如何在C中的malloc分配的内存区域中存储指向偏移位置的指针？

Question

I'm trying to make a very simple Hashtable with libraries I got. The row that I try to make contains either C-string or int as key. Value can be of 4 different types. If it's struct I try to store a pointer in malloced memory area. However in my world this does not follow the same logic as non-pointer fields and I get:

error C2106: '=' : left operand must be l-value when I try to store pointer as value:

Here's the code

static void MakeRow(UtiHashtable_t *Hashtable, void *Key, void *Value, void *Row)
{
    int     KeySize, ValueSize;

    KeySize = sizeof(char) * Hashtable->KeyStringLength;
    ValueSize = GetValueSize(Hashtable);
    Row = malloc(KeySize + ValueSize);

    if (Hashtable->KeyType ==  MY_HASHTABLE_TYPE_STRING)  
        MyStrcpy((char *)Row, (char *)Key, Hashtable->KeyStringLength); 
    else if (Hashtable->KeyType ==  MY_HASHTABLE_TYPE_INT)
        ((int *)Row)[0] =  *(int *)Key;
    else
        MyAssert(0);

    switch (Hashtable->ValueType)
    {
    case MY_HASHTABLE_TYPE_STRING:
        MyStrcpy((char *)Row + KeySize, (char *)Value, Hashtable->ValueStringLength); 
        break;
    case MY_HASHTABLE_TYPE_INT: 
        *(int *)((char *)Row + KeySize) =  *(int *)Value;
        break;
    case MY_HASHTABLE_TYPE_DOUBLE: 
        *(double *)((char *)Row + KeySize) =  *(double *)Value;
        break;
    case MY_HASHTABLE_TYPE_STRUCT: 
        (int *)((char *)Row + KeySize) = (int *)Value;
        break;
    default:
        MyAssert(0);
    }
}

I know this is really basic and will get few down votes but besides the answer I would like to have explanation why this taking * away does not make it pointer.

The compiler allows:

(int *)Value = (int *)Key;

so why not:

(int *)((char *)Row + KeySize) = (int *)Value;

Thanks & BR -Matti

Answer 1

Using casts on an lvalue is deprecated.

Also, some systems don't allow int s or pointers to be placed in arbitrary memory locations due to alignment issues.

What you should do is to memcpy the pointer value to where you want it. And when you want to use the pointer you need to memcpy it back to a pointer variable.

Answer 2

您取消了左侧的取消引用*，因此分配不再有意义。