如何将sockaddr_in结构体的in_addr赋值给uint8_t数组？

Short answer: No, this won't work—here's why, and what to do instead.

Why the original line fails #

First, there's a fundamental type mismatch:

• addr[0] is a single uint8_t (1 byte)
• help->sin_addr is a struct in_addr, which on most systems contains a uint32_t field (s_addr) holding the 4-byte network-order IP address.

Assigning a struct to a single byte will either:

• Throw a compiler error (since you can't implicitly convert a struct to a uint8_t), or
• If your compiler allows it (via non-standard extensions), only copy the first byte of the struct into addr[0], leaving the other 3 elements of addr uninitialized. Either way, you won't get the full IP address into the array.

Correct ways to fill addr from sin_addr #

Since sin_addr.s_addr is already in network byte order (same format as what inet_pton outputs), you have a few safe options:

• Use memcpy (most portable)
  This copies the 4 bytes directly from the s_addr field into the addr array, regardless of system endianness:

  #include <string.h> // For memcpy
  
  memcpy(addr, &help->sin_addr.s_addr, sizeof(addr));
  

• Manually extract each byte (explicit and endian-safe)
  If you want to avoid memcpy, you can shift the 32-bit network-order value to get each octet:

  addr[0] = (help->sin_addr.s_addr >> 24) & 0xFF;
  addr[1] = (help->sin_addr.s_addr >> 16) & 0xFF;
  addr[2] = (help->sin_addr.s_addr >> 8) & 0xFF;
  addr[3] = help->sin_addr.s_addr & 0xFF;
  

  This works because s_addr is stored in big-endian (network order), so shifting right by 24 gives the first octet, and so on.

• Cast the array to a uint32_t* (use with caution)
  You can assign the s_addr value directly to the array treated as a 32-bit integer, but this depends on your system's endianness matching network order (big-endian) to get the bytes in the correct order. This is less portable, but works if you're targeting systems where this is consistent:

  *(uint32_t*)addr = help->sin_addr.s_addr;
  

  Note: This may trigger strict aliasing warnings in some compilers—memcpy is safer for cross-platform code.

内容的提问来源于stack exchange，提问作者Luk