“无法将 vector<int>* 转换为 int”是什么意思？答案

【问题标题】：What does "cannot Convert vector<int>* to int" mean?“无法将 vector<int>* 转换为 int”是什么意思？
【发布时间】：2016-05-17 14:43:11
【问题描述】：

我有错误：

错误 1 错误 C2664: 'int practiceMergeSort(int,int,std::vector>)' : 无法将参数 1 从 'std::vector> *' 转换为 'int'

我不知道错误可能是什么原因，或者错误告诉我什么。

编辑：

Function Prototypes
int practiceMergeSort(int, int, vector<int>);
int practiceMerge(vector<int>, int, int, int);

int main()
{
    int numOfItems;
    srand(int(time(0))); // Initialize the random number generator
    printf("Algorithm Comparison\n\n");

    printf("Selection Sort\n");
    //selectionSort();
    practiceSelectionSort(); // Fully functioning version

    //------------------------------------------------------------------
    cout << "\nMerge Sort\n";

    cout << "Enter the number of items to be sorted: ";
    cin >> numOfItems;

    vector<int> mergeArray(numOfItems);

    cout << "Value of numOfItems: " << numOfItems << "\n";

    cout << "Array values: \n";
    for (int x = 0; x < numOfItems; x++)
    {
        mergeArray[x] = rand();
        cout << mergeArray[x] << "\n";
    }

    practiceMergeSort(&mergeArray, 0, numOfItems);
    //------------------------------------------------------------------

    // Testing of the Array Filler
    //printf("\nArray Filler\n");
    //arrayFiller();

    cout << "\n\n";

    system("pause");

    return 0;
}

int practiceMergeSort(vector<int> mergeArray[], int low, int high)
{
    if (low < high) {
        int mid = (high + low) / 2;
        practiceMergeSort(mergeArray, low, mid);
        practiceMergeSort(mergeArray, mid + 1, high);
        practiceMerge(mergeArray, low, mid, high);
    }
    return 0;
}

int practiceMerge(vector<int> mergeArray[], int low, int mid, int high)
{
    vector<int> b[10000];
    int i = low, j = mid + 1, k = 0;

    while (i <= mid && j <= high) {
        if (mergeArray[i] <= mergeArray[j])
            b[k++] = mergeArray[i++];
        else
            b[k++] = mergeArray[j++];
        }
    while (i <= mid)
        b[k++] = mergeArray[i++];

    while (j <= high)
        b[k++] = mergeArray[j++];

    k--;
    while (k >= 0) {
        mergeArray[low + k] = b[k];
        k--;
    }
    return 0;
}

【问题讨论】：

我们不知道这意味着什么，因为您没有发布导致错误的代码。
贴出相关代码，不然没办法帮你。
请发帖minimal reproducible example。此外，该错误消息非常清楚。您是否正在尝试通过反复试验来学习 C++？如果是这样，请停下来，转而从一本好书或教程中学习。
当编译器告诉您以错误的顺序将参数传递给函数时，您怎么能声称自己有 C 经验，却又感到困惑？
错误告诉您您尝试将std::vector<int>* 传递给采用int 的参数。它甚至会告诉你哪个参数。

标签： c++ visual-studio vector error-handling mergesort

【解决方案1】：

您的问题是函数原型与实际的函数定义不匹配：

//Function Prototypes
int practiceMergeSort(int, int, vector<int>);
int practiceMerge(vector<int>, int, int, int);

// Functions
int practiceMergeSort(vector<int> mergeArray[], int low, int high)
{
//...
}

int practiceMerge(vector<int> mergeArray[], int low, int mid, int high)
{
//...
}

将原型更改为：

int practiceMergeSort(vector<int> mergeArray[], int low, int high);
int practiceMerge(vector<int> mergeArray[], int low, int mid, int high);

或者，如果您想继续使用带有未命名参数的原型：

int practiceMergeSort(vector<int> [], int, int);
int practiceMerge(vector<int> [], int, int, int);

这将使它编译。

【讨论】：