CODEWITHSUNDEEP
c++visual-c++struct
Anwar Mohamed
Anwar Mohamed

Reputation: 653

how to copy struct contents to another array struct

I have this c++ struct

struct PACKET
{
    BOOL isTCPPacket;
    BOOL isUDPPacket;
    BOOL isICMPPacket;
    BOOL isIGMPPacket;
    BOOL isARPPacket;
    BOOL isIPPacket;

    struct PETHER_HEADER
    {
        string  DestinationHost;
        string  SourceHost;
        struct PROTOCOL_TYPE
        {
            string  Name;
            WORD    Identifier;
        } ProtocolType;
    } EthernetHeader;
};

and i have

PACKET* Packet;
PACKET* Packets[6];

how can i copy contents of Packet into Packets[3] for example, knowing that the Packet contents will vary for each array in Packets[INDEX]

I have tried memcpy as

memcpy((void*)&Packets[i],(void*)&Packet,sizeof(PACKET));

with no luck

Upvotes: 0

Views: 2949

Answers (3)

Ajay
Ajay

Reputation: 18451

Ah! Just one thing:

PACKET* Packets[6];

is different than:

PACKET (*Packets)[6];

Former is an array of pointer (i.e. 6-pointers), but latter is a pointer to hold address-of a PACKET[6] array!

In either case, you must allocate memory for your Packet variable. memcpy or assignment will not work, unless some heap or stack memory is allocated.

Upvotes: 0

AgA
AgA

Reputation: 2126

Hope Packet contains valid object. Also Packets[i] point to allocated objects.

Do this: Packet[i] = Packet;

As string objects store pointers in their objects and can only be duplicated using assignment operator.

Upvotes: 0

Luchian Grigore
Luchian Grigore

Reputation: 258678

How about Packets[3] = Packet;? This does exactly what you asked (copies the content of Packet, which is a pointer, into Packets[3]).

If you want to copy the values, you can use assignment as well: *(Packets[3]) = *Packet;, assuming the pointers are valid. The compiler generated assignment operator for PACKET should work just fine.

I'd avoid memcpy in this case, since some of the members are std::string - you're bound to run into trouble.

Upvotes: 5

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