Question

I have a file that can store up to 8 bytes of data.

I have a system(let’s call is Sys1) which can bring me 8 bytes from the file and save it in a buffer inside the ram.

I can use these 8 bytes and copy some stuff in it, and then use this system to say “OK I’m done with it, just send these data back to the file”.

I want to bring the 8 bytes to the buffer, copy an integer to it, which takes 4 bytes, and on the remaining 4 bytes copy another integer. I just want two integers in 8 bytes, which I believe can be done.

I have managed to copy 4 bytes only, which is just 1 integer and managed to send it back to the file. So I have something like this:

     char *data;
     sys1.bring_the_8_bytes(&data);
    //now the 8 bytes are in the buffer,
    // data points to the first byte of the 8 byte  sequence
    int firstInteger = 10;
    int secondInteger = 20;
    //the two integers 
    memcpy(data,&firstInteger,sizeof(int));
    //now the values of the first integer have been copied successfully.
    sys1.send_the_8_bytes_back();

now this works fine! However I’m not sure how to copy the first integer and then the second immediately.

I should actually know the address of the first byte that is exactly after the last byte that is taken by the first integer, and then use this address as a destination address in the memcpy call, is my thinking correct? if yes,

how can I achieve this?? How can I find out this address that I want?

thanks in advance

Answer 1

The address you are having problems calculating is simply sizeof(int) bytes past the data pointer you already have. You can just cast the char* to an int* and work with that seeing as you know you always have a valid 8-byte region to write to.

char *data;
sys1.bring_the_8_bytes(&data);
int *tmp = (int*)data;
tmp[0] = 10;
tmp[1] = 20;

Or you can just add to the char* (add sizeof(int)) and do a second memcpy.