c++ - 是否需要在原始指针位置调用delete[]

Question

根据这个问题的答案:“How does delete[] "know" the size of the operand array?”

your allocator will keep track of how much memory you have allocated

和

How free knows how much memory to deallocate"

Yes, the memory block size is stored "somewhere" by malloc (normally in the block itself), so that's how free knows. However, new[]/delete[] is a different story. The latter basically work on top of malloc/free. new[] also stores the number of elements it created in the memory block (independently of malloc), so that later delete[] can retrieve and use that number to call the proper number of destructors

这是否意味着delete[] 与指针指向的位置无关？以下代码有效还是会导致内存泄漏？

void createArray(){
    char* someArray = new char[20];
    readData(someArray);
    //Is this delete still valid after moving the pointer one position?
    delete[] someArray;
}

char readData(char* &arr){
    char value = *arr;
    //Let it point to the next element
    arr += 1;
    return value;
}

Answer 1

是的，确实如此。如果您更改 new[]-ed 指针值，然后对其调用 delete[] 运算符，您将调用未定义的行为:

char* someArray = new char[20];
someArray++;
delete[] someArray; // undefined behavior

而是将原始值存储在不同的指针中并对其调用delete[]:

char* someArray = new char[20];
char* originalPointer = someArray;
someArray++; // changes the value but the originalPointer value remains the same
delete[] originalPointer; // OK