c++ - c++溢出、归并排序、反演计算器

Question

我的代码是计算数组中的反转数。计算工作正常。问题是通常 unsigned int 的范围是 4,294,967,295 但它只接受 2,147,483,647 的范围，等效地，它不识别无符号数据类型，而是接受它作为有符号。

unsigned int _mergeSort(int arr[], int temp[], unsigned int left, unsigned int right);
unsigned int merge(int arr[], int temp[], unsigned int left, unsigned int mid, unsigned int right);

unsigned int mergeSort(int arr[], int array_size) {
    int *temp = (int *)malloc(sizeof(int)*array_size);
    return _mergeSort(arr, temp, 0, array_size - 1);
}

unsigned int _mergeSort(int arr[], int temp[], unsigned int left, unsigned int right) {
    unsigned int mid;
    unsigned int inv_count = 0;
    if (right > left)   {

        mid = (right + left) / 2;

        inv_count = _mergeSort(arr, temp, left, mid);

        inv_count += _mergeSort(arr, temp, mid + 1, right);

        inv_count += merge(arr, temp, left, mid + 1, right);
    }
    return inv_count;
}

unsigned int merge(int arr[], int temp[], unsigned int left, unsigned int mid, unsigned int right) {
    unsigned int i, j, k;
    unsigned int inv_count = 0;
    i = left;
    j = mid; 
    k = left; 
    while ((i <= mid - 1) && (j <= right)) {
        if (arr[i] <= arr[j]) {
            temp[k++] = arr[i++];
        } else {
            temp[k++] = arr[j++];
            inv_count = inv_count + (mid - i);
        }
    }
        while (i <= mid - 1)
            temp[k++] = arr[i++];
        while (j <= right)
            temp[k++] = arr[j++];
        for (i = left; i <= right; i++)
            arr[i] = temp[i];
    return inv_count;
}

int main(int argv, char** args) {
    const int size = 100000;
    int arr[size];
    unsigned int i = 0;
    int converted;

    string input;
    fstream dataFile("IntegerArray.txt", ios::in);
    if (dataFile) {
        getline(dataFile, input);
        while (dataFile) {
            converted = atoi(input.c_str());
            arr[i++] = converted;           
            getline(dataFile, input);
        }
        dataFile.close();
    } else {
        cout << "error" << endl;
    }

    printf("Number of inversions are %d \n", mergeSort(arr, i));
    getchar();
    return 0;
}

Answer 1

在您致电给printf()您时，请使用格式说明符，%d其表示有符号整数。传递的实际值是一个无符号整数，正确的格式说明符是%u. 修复printf()呼叫并再次测试。

这是一个参考printf()及其格式选项。