Compile-time evaluation of constexpr functions in debug mode

I have been playing with the following code snippet to understand constexpr.

#include <stdlib.h>

///////////////////
bool runtimeIsPalindrome(const char* s, int len)
{
  if(len < 2)
    return true;
  else
    return s[0] == s[len-1] && runtimeIsPalindrome(&s[1], len-2);
}

///////////////////
constexpr bool compileTimeIsPalindrome(const char* s, int len)
{
    return len < 2 ? true : s[0] == s[len-1] && compileTimeIsPalindrome(&s[1], len-2);
}

///////////////////
int main()
{
    constexpr char c[] = "helloworlddlrowolleh";
    for(size_t nn=0;nn<1e8; ++nn) {
        // static_assert(compileTimeIsPalindrome(c, sizeof(c)-1 ), "Blah");
        // compileTimeIsPalindrome(c, sizeof(c)-1 );
        // runtimeIsPalindrome( c, sizeof(c)-1 );
    }
}

With the runtimeIsPalindrome version ...

clear; g++ -std=c++11 plaindrome.cpp; time ./a.out 
real    0m8.333s
user    0m8.322s
sys     0m0.005s

With the compileTimeIsPalindrome version ...

clear; g++ -std=c++11 plaindrome.cpp; time ./a.out
real    0m8.257s
user    0m8.247s
sys     0m0.004s

... but with the static_assert(compileTimeIsPalindrome version I actually appear to observe some compile time magic ...

clear; g++ -std=c++11 plaindrome.cpp; time ./a.out
real    0m0.265s
user    0m0.263s
sys     0m0.001s

Why does compile-time evaluation only work when I try the assertion in this example?

Note: Profiling with any optimisation seems pointless for this example, as it appears the compiler spots that the result is constant irrespective of the function called in the loop giving similar timings to the fastest profile time above.

constexpr doesn't guarantee compile time evaluation, unless used in a static_assert, template argument or any other place where the value has to be known at compile time by the language rules.

The fibonacci series f(n) = f(n - 1) + f(n - 2), f(0) = f(1) = 1 is a great example for that. On my machine with gcc, for n <= 10, this get's evaluated at compile time. For any other argument, the compiler is allowed to -- and indeed does -- decide that it is too computationally intensive and default to runtime evaluation.