lambda 函数可以递归吗?
可能的重复:
c++0x 中的递归 lambda 函数
这是一个普通的旧递归函数:
Here is a plain old recursive function:
int fak(int n)
{
return (n <= 1) ? 1 : n * fak(n - 1);
}
我将如何编写像 lambda 函数这样的递归函数?
How would I write such a recursive function as a lambda function?
[](int n) { return (n <= 1) ? 1 : n * operator()(n - 1); }
// error: operator() not defined
[](int n) { return (n <= 1) ? 1 : n * (*this)(n - 1); }
// error: this wasn't captured for this lambda function
是否有任何表达式表示当前的 lambda，以便它可以递归地调用自己?
Is there any expression that denotes the current lambda so it can call itself recursively?
推荐答案
是的，他们可以.您可以将其存储在变量中并引用该变量(尽管您不能将该变量的类型声明为 auto，但您必须使用 std::function 对象代替).例如:
Yes, they can. You can store it in a variable and reference that variable (although you cannot declare the type of that variable as auto, you would have to use an std::function object instead). For instance:
std::function<int (int)> factorial = [&] (int i)
{
return (i == 1) ? 1 : i * factorial(i - 1);
};
否则，不，您不能从 lambda 的主体内部引用 this 指针.
Otherwise, no, you cannot refer the this pointer from inside the body of the lambda.
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