编译C ++循环的时间评估 [英] Compile time evaluation of a C++ loop
问题描述
我想写一个C ++循环,我想在编译时评估。这是因为我需要使用循环变量作为模板参数来初始化一个类。在非常简单的情况下,它看起来像:
I am trying to write a C++ loop that I want evaluated at compile-time. This is because I need to use the looping variable as a template argument to initialize a class. In the very simple case it would look something like:
for (unsigned int i = 1; i < 8; i++) {
Vector<i> foo;
// do something with foo
}
经过一些类似的StackOverflow 问题,我发现了一种递归编写静态循环的方法。现在我的代码看起来像这样:
After going through some similar StackOverflow questions, I found a way to write a static for loop recursively. Now my code looks something like this:
template <unsigned int i, unsigned int end>
struct static_for {
template <typename Lambda>
void operator()(const Lambda& function) const {
if (i < end) {
function(i);
static_for<start + 1, end>(function);
}
}
};
template <int N>
struct static_for<N, N> {
template <typename Lambda>
void operator()(const Lambda& function) const {
// This is just to avoid the unused variable warning - does nothing.
(void)function;
}
};
这是完美的,如预期。为了调用 static_for
循环,我可以简单地写:
This works perfectly and as expected. In order to call this static_for
loop I can simply write:
static_for<0, 8>()([](int size) {
// some code here.
});
但是,问题是我还是不能评价 i
在编译时。由于在我无法在lambda表达式中创建模板对象:
The problem, however, is that I can still not evaluate i
at compile time. As in I cannot create a templated object within the lambda expression:
static_for<0, 8>()([](int size) {
Vector<size> var; // does not compile as size is not a constexpr.
});
如何使这项工作?我的 static_for
循环有 i
变量作为模板参数,在编译时可用。我希望我的lambda表达式在编译时接收这个参数( size
),而不是作为运行时变量传递。这将如何工作?
How can I make this work? My static_for
loop has the i
variable as a template argument which is available to it at compile time. I would like my lambda expression to receive this argument (size
) also at compile time rather than being passed in as a runtime variable. How would this work? Conceptually this seems simple and useful as a more general language feature as well.
推荐答案
如果你可以使用C ++ 14,可以将 std :: integral_constant
传递给通用lambda,而不是 int
。 gcc.godbolt.org示例
If you can use C++14, you could pass an std::integral_constant
to a generic lambda instead of an int
. gcc.godbolt.org example
template <unsigned int i, unsigned int end>
struct static_for {
template <typename Lambda>
void operator()(const Lambda& function) const {
if (i < end) {
function(std::integral_constant<int, i>{});
static_for<start + 1, end>(function);
}
}
};
static_for<0, 8>()([](auto size) {
Vector<decltype(size)::value> var; // does not compile as size is not a constexpr.
});
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