C ++-传递未知大小的数组

问题描述

尝试传递以1开头的连续数字的int数组，但假设接收此数组的函数不知道其长度。尝试计算函数内部的长度时，它只会给我1，因为它仅在计算sizeof（arrayName）时会找到第一个元素。

Trying to pass an int array of consecutive numbers starting with 1 but assuming the function receiving this array does not know it's length. When trying to calculate the length inside the function it just gives me 1 since it only finds the first element when calculating sizeof(arrayName).

#include <iostream>
using namespace std;

int Sum(int intArray[]) {
    int n = sizeof(intArray) / sizeof(*intArray);
    cout << "Array size in function: " << n << endl;
    return n * (n + 1) / 2;
}

int main() {
    int anArray[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    int arraySum = Sum(anArray);

    cout << "Array size in main: " << sizeof(anArray) / sizeof(*anArray) << endl;
    cout << "Sum is: " << arraySum;
    int a;
    cin >> a;
    return 0;
}

推荐答案

您的函数正在使用指针诠释。当您将数组传递到指针时，所有大小信息都会丢失。但是您可以使用函数模板来实例化与正确的数组大小匹配的函数：

Your function is taking a pointer to int. All size information is lost as you pass the array into the pointer. But you can use a function template to instantiate functions that match the right array size:

template <size_t N>
int Sum(const int (&intArray)[N])
{
  cout << "Array size in function: " << N << endl;
  return std::accumulate(std::begin(intArray), std::end(intArray), 0);
}

此 Sum 函数将接受编译时已知的纯数组或大小。但是，在这些情况下使用 std :: array 更为合理，在以下情况下使用 std :: vector

This Sum function will accept plain arrays or size known at compile time. However, it makes more sense to use std::array for these cases, or std::vector for cases when the size is chosen at runtime.

请注意，调用 std :: accumulate 只是解决该问题的一个示例总和问题。它不需要 N 的知识，并且可以完全替换该功能。大小由 std :: begin 和 std :: end 照顾。您需要标题< numeric> 和< iterator> 标头，用于累积和开始/结束。

Note that the call to std::accumulate is just an example that solves the sum problem. It does not require knowledge of N, and could replace the function entirely. The size is taken care of by std::begin and std::end. You would need headers <numeric> and <iterator> for accumulate and begin/end respectively.

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