问题
I'm trying the fold expression with clang 3.6 '--std=c++1z', but something I don't quite get. The function that I'm testing is:
auto minus = [](auto... args) { return (args - ...); };
...
std::cout << minus(10, 3, 2) << std::endl;
according to n4191, I'm expecting it expands as a left fold to
(10 - 3) - 2
which gives result 5, however, the result is 9, which seems to be a right fold expansion, i.e.
10 - (3 - 2)
Am I missing anything or mis-understand n4191? Thanks
回答1:
n4191 was revised by n4295. According to that, an expression of the form (e op ...) is a unary right fold, and that is expanded as: E1 op (... op (EN-1 op EN)), i.e. as a right fold expansion.
This does seem to be the reverse of what n4191 stated in terms of the fold direction. Clang 3.6 implements the n4295 proposal, as shown here.
... - args would be a unary left fold and expand in the direction you want.
来源：https://stackoverflow.com/questions/28805431/clang-3-6-fold-expression-left-right