why is a const array not accessible from a constexpr function?

3
votes

i have a constexpr function named access, and i want to access one element from an array: 

char const*const foo="foo";
char const*const bar[10]={"bar"};

constexpr int access(char const* c) { return (foo == c); }     // this is working
constexpr int access(char const* c) { return (bar[0] == c); }  // this isn't
int access(char const* c) { return (bar[0] == c); }            // this is also working

i get the error:

error: the value of 'al' is not usable in a constant expression

why can't i access one of the elements from access? or better how do i do it, if it is even possible?

1
c++c++11constexpr
Does it work? or not? – Grijesh Chauhan
@MikeSeymour ok I don't know C++ Should I delete comment? – Grijesh Chauhan
@GrijeshChauhan sorry, no it's not working. my approach is working without constexpr. – user1810087

1 Answers

8
votes

The array needs to be declared constexpr, not just const.

constexpr char const* bar[10]={"bar"};

Without that, the expression bar[0] performs an lvalue-to-rvalue conversion in order to dereference the array. This disqualifies it from being a constant expression, unless the array is constexpr, according to C++11 5.19/2, ninth bullet:

an lvalue-to-rvalue conversion unless it is applied to

  • a glvalue of literal type that refers to a non-volatile object defined with constexpr

(and a couple of other exceptions which don't apply here).