[英]How to expand into initializer list with fold expressions?
I would like to insert as many zeros into a vector as there are arguments to a variadic templated function (ie number of arguments zeros into a vector).我想在向量中插入尽可能多的零,因为 arguments 到可变参数模板化的 function (即向量中的 arguments 零的数量)。 I am trying to use fold expressions to achieve this and it works when using
(vec.push_back(zeroHelper(args)), ...);
我正在尝试使用折叠表达式来实现这一点,并且在使用
(vec.push_back(zeroHelper(args)), ...);
. .
What I do not understand: Why does it not work when I try to initialize the vector directly by "unfolding" into an initializer list like follows:我不明白的是:为什么当我尝试通过“展开”到初始化器列表中直接初始化向量时它不起作用,如下所示:
std::vector<int> vec = { (zeroHelper(args), ...) };
? ?
Full source code:完整源代码:
template <typename T>
T zeroHelper (T a) { return T{0}; }
template<typename ...Args>
void printer(Args&&... args) {
std::vector<int> vec; // = { (zeroHelper(args), ...) };
(vec.push_back(zeroHelper(args)), ...);
for (auto i : vec) {
std::cout << i << '\n';
}
}
int main()
{
printer(1, 2, 3, 4);
return 0;
}
And here's the source on OnlineGDB.这是 OnlineGDB 上的源代码。
Expected output:预期 output:
0
0
00
00
00
Output with the initializer list approach: Output 与初始化列表方法:
0
0
Why?为什么?
Because in因为在
std::vector<int> vec = { (zeroHelper(args), ...) };
the parentheses return only one element, the last one;括号只返回一个元素,最后一个; the comma operator discard the precedings.
逗号运算符丢弃前面的内容。
You should write你应该写
std::vector<int> vec = { zeroHelper(args) ... };
to maintains all elements.维护所有元素。
In在
(vec.push_back(zeroHelper(args)), ...);
the parentheses activate the folding, the comma discards all elements except the last one but the push_back()
operation is applied to vec
for every args...
elements.括号激活折叠,逗号丢弃除最后一个之外的所有元素,但
push_back()
操作适用于每个args...
元素的vec
。
Observe also that, using the discarding propriety of the comma operator, you don't need zeroHelper()
at all还要注意,使用逗号运算符的丢弃特性,您根本不需要
zeroHelper()
You can write你可以写
std::vector<int> vec { ((void)args, 0)... };
or或者
(vec.push_back(((void)args, 0)), ...);
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