How to insert, prepend new elements to char array - arrays

Question

I have problem with push to front new elements to char array. I have in my program a function which takes two arguments: a pointer to a char array, and a pointer to an int. This function fills an array with some data, and returns the count of elements in this array.
int function1( char* buffer , int* outSize);
buffer[1000]
int size;
result = function1(buffer , &size);
// after this a get some data in buffer and count how many data is in buffer. let's say 456 elements
// now a try push to front two new bytes as size this array
finallBuffer *char = new char[size+2];
memcpy(finallyBuffer, (char*)&size, 2); // here a get size in two bytes and put to finnalBuffer
// next I try just copy to finnalyBuffer received data from Buffer
strcat(finallyBuffer , buffer);
doSomething(finallyBuffer);
delete []finallyBuffer;
After this in finnalyBuffer I have only saved two bytes with size. Why don't I see data from buffer?
In finally I want to achieve new table with two new bytes in front. This two bytes is size old table. Let's say received data heve only 5 characters
char table[5] = {'a','b','b','c','t'};
So size is 5. 5 in two bytes is char size = {'5','0'};
Result should be looks like.
char table[5] = {'5','0','a','b','b','c','t'};

Answer 1

I am not sure to understand exactly what you want to do but i have tried to figure it. See comments in code bellow i hope it will help you.
int function1( char* buffer , int* outSize);
buffer[1000]
int size;
result = function1(buffer , &size);
// after this a get some data in buffer and count how many data is in buffer. let's say 456 elements
// now a try push to front two new bytes as size this array
char *finallBuffer = new char[size*sizeof(char)+2];
short ssize = (short)size;
memcpy(finallyBuffer, &ssize, 2); // here a get size in two bytes and put to finnalBuffer
// next I try just copy to finnalyBuffer received data from Buffer
/* NOTE: strcat is for null terminated string. If i understand what you want
to do you want store int in 4 bytes ?
I assume size is the actual size of the buffer
*/
//strcat(finallyBuffer , buffer);
memcpy(&finallyBuffer[2], buffer, sizeof(char)*size);
doSomething(finallyBuffer);
delete []finallyBuffer;

Answer 2

Those lines are wrong:
finallBuffer *char = new char[size+2];
memcpy(finallyBuffer, (char*)&size, 2);
They should be:
char * finalBuffer = new char[size+2];
memcopy(finallyBuffer, buffer, size);
But I really understand few things of what You are trying to do.
Edit:
Now I see what you are doing and Olivier gave you the answer. If you don't want (or can't) use std::vector for your buffer you can save the cost of reallocating the array, anyway.
Just start allocating for buffer 2 more bytes: char buffer[1002];.
Then to store values you can pass the pointer to the third element: char* ptr = buffer + 2;.
When you know the real size of the data, just put the bytes in buffer[0] and buffer[1].
A struct should be advised.