我正在看一下C 17
fold expressions,我想知道为什么以下程序输出
4 5 6 4 5 6@H_403_4@对于两个for_each调用
template<typename F,typename... T>
void for_each1(F fun,T&&... args)
{
(fun (std::forward<T>(args)),...);
}
template<typename F,typename... T>
void for_each2(F fun,T&&... args)
{
(...,fun (std::forward<T>(args)));
}
int main()
{
for_each1([](auto i) { std::cout << i << std::endl; },4,5,6);
std::cout << "-" << std::endl;
for_each2([](auto i) { std::cout << i << std::endl; },6);
}
@H_403_4@Live Example
@H_403_4@我认为第二个折叠表达式是为了以相反的顺序输出数字
6 5 4@H_403_4@结果如何相同？
解决方法
根据§14.5.3 / 9
@H_403_4@The instantiation of a fold-expression produces: @H_403_4@(9.1) — ((E1 op E2) op · · · ) op EN for a unary left fold, @H_403_4@(9.2) — E1 op (· · · op (EN-1 op EN )) for a unary right fold, @H_403_4@(9.3) — (((E op E1) op E2) op · · · ) op EN for a binary left fold,and @H_403_4@(9.4) — E1 op (· · · op (EN-1 op (EN op E))) for a binary right fold @H_403_4@In each case,op is the fold-operator,N is the number of elements in the pack expansion parameters,and each Ei is generated by instantiating the pattern and replacing each pack expansion parameter with its ith element.@H_403_4@在上面的代码中,它们是一元折叠表达式,它们的扩展是
template<typename F,T&&... args) {
// Unary right fold (fun(args_0),(fun(args_1),(fun(args_2),...)))
(fun (std::forward<T>(args)),T&&... args) {
// Unary left fold ((fun(args_0),fun(args_1)),fun(args_2)),...
(...,fun (std::forward<T>(args)));
}
@H_403_4@所以表达式具有与comma operator定义的相同的评估顺序,因此输出相同.
@H_403_4@信用：感谢我的朋友Marco首先提出了原来的问题,并让我有机会解决这个潜在的误导性问题.
