I\'m having some trouble with constexpr
. The book C++ Primer shows a line of code:
constexpr int sz = size(); // only size() is a con
A constant expression function must be defined before its first use. See this paper, end of section 4.1.
A constexpr
function does NOT have to be defined before its first use, however the result of any call made prior to definition is not a constant expression.
Source: C++ Standard draft n4296, section 5.20:
A conditional-expression
e
is a core constant expression unless the evaluation ofe
, following the rules of the abstract machine, would evaluate one of the following expressions:
this
, except in aconstexpr
function or aconstexpr
constructor that is being evaluated as part ofe
;- an invocation of a function other than a
constexpr
constructor for a literal class, aconstexpr
function, or an implicit invocation of a trivial destructor [ Note: Overload resolution is applied as usual — end note ];- an invocation of an undefined
constexpr
function or an undefinedconstexpr
constructor;- ...
version from draft 3485 (section 5.19):
A conditional-expression is a core constant expression unless it involves one of the following as a potentially evaluated subexpression, but subexpressions of logical AND, logical OR, and conditional operations that are not evaluated are not considered [ Note: An overloaded operator invokes a function. — end note ]:
this
[ Note: when evaluating a constant expression, function invocation substitution replaces each occurrence ofthis
in aconstexpr
member function with a pointer to the class object. — end note ];- an invocation of a function other than a
constexpr
constructor for a literal class or aconstexpr
function [ Note: Overload resolution is applied as usual — end note ];- an invocation of an undefined
constexpr
function or an undefinedconstexpr
constructor- ...
The example int x2 = s. t();
in n2235 actually became valid due to the changes made prior to Standardization. However, constexpr int x2 = s. t();
remains an error.