如何将不同大小的数组传递给我的c ++函数?

问题描述：

我不确定这是否可行，但我愿意接受.

I'm not sure if this is possible but i'm open to ideas.

我有一个c ++函数.它会对用户定义为输入的数组进行一些计算，并将生成两个输出值.我们称之为"myfunction"，它看起来像这样:

I have a c++ function. It does some calculation to arrays that are being defined by users as inputs and will generate two output values. Let's call it "myfunction" and it looks something like this:

void myfunction(double array1 [18]，double array [9]，double array [2]，double * output1，double * output2)

void myfunction(double array1[18],double array[9],double array [2],double *output1, double *output2)

我想使此函数更通用，以便能够接受不同大小的数组.意思是，我希望该函数的用户能够定义其他大小的数组输入.

I want to make this function more generic so that it is able to take in arrays of different size. Meaning, I want users of this function to be able to define array inputs of other size.

例如，

void myfunction(double array1 [81]，double array [27]，double array [3]，double * output1，double * output2)

void myfunction(double array1[81],double array[27],double array [3],double *output1, double *output2)

从数学上讲，尽管数组大小，"myfunction"仍能够计算出准确的输出.我不想复制"myfunction"，因为用户可以将总共5组不同的数组大小定义为输入.

Mathematically, "myfunction" is able to calculate accurate outputs despite the array size. I don't want to make a duplicate of "myfunction" because there are a total of 5 different sets of array size that the user can define as inputs.

有什么想法吗?预先谢谢你

Any ideas? thank you in advance

答

您可以传递std::vector或使用template:

template <std::size_t N1, std::size_t N2, std::size_t N3> // and so on...
void work_with_arrays(double (&arr1)[N1], double (&arr2)[N2], double (&arr3)[N3]){
    std::cout << "First arr of a size: " << N1 << "\n";
    std::cout << "Second arr of a size: " << N2 << "\n";
    std::cout << "Third arr of a size: " << N3 << "\n";
};

int main() {
    double something[32];
    double fewer[13];
    double maybe_byte[8];

    work_with_arrays(something, fewer, maybe_byte);
}

代码输出:

First arr of a size: 32
Second arr of a size: 13
Third arr of a size: 8

说明:

应该知道类型T[](T是任何类型)的自变量会衰减到T*.如果我们要处理的是指向数组的指针，那么我们就没有关于其长度的信息，这是非常不幸的，因为固定长度数组的长度在编译时是已知的，并且在我们使用它们的任何地方都可以看到.

One should know that the argument of type T[] (T being any type) decays to T*. If we're dealing with a pointer to an array, we have no information about its length, which is quite unfortunate, given the fact that lengths of fixed-length arrays are known at compile time and could be visible everywhere we work with them.

还应该知道功能模板不是功能.它是用于创建函数的模板.对于以上示例中使用的每个N#组，都会生成一个供我们使用的函数.

One should also know that a function template is not a function. It is a template used to create functions. For every different set of N# used in the example above, there will be a function generated for our use.

该衰减问题的解决方法是什么?

我们应该将 reference 传递给T[]，而不是传递原始的T[].这样，类型不会衰减，并且数组的大小将是已知的.

Instead of passing a raw T[], we should pass a reference to T[]. That way the type does not decay and the size of the array will be known.

语法?

人们可能注意到T (&name)[some_size]看起来至少很奇怪.括号是必需的，因为简单的T &name[size]将被解释为T& name[size]，它是引用的数组，而不是对数组的引用.

One could notice that T (&name)[some_size] looks at least bizarre. The parentheses are required, since plain T &name[size] would be interpreted as T& name[size], which is an array of references, not a reference to an array.

结论:

为了能够检测传递参数数组的大小，我们不想将自己限制在一种或两种情况下-我们想将其全部覆盖，因此我们使用template生成具有所需值N#的函数.

Being able to detect the size of a passed-as-argument array, we don't want to limit ourselves with one or two cases - we want to cover them all, so we use a template to generate functions with N# of needed value.

如何将不同大小的数组传递给我的c ++函数?

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