int main(int argc,char**)
{
constexpr int a = argc * 0;
(void)a;
constexpr int b = argc - argc;
(void)b;
return 0;
}
argc不是常量表达式,但在两种情况下,编译器仍然能够在编译时(即0)计算a和b的结果.
标准是否允许编译器在constexpr方面与g一样聪明？
解决方法
A conditional-expression is a core constant expression unless it involves one of the following as a potentially
evaluated subexpression […]
并且有几个子弹,包括以下关于左值到右值的转换：
an lvalue-to-rvalue conversion (4.1) unless it is applied to
a glvalue of integral or enumeration type that refers to a non-volatile const object with a preceding
initialization,initialized with a constant expression,ora glvalue of literal type that refers to a non-volatile object defined with constexpr,or that refers
to a sub-object of such an object,ora glvalue of literal type that refers to a non-volatile temporary object whose lifetime has not
ended,initialized with a constant expression;
有趣的是,gcc不接受以下代码(see it live)：
constexpr int a = argc * 2;
所以看起来gcc说我知道结果将为零,因此它执行常量折叠,并且不需要执行argc的左值到右值转换来确定结果.
不幸的是,我认为第5.19节中没有任何条款允许这种短路.这看起来非常类似于int a=1,is a || 1 a constant expression?中的情况,其中有一个错误报告,但gcc团队中没有人回复那个.我在该错误报告中添加了一个comment,表明这似乎是相关的.
There is a whole c++-delayed-folding branch on which some gcc developers are working,which will delay a number of optimizations and might fix this. It is important for other reasons,rejecting the code in this question is very low priority
常量表达式是否严格要求每个子表达式都是常量表达式？不,例如5.19它说:(强调我的)
A conditional-expression is a core constant expression unless it involves one of the following as a potentially
evaluated subexpression (3.2),but subexpressions of logical AND (5.14),logical OR (5.15),and conditional
(5.16) operations that are not evaluated are not considered […]
所以以下是一个常量表达式：
constexpr int a = false && argc * 0;
因为argc * 0未评估,因为&&评估从左到右和短路.
