Bitwise operators: How to check if source port is great than 32767 (0x7FFF)

Question

I have a program that reads a binary file of 20 bytes, modifies the data and then write on another binary file. I am trying to check if the source port is bigger than 32767 or(0x7FFF). If it is bigger then I must zero out the most significant two bits of it. I am only allowed to use bitwise operators. Anybody has any idea how could I do it? Thanks.

Data: 32 27 A9 49 03 8C BD 89 01 87 9B 8D 50 13 00 00 FF FF 00 00
Source port: 32 27 (16 bits)

void modify(const unsigned char oldData [], unsigned char newData []){

/*Accessing the souce port oldData[0] and oldData[1].*/


}

Answer 1

Just pick out the port value from the data. This depends on endianness. little or big.

You will need to know how your data is organized - if the 32 is the high or low part.

Assuming your data is little-endian, you can read it like this:

#include <endian.h>

void modify()
{
    uint16_t port = le16toh( *((uint16_t*)oldData) );
    if(port>32767)
        {/*whatever*/}

For the particular case of checking against >=32768 , which is >=0x8000, can test the high-bit as already commented:

if(oldData[1]&0x80) // [1] is highbyte if using little-endian data.
    {/*whatever*/}