send() function returns more bytes that it was required c++

Question

I am doing a socket program and, after my server is connected with the device, I am trying to send a message to him. But the send() function returns a number of bytes greater th

Answer 1

The type of HEX_bufferMessage is pointer to CHAR. You are using sizeof( HEX_bufferMessage ) that is equal to 8 in your paltform.

Answer 2

You did not mean sizeof(HEX_bufferMessage). The type of HEX_bufferMessage is [presumably] char*, and the size of a char* is 8 bytes on your [64-bit] system.

Pass the number 7 instead, preferably using a constant, and get rid of that dynamic allocation if the value 7 really is fixed.

const int BUF_SIZE = 7;
char HEX_bufferMessage[BUF_SIZE];

// ...

retorno = send(sckSloMo, &HEX_bufferMessage[0], BUF_SIZE, 0); 

Answer 3

Assuming you've got

TYPE* myPointer;

then sizeof(myPointer) will give you the size of the pointer (i.e. 4 bytes on a 32 bit system, 8 on a 64 bit system), not the size of the array.

You want to do

const int bufferSize = 7;
HEX_bufferMessage = new CHAR[bufferSize]; 

...

retorno = send(sckSloMo, HEX_bufferMessage, sizeof(CHAR) * bufferSize, 0); 

Answer 4

It seems you assume that sizeof(HEX_buffer_message) yields the number of elements in HEX_buffer_message. It doesn't. It produces the size of the type of HEX_buffer_message which seems to be a 64 bit pointer yielding 8 rather than 7. There is no way to determine the size of an array allocated with new T[n] from the returned pointer. You'll need to use the n passed somehow.