Unsigned int using ntohs

Question

I am getting some data from a DNS server and I am trying to convert a certain part into an unsigned int (which represents the refresh interval, expiration etc). By conversion, I mean from big endian to little endian. The problem is that the ntohs places only 2 bytes of data instead of 4.

memcpy(&number, data, 4);
printf("%x ",number);
number = ntohs(number);
printf("%x ",number);

Output:
b6fc0b78 780b

About the types:

:t data
char*
:t number
unsigned int

The weird thing is that even by trying using bit shifts to reconstruct the number, the values are pretty different: like 0, for example.

How could I get 780bfcb6 from b6fc0b78 into an unsigned int?

Answer 1

Notice that 780b is the byte-swap of 0b78, which is the low 16 bits of the number you passed in.

The s in ntohs is for short. It operates on uint16_t quantities. The argument and return type of ntohs are unsigned int only because it predates C89 prototypes; in K&R C it was impossible to express that a function took an argument shorter than int. (The latest POSIX specification seems to have decided that this historical wart is no longer necessary. I can persuade GCC and clang to warn about this mistake, but only by using -Wconversion, which is not turned on by -Wall, -Wextra, nor -pedantic.)

The function you are looking for is ntohl. It operates on uint32_t quantities, which is what you want. (The l is for long, but it dates to a time when 64-bit long was unheard of.)