]/2 in C++14) and certain things that values have to result in ([expr.const]/4 in C++14). This list changes from standard to standard, becoming more permissive with time.
In trying to evaluate:
constexpr bool result = (0 == ("abcde"+1));
there is nothing that we're not allowed to evaluate, and we don't have any results that we're not allowed to have. No undefined behavior, etc. It's a perfectly valid, if odd, expression. Just one that gcc 6.3 happens to disallow - which is a compiler bug. gcc 7+, clang 3.5+, msvc all compile it.
There seems to be a lot of confusion around this question, with many comments suggesting that since the value of a string literal like "abcde" is not known until runtime, you cannot do anything with such a pointer during constant evaluation. It's important to explain why this is not true.
Let's start with a declaration like:
constexpr char const* p = "abcde";
This pointer has some value. Let's say N. The crucial thing is - just about anything you can do to try to observe N during constant evaluation would be ill-formed. You cannot cast it to an integer to read the value. You cannot compare it to a different, unrelated string† (by way of [expr.rel]/4.3):
constexpr char const* q = "hello";
p > q; // ill-formed
p <= q; // ill-formed
p != q; // ok, false
We can say for sure that p != q because wherever it is they point, they are clearly different. But we cannot say which one goes first. Such a comparison is undefined behavior, and undefined behavior is disallowed in constant expressions.
You can really only compare to pointers within the same array:
constexpr char const* a = p + 1; // ok
constexpr char const* b = p + 17; // ill-formed
a > p; // ok, true
Wherever it is that p points to, we know that a points after it. But we don't need to know N to determine this.
As a result, the actual value N during constant evaluation is more or less immaterial.
"abcde" is... somewhere. "abcde"+1 points to one later than that, and has the value "bcde". Regardless of where it points, you can compare it to a null pointer (0 is a null pointer constant) and it is not a null pointer, hence that comparison evaluates as false.
This is a perfectly well-formed constant evaluation, which gcc 6.3 happens to reject.
†Although we simply state by fiat that std::less()(p, q) provides some value that gives a consistent total order at compile time and that it gives the same answer at runtime. Which is... an interesting conundrum.