-1
所以我正在處理這段代碼,並且我得到一個錯誤「無效的操作數到二進制表達式'int'到'int *'」我會告訴你我試過的測試方法,我試圖去上班,這給了我一個錯誤。Xcode運行到C++指針錯誤
//assume aptr has been declared in a base class
template<class T>
void SortableVector<T>::sortDescending() {
for(int x = 0; x < this->arraySize; x++)
cout << *(this->aptr + x) << ' '; // this line works just fine.
//everything beyond this line does not work
for (int i = 0; i < this->arraySize; i++)
{
for (int j = i+1; j < this->arraySize; j++)
{
if (*(this->aptr + i) < *(this->aptr + j))
{
int temporary = *(this->aptr+i);
*(this->aptr + i) = *(this->aptr + j)
*(this->aptr + j) = temporary; // here is where the errors appear
// also, it doesn't appear anywhere else
// e.g. on the line above it.
}
}
}
}
請,我真的很感激,如果有人能告訴我,如果我在這裏失去了一些東西。我試圖在Xcode中做到這一點,我會嘗試「強制運行」它,但我不知道該怎麼做,也不知道如何在Xcode中的功能
是我剛纔讀的頭說的xD謝謝了! :d –