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_301_3@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_301_3@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_301_3@结果如何相同？
解决方法
根据§14.5.3 / 9
The instantiation of a fold-expression produces:
(9.1) — ((E1 op E2) op · · · ) op EN for a unary left fold,
(9.2) — E1 op (· · · op (EN-1 op EN )) for a unary right fold,
(9.3) — (((E op E1) op E2) op · · · ) op EN for a binary left fold,and
(9.4) — E1 op (· · · op (EN-1 op (EN op E))) for a binary right fold
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.
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_301_3@
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.
