[英]Deduce one dimension of 2D array at compile time
I have the following function: 我有以下功能:
template <int size>
double** writeArray(double input[size][2]) {
double** Points = new double*[size];
for (int i = 0; i < size; ++i) {
Points[i] = new double[2];
}
for (int i = 0; i < size; ++i) {
Points[i][0] = input[i][0];
Points[i][1] = input[i][1];
}
return Points;
}
which writes from a double[size][2]
array into a dynamically allocated double **
pointer. 它从
double[size][2]
数组写入动态分配的double **
指针。
Is there any way to deduce the size
automatically, so that I could use it like that: 有没有办法自动推断出
size
,以便我可以像这样使用它:
double** Points = writeArray(Test1);
instead of: 代替:
double** Points = writeArray<2>(Test1);
Yes! 是! You can actually deduce both dimensions at compile-time.
实际上,您可以在编译时推导出两个维度。 The idea is to write a template function that takes as its argument an array by reference .
我们的想法是编写一个模板函数,该函数通过引用将数组作为参数。 This prevents C++ from decaying the array type to a pointer type, which causes it to lose the array size.
这可以防止C ++将数组类型衰减为指针类型,从而导致它丢失数组大小。 Here's an example:
这是一个例子:
template <typename T, size_t M, size_t N>
void howBigAmI(T (&array)[M][N]) {
std::cout << "You are " << M << " x " << N << " in size." << std::endl;
}
You should be able to adapt this to fit your needs if you'd like. 如果您愿意,您应该能够根据自己的需要进行调整。
You should pass array by reference like: 您应该通过引用传递数组,如:
template <int size>
double** writeArray(const double (&input)[size][2])
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