[英]Setting the size of array in a struct by passing as a const to a function - Non-type template argument is not a constant expression
User sets k
at running time.用户在运行时设置
k
。 This number will be constant
for the rest of the code.对于代码的 rest,此数字将保持
constant
。 I want to create a function to pass and create a struct that includes an array of size k with that number.我想创建一个 function 来传递并创建一个结构,该结构包含一个大小为 k 的数组,该数组具有该数字。 However, the compiler returns this error:
但是,编译器返回此错误:
Non-type template argument is not a constant expression
Any recommendation will be appreciated.任何建议将不胜感激。
The code is like:代码如下:
template <int N>
struct UL {
unsigned long ul [N];
};
void func(const int k){
UL<k> x; //problem is here
}
int main () {
int k;
cin >> k;
func(k);
return 0;
}
Variable k
is set at runtime, so when the code is being compiled, the compiler doesn't know what the value of k
is.变量
k
是在运行时设置的,所以在编译代码时,编译器不知道k
的值是什么。 You can't do this.你不能这样做。 But if you know what your
k
value is going to be and the range of values it can be is limited, you can create your struct for each possible value of k
and choose the matching class at runtime.但是,如果您知道您的
k
值将是什么并且它的值范围是有限的,您可以为每个可能的k
值创建您的结构,并在运行时选择匹配的 class。 Of course this is not probably what you want.当然,这可能不是您想要的。 You just need to be able to distinguish between what is known at compile-time and at runtime.
您只需要能够区分编译时和运行时已知的内容。 Templated literals (I hope I used the right name) are a compile-time feature of C++.
模板文字(我希望我使用了正确的名称)是 C++ 的编译时特性。
Templates are processed at compile-time only.模板仅在编译时处理。 You can't pass a run-time variable, like a function parameter, to a template.
您不能将运行时变量(例如 function 参数)传递给模板。 For what you are trying to do, you will have to use
std::vector
instead, eg:对于您要执行的操作,您必须改用
std::vector
,例如:
#include <vector>
struct UL {
std::vector<unsigned long> ul;
};
void func(const int k){
UL x;
x.ul.resize(k);
}
int main () {
int k;
cin >> k;
func(k);
return 0;
}
A fundamental principle about templates is that:关于模板的一个基本原则是:
Any template argument must be a quantity or value that can be determined at compile time .
任何模板参数都必须是可以在编译时确定的数量或值。
This has dramatic advantages on the runtime cost of template entities.这对模板实体的运行时成本具有显着优势。
But in your example, k
is not a compile time constant and you're using it as a template argument and so as a consequence of the above quoted statement you get the error.但是在您的示例中,
k
不是编译时间常数,并且您将其用作模板参数,因此由于上述引用的语句,您会收到错误。
To solve your problem you can use a std::vector
as shown below:要解决您的问题,您可以使用
std::vector
,如下所示:
#include <iostream>
#include <vector>
struct UL {
std::vector<unsigned long> ul;
//constructor
UL(int k): ul(k) //this creates vector ul of size k
{
std::cout<<"size of vector set to: "<<ul.size()<<std::endl;
}
};
void func(const int k){
UL x(k); //pass k as argument to constructor
}
int main () {
int k;
std::cin >> k;
func(k);
return 0;
}
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