你能看看我所面临的：http ://sdrv.ms/WgafvN
另一个截图：http ://sdrv.ms/UZIp6H
我的函数的文本是：
bool print_all_points(POINT** pointer)
{
if (pointer == NULL||is_array_empty(pointer))
{
cout << "The array of points is empty." << endl << endl;
return false;
}
else
{
int n = _msize(pointer)/sizeof(pointer[0]);
cout << "The list of points: " << endl<< endl;
cout << "id (x, y)" << endl;
cout << "----------" << endl;
for (int i = 0; i < n; i++)
{
cout << (*pointer[i]).id << " (" << (*pointer[i]).x << ", " << (*pointer[i]).y << ")" << endl;
}
}
return true;
}
该函数预计会打印出数组中的所有点。我的问题是它完美地打印了 3 个点的数组而不是 4 个点的数组。在第四点，它咬住了灰尘。
我不明白是什么问题。从图中可以看出： 1. 数组的所有 4 个元素都存在。2.正确判断有4个。
问题是什么？你能在这里踢我一脚吗？
稍后添加。
调用它的函数：
POINT** new_point(POINT** pointer, int occup)
{
char x;
char y;
system("cls");
cout << "INPUT A NEW POINT" << endl << endl;
cout << "Input x: ";
cin >> x;
cout << "Input y: ";
cin >> y;
size_t m;
if (pointer != NULL)
{
m = _msize(pointer);
}
POINT * tmp_point = new POINT();
(*tmp_point).id = occup;
(*tmp_point).x = x-48;
(*tmp_point).y = y-48;
POINT** pn = new POINT * [occup];
int necessary_memory = occup * 4; // ???? 4 is the size of a pointer.
if (occup !=1)
{
memcpy(pn, pointer, necessary_memory);
}
POINT ** tmp = new POINT * [occup];
pn[occup - 1] = tmp_point;
memcpy(tmp, pn, occup * sizeof(POINT));
delete[] pn;
pn = tmp;
size_t n = _msize(pn);
cout << endl;
print_all_points(pn);
return pn;
}