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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?

user1810087
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1 Answers1

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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).

Community
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Mike Seymour
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    Now the interesting question is, why does it work for the non-array version even if that is just `const` but not `constexpr`? – us2012 Sep 18 '13 at 17:31
  • @us2012: I'm not sure of the exact rules; I think you can use `const` values (including pointers like `foo`), but not the results of expressions (like `bar[0]`) unless all the subexpressions are `constexpr`. Or something. – Mike Seymour Sep 18 '13 at 17:33
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    @us2012: I think I've found the relevant rule now. – Mike Seymour Sep 18 '13 at 17:40
  • So then how the array element is accessed - this is so hilarious. – AnArrayOfFunctions Mar 20 '16 at 10:46