c++ - 合并两个数组后得到不正确的数组大小

Question

我获取了两个数组，然后将这两个数组合并到新创建的第三个数组中，它起作用了，但是当我输出数组的大小时，我得到的大小为“1”。我不明白为什么该数组的大小为“1”，即使其中有 5 个元素。

#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
    int arr1[] = { 1,2,3 };
    int arr2[] = { 9,4 };
    int size1 = sizeof(arr1) / sizeof(int);
    int size2 = sizeof(arr2) / sizeof(int);

    int *arr = new int[size1 + size2];

    //merging the two arrays by transferinng the elements into the third array
    for (int i = 0; i < size1; i++)
    {
        arr[i] = arr1[i];
    }
    for (int i = size1; i < (size1 + size2); i++)
    {
        arr[i] = arr2[i - size1];
    }

    //sorting the array
    sort(arr, arr + (size1 + size2));
    cout << endl;

    //finding the size of newly merged array
    int mergeSize = sizeof(arr) / sizeof(int);
    cout << "The size of the array is " << mergeSize << endl;   //why am I getting the size of the array as '1'
    return 0;
}

Answer 1

sizeof(arr) 为您提供指针 arr 的大小，它不取决于您为其分配的元素数量。

使用 std::array 可以避免这个问题。它没有 std::vector 的开销，而且比 C 风格的数组更容易使用。

int main()
{
    array<int, 3> arr1 = { 1, 2, 3 };
    array<int, 2> arr2 = { 9, 4 };

    array<int, arr1.size() + arr2.size()> arr;

    //merging the two arrays by transferinng the elements into the third array
    for (int i = 0; i < arr1.size(); i++)
    {
        arr[i] = arr1[i];
    }
    for (int i = 0; i < arr2.size(); i++)
    {
        arr[i + arr1.size()] = arr2[i];
    }

    //sorting the array
    sort(arr.begin(), arr .end());
    cout << endl;

    //finding the size of newly merged array
    int mergeSize = arr.size();
    cout << "The size of the array is " << mergeSize << endl;   //why am I getting the size of the array as '1'
    return 0;
}