我的教授希望我只使用指针，任何函数都不允许下标 我想让用户将两个大小和元素输入到两个单独的数组中，然后创建一个联合函数，它找到所有相同的元素（在任何一个数组中都没有重复的值）
我想要做的就是将我的数组元素从第一个数组移到union数组中，但是当我执行程序时，我只是得到随机数
void get_union(short *set1,short size1,short *set2,short size2,short *union_array,short size_union) // look at call statement to assist incompleting this statement
{
short *end1=(set1+size1), //*end2=(set1+size1+size2);
for( ;set1<end1;set1++)
{
union_array=set1;
cout<<"Union array value number "<<count++<<" "<<*union_array++<<endl;
}
}
我也试过memcpy，但我以前从未使用它，特别是没有使用指针
memcpy(union_array,set1,sizeof(union_array));
这是我的前两个函数，它们允许用户输入数据，然后显示它。
void inputData(short *data, short size) // function to enter data into the array
{
short count=1;
short *end=(data+size);
for( ;data<end;data++)
{
cout<<"Enter Number "<<count++<<" : ";
cin>>*data;
cout<<endl;
}
}
和
void displayData(short *data, short size) // function to display data in an array
{
short count;
short *end=(data+size);
for( ;data<end;data++)
{
cout<<"Number "<<count++<<" : "<<*data<<endl;
}
}
这是我运行程序时得到的结果。整个过程贯穿但联合数组得到随机数
enter the number of values to store in the data set 1
or zero to terminate the program
3
Enter Number 1 : 2
Enter Number 2 : 4
Enter Number 3 : 6
there are 3 values in the array set1
Number 0 : 2
Number 1 : 4
Number 2 : 6
enter the number of values to store in the data set 2
3
Enter Number 1 : 10
Enter Number 2 : 11
Enter Number 3 : 12
there are 3 values in the array set2
Number 0 : 10
Number 1 : 11
Number 2 : 12
Union array value number 1 -5245
the union array contains 0 values
the intersection array contains -1 values
答案 0 :(得分：1)
我不确定你的问题，但让我试着帮助你：
我想要做的就是将我的数组元素从第一个数组移到union数组中，但是当我执行程序时，我只是得到随机数
using namespace std;
void get_union(short *set1,short size1,short *union_array,short size_union) {
for (; size1; --size1) {
*union_array++ = *set1++;
}
}
int main () {
short set1[] = {1, 0, 15, 35, 200, 12};
size_t size = sizeof(set1) / sizeof(*set1);
short union_arr[size];
get_union(set1, size, union_arr, size);
for (size_t i = 0; i < size; i++) {
cout << union_arr[i] << " ";
}
cout << endl;
return 0;
}
创建一个union函数，它找到所有相同的元素（在任何一个数组中都没有重复的值）
size_t get_union(short *set1, unsigned size1, short *set2, unsigned size2, short *union_array, unsigned size_union) {
size_t count = 0;
// TODO:: Make sure that the arrays are already sort
// todo:: or use the following two commands:
qsort(set1, size1, sizeof(*set1), [](const void * a, const void * b) -> int{
return ( *(short*)a - *(short*)b );
});
qsort(set2, size2, sizeof(*set2), [](const void * a, const void * b) -> int{
return ( *(short*)a - *(short*)b );
});
while (size1 && size2) {
if (*set1 > *set2) {
++set2;
--size2;
} else if (*set1 < *set2) {
++set1;
--size1;
} else {
*union_array++ = *set1++;
--size1;
++set2;
--size2;
++count;
}
}
return count;
}
int main () {
short set1[] = {1, 0, 15, 35, 200, 12};
short set2[] = {50, 0, 15, 0, 200, 12};
size_t size = sizeof(set1) / sizeof(*set1);
short union_arr[size];
size_t count;
count = get_union(set1, size, set2, size, union_arr, size);
cout << count << endl;
for (size_t i = 0; i < count; i++) {
cout << union_arr[i] << " ";
}
cout << endl;
return 0;
}
答案 1 :(得分：1)
他想从我们这里写一个函数来比较两个数组（我拥有的set1和set2）并将所有数字都放在其中。所以如果第一个数组中有0个元素而另一个中有5个元素，那么union数组应该有5个元素
size_t get_unique_union(short *arr, size_t size, short *target) {
size_t target_bigger = 0;
short *curr, *curr_test;
//size_t dup = 0; (1)
if (!size) {
return target_bigger;
}
curr = arr + 1; // Current place in the array.
curr_test = curr; // Current place with the offset of the duplicate elements.
while (curr_test < arr + size) {
while (curr_test < arr + size && *arr == *curr_test) {
curr_test++;
//dup++; // | (1) Update global size. see comment.
}
*curr++ = *curr_test++;
}
size -= curr_test - curr; // Update global size | (1) size -= dup;
if (curr == curr_test) { // If it is a unique appearance (If there were no founded duplicates).
*target = *arr; // Set target in the current place the appearance.
target_bigger = 1; // Mention that in the next recursive calling,
// it will be called from the next place in the unique array.
}
for (size_t i = 0; i < size; i++) { // Display the current checked array (Without the doubles of the current appearance).
cout << arr[i] << " ";
}
cout << endl;
return target_bigger + get_unique_union(arr + 1, size - 1, target + target_bigger); // Recursive call with the next element in the array.
}
size_t get_union(short *set1, unsigned size1, short *set2, unsigned size2, short *uniq_arr) {
size_t global_size = size1 + size2;
size_t uniq_size = 0;
short union_array[global_size];
for (size_t i = 0; i < size1; i++) {
union_array[i] = set1[i];
}
for (size_t i = 0; i < size2; i++) {
union_array[i + size1] = set2[i];
}
for (size_t i = 0; i < global_size; i++) {
cout << union_array[i] << " ";
}
cout << endl;
return get_unique_union(union_array, global_size, uniq_arr);
}
int main () {
short set1[] = {12, 0, 2, 1, 12, 12, 6, 8};
short set2[] = {3, 0, 300, 12, 12, 12};
size_t size1 = sizeof(set1) / sizeof(*set1);
size_t size2 = sizeof(set2) / sizeof(*set2);
short union_arr[size1 + size2];
size_t count;
count = get_union(set1, size1, set2, size2, union_arr);
cout << "Results:" << endl;
cout << "Count: " << count << endl;
cout << "Arr: [";
for (size_t i = 0; i < count; i++) {
cout << union_arr[i] << ((i < count - 1) ? ", " : "");
}
cout << "]" << endl;
return 0;
}
此代码甚至可以处理同一阵列中存在重复的情况。它将数组组合到同一个大数组中，删除重复项，并仅插入完全没有重复项的元素。 注意函数“get_unique_union”是递归的，并返回唯一元素的计数，而不是唯一的union数组的实际大小。 代码格式是C one，而不是C ++（它适用于C ++，已经用cpp文件编写并用g ++编译器编译。但在C ++中，它将使用向量编写，并且它将在唯一的联合数组中保存更多无用的位置）。