CODEWITHSUNDEEP
c++buffervariable-assignmentarrays
user85917
user85917

Reputation: 1005

copy pointer value c++

I have a unsigned c++ char buffer 

unsigned char* pBuffer =  new unsigned char [1024];

I want to save the pBuffer pointer in the first few bytes of the newly allocated buffer by using an assignement rather than memcpy- ie 

*(unsigned char*) &pBuffer[0] = pBuffer;

I am not able to get the sytax right. Appreciate help with the correct syntax for doing this.

Visual Studio 2010

Upvotes: 0

Views: 671

Answers (3)

Mooing Duck
Mooing Duck

Reputation: 66922

I don't know why you would want to do this, but I can think of a few oddball reasons, so:

reinterpret_cast<unsigned char**>(pBuffer)[0] = pBuffer;

You want the compiler to reinterpret the pointer as a unsigned char**, and store the value of pBuffer in the 0-index slot.

You said you're indexing a "linked list" with multiple pBuffers. In that case, you should not use reinterpret_cast. Instead:

struct node {
     node* prev;
     std::unique_ptr<node> next;
     unsigned char buffer[1024];
     node(node* prev_) : prev(prev_) {}
};

std::unique_ptr<node> list(new node(nullptr)); //tada

Upvotes: 3

hochl
hochl

Reputation: 12930

Not sure why you are doing this, but this works with my g++ compiler:

*(unsigned char**)pBuffer=pBuffer;

Upvotes: 0

Some programmer dude
Some programmer dude

Reputation: 409176

How about:

*((unsigned char**) pBuffer) = pBuffer;

It casts the pBuffer pointer to a pointer to a pointer to a char, and then dereferences that to make it a pointer to a char that is then assigned the pBuffer pointer.

It might be easier to understand if we write it like this:

unsigned char **ppBuffer = &pBuffer;
*ppBuffer = pbuffer;

Upvotes: 1

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