memcpy with char * not working

Question

char recBuffer[8024];
char* temp = (char*)malloc(65536);
ZeroMemory(recBuffer, 8024);
ZeroMemory(temp, 65536);

bytesRead = recv(_socket, recBuffer, sizeof(recBuffer), 0);

memcpy(temp , &recBuffer, bytesRead );

memcpy doesn't work here. It copies a random char into temp. And if I play around with pointers I can get it to copy the first char of the data received. How do I do this properly?

I want the data recieved recBuffer to be copied into the temp buffer.

---

Edit: Working code from comment:

#include <string.h>
#include <stdio.h>

int main()
{
    char recBuffer[8024];
    char* temp = (char*)malloc(65536);

    strcpy(recBuffer, "Hello\n");

    int bytesRead = 7;
    memcpy(temp , &recBuffer, bytesRead );

    printf("%s\n", temp);

    return 0;
}

EDIT 2 Why this fails?:

#include <stdio.h> 

void Append(char* b, char data, int len)
{
memcpy(b , &data, len ); 
}

int main() { 
int bytesRead = 7; 
char recBuffer[8024]; 
char* temp = (char*)malloc(65536); 
strcpy(recBuffer, "Hello\n"); 
Append(temp, recBuffer, bytesRead);    
printf("%s\n", temp); 
    return 0; 
}

Answer 1

In example 2 char data needs to be changed to char const * data and passed to memcopy as data not &data. Bad question I know.. did not realize..

Answer 2

I have a feeling you are seeing the problem because you don't have

#include <stdlib.h>

I get the same result whether I use:

memcpy(temp , &recBuffer, bytesRead );

or

memcpy(temp , recBuffer, bytesRead );

Answer 3

Try

memcpy(temp, &recBuffer[0], bytesRead);

Answer 4

Change:

memcpy(temp , &recBuffer, bytesRead );

To:

memcpy(temp , recBuffer, bytesRead );