为什么在Socket Programming-C中将位移位8？

Question

I am studying the FreeBSH Developers Handbook and there I am seeing this code for Socket programming:

struct sockaddr_in sa;
sa.sin_addr.s_addr = htonl((((((192 << 8) | 43) << 8) | 244) << 8) | 18);

Why is the bitwise shift and or is used here?

I know what bitwise shift and xor is, I made a very small test program:

int c = 5;
printb(c);

int d = 5<<2;
printb(d);

int e = d | c;
printb(e);

which prints:

00000000000000000000000000000101
00000000000000000000000000010100
00000000000000000000000000010101

however I do not understand why for sa.sin_addr.s_addr I need to shift the ip and or it with the following value. Can someone explain?

Answer 1

It is creating the binary representation for the IP 192.43.244.18.

Let's analyse each operation one by one.

Start from the binary representation of those constants:

192 = 11000000
 43 = 00101011
244 = 11110100
 18 = 00010010

Innermost operation:

(192 << 8) =

= 11000000 << 8 =

= 1100000000000000

Next one:

(192 << 8) | 43 =

= 1100000000000000 |
          00101011 =

  1100000000101011

Next one:

((192 << 8) | 43) << 8) =

= 1100000000101011 << 8 =

= 110000000010101100000000

Next one:

(((192 << 8) | 43) << 8) | 244 =

= 110000000010101100000000 | 244 =

= 110000000010101100000000 |
                  11110100 =

= 110000000010101111110100

Next one:

(((192 << 8) | 43) << 8) | 244) << 8 =

= 110000000010101111110100 << 8 =

= 11000000001010111111010000000000

And finally:

(((((192 << 8) | 43) << 8) | 244) << 8) | 18 =

= 11000000001010111111010000000000 | 18 =

= 11000000001010111111010000000000 |
                          00010010 =

= 11000000001010111111010000010010

Answer 2

Let's start by looking at the actual structure of sock_addr_in :

#include <netinet/in.h>

struct sockaddr_in {
    short            sin_family;   
    unsigned short   sin_port;    
    struct in_addr   sin_addr;     
    char             sin_zero[8];  
};

And the struct of in_addr :

struct in_addr {
    unsigned long s_addr; 
};

Now, usually, when we utilisize this s_addr, we write :

inet_aton("89.161.169.137", &myaddr.sin_addr.s_addr); //dummy ip

OK. How about we check up the inet_aton man ?

inet_aton() converts the Internet host address cp from the IPv4 numbers-and-dots notation into binary form (in network byte order) and stores it in the structure that inp points to. inet_aton() returns nonzero if the address is valid, zero if not.

Now we clearly know that the unsigned long s_addr contains the binary form of the ip in network byte order .

You don't need to shift the ip youreslf, instead just use inet_aton() and inet_ntoa() for manipulating ip adresses.

Answer 3

First, lets begin with the binary operations. The shift ' a << b ' as you have correctly assumed moves the bits in a by an amount of b in the left direction. Shifting by 1 can be seen as a multiplication by 2. (This is actually how optimization can happen, as a shift operation is faster than multiplication, but most modern compilers take that into account.) The or-statement is ' a|b '. In or, if one of the two elements is 1, the result is 1. This is a standard method used in bitwise operations and actually all the bits inside a and b are or-ed. Eg a = 0010 1011, b = 1111 0000, a|b = 1111 1011.

Now back to the problem. In your case sa.sin_addr.s_addr is 0 at the beginning, so when it's or-ed with 192, it adds the bit representation of 192 to the variable. Then it is shifted by 8 bits, in order to free space for the next element in the ip address - each can have values between 0-255, which is exactly 8 bits. This is then repeated for the rest of the segments.