C ++：将int转换为unsigned char

Question

I declared an IP address:

int test_ip = (192) | (168<<8) | (33<<16) | (255<<24);

Then I need to convert it to (unsigned char*). I trying to do this:

(u_char*)test_ip

I use VS 2010 C++ debugger in order to evaluate the last expression. Debugger's output for this expression:

0xff21a8c0 <Bad Ptr>

What am I doing wrong? Why Bad Ptr?

Answer 1

Your conversion better be done by in_addr data type:

This is the windows in_addr so if you want to use it on linux you'll have to re-declare it with other name.

typedef struct in_addr {
  union {
    struct {
      u_char s_b[4];
    } S_un_b;
    struct {
      u_short s_w[2];
    } S_un_w;
    u_long S_addr;
  } S_un;
};

like this:

in_addr address;
address.S_addr = (192) | (168<<8) | (33<<16) | (255<<24);

address.S_un_b.s_b is ready now, with array of 4 bytes representing the address...

Answer 2

Attempting a c-style cast will give you undefined behaviour as the pointer types are unrelated. Your best bet is to use a static cast of the int, modulo 256.

unsigned char c = static_cast<unsigned char>(test_ip % 256)

This will be safe since the standard guarantees the size of an unsigned char as being 8 bits.

Once you've taken this cast to unsigned char, take the address of that as your pointer &c.

If you want other parts of test_ip then combine with integral division etc. Just take care not to overflow the unsigned char when taking the cast.

Answer 3

Presumably, you need it as an unsigned char* to pass it to some protocol. In which case, you can't just cast, since the bit representation of your int is not necessarily what the protocol expects. You have to extract it byte by byte, just as you constructed it, placing the bytes in an unsigned char buffer.