clang version 5.0.0 (trunk 305664) Target: x86_64-unknown-linux-gnu
以下代码成功编译：
template <int... A>
void f() {
([](auto) {
if constexpr (A == 0)
return 42;
else
return 3.14;
}(0),...);
}
int main() {
f<0,1>();
}
……但是这个没有：
template <int... A>
void f() {
([](auto...) { // Variadic lambda
if constexpr (A == 0)
return 42;
else
return 3.14;
}(),...); // No argument
}
int main() {
f<0,1>();
}
…屈服：
<source>:7:13: error: 'auto' in return type deduced as 'double' here but deduced as 'int' in earlier return statement
return 3.14;
^
<source>:3:6: note: in instantiation of function template specialization 'f()::(anonymous class)::operator()<>' requested here
([](auto...) { // Variadic lambda
^
<source>:12:5: note: in instantiation of function template specialization 'f<0,1>' requested here
f<0,1>();
^
我不希望在空参数包和伪参数之间有不同的行为.
是否存在这种差异的原因,或者这是编译器错误？
解决方法
我相信这是一个铿锵的错误.
[dcl.spec.auto]中的规则强调我的：
If the declared return type of the function contains a placeholder type,the return type of the function is deduced from non-discarded
returnstatements,if any,in the body of the function ([stmt.if]).[…]
If a function with a declared return type that contains a placeholder type has multiple non-discarded
returnstatements,the return type is deduced for each such return statement. If the type deduced is not the same in each deduction,the program is ill-formed.
lambda中的一个或另一个return语句被丢弃(如果constexpr被称为废弃语句,则取非分支),只留下一个非废弃的return语句,因此lambda的返回类型应该简单地从中推断出来.一个遗留下来.
此外,clang还可以这样：
template <int A>
void f() {
[](auto...) {
if constexpr (A == 0)
return 42;
else
return 3.14;
}();
}
int main() {
f<0>();
f<1>();
}
所以这可能与lambdas在pack表达式中的工作方式有一些不良的交互.
