c - GCC 提示缺少对同一文件中函数的引用

Question

沿着 makefile 运行，我得到了这个:

cc client.o MurmurHash3.o libstorage.a -Wall -lreadline -pthread  -o client
MurmurHash3.o: In function `MurmurHash3_x64_128':
/home/evantandersen/mount/src/MurmurHash3.c:59: undefined reference to `rotl64'
/home/evantandersen/mount/src/MurmurHash3.c:106: undefined reference to `fmix'

并且，MurmurHash3.c:

inline uint64_t rotl64 ( uint64_t x, int8_t r )
{
  return (x << r) | (x >> (64 - r));
}

//-----------------------------------------------------------------------------
// Finalization mix - force all bits of a hash block to avalanche

inline uint64_t fmix ( uint64_t k )
{
  k ^= k >> 33;
  k *= 0xff51afd7ed558ccd;
  k ^= k >> 33;
  k *= 0xc4ceb9fe1a85ec53;
  k ^= k >> 33;

  return k;
}

//-----------------------------------------------------------------------------

void MurmurHash3_x64_128 ( const void * key, const int len,
                           const uint32_t seed, void * out )
{

  const uint8_t * data = (const uint8_t*)key;
  const int nblocks = len / 16;

  uint64_t h1 = seed;
  uint64_t h2 = seed;

  const uint64_t c1 = 0x87c37b91114253d5;
  const uint64_t c2 = 0x4cf5ad432745937f;

  //----------
  // body

  const uint64_t * blocks = (const uint64_t *)(data);

  for(int i = 0; i < nblocks; i++)
  {
    uint64_t k1 = blocks[i*2+0];
    uint64_t k2 = blocks[i*2+1];

    k1 *= c1; k1  = rotl64(k1,31); k1 *= c2; h1 ^= k1;

    h1 = rotl64(h1,27); h1 += h2; h1 = h1*5+0x52dce729;

    k2 *= c2; k2  = rotl64(k2,33); k2 *= c1; h2 ^= k2;

    h2 = rotl64(h2,31); h2 += h1; h2 = h2*5+0x38495ab5;
  }

  //----------
  // tail

  const uint8_t * tail = (const uint8_t*)(data + nblocks*16);

  uint64_t k1 = 0;
  uint64_t k2 = 0;

  switch(len & 15)
  {
  case 15: k2 ^= ((uint64_t)tail[14]) << 48;
  case 14: k2 ^= ((uint64_t)tail[13]) << 40;
  case 13: k2 ^= ((uint64_t)tail[12]) << 32;
  case 12: k2 ^= ((uint64_t)tail[11]) << 24;
  case 11: k2 ^= ((uint64_t)tail[10]) << 16;
  case 10: k2 ^= ((uint64_t)tail[ 9]) << 8;
  case  9: k2 ^= ((uint64_t)tail[ 8]) << 0;
           k2 *= c2; k2  = rotl64(k2,33); k2 *= c1; h2 ^= k2;

  case  8: k1 ^= ((uint64_t)tail[ 7]) << 56;
  case  7: k1 ^= ((uint64_t)tail[ 6]) << 48;
  case  6: k1 ^= ((uint64_t)tail[ 5]) << 40;
  case  5: k1 ^= ((uint64_t)tail[ 4]) << 32;
  case  4: k1 ^= ((uint64_t)tail[ 3]) << 24;
  case  3: k1 ^= ((int64_t)tail[ 2]) << 16;
  case  2: k1 ^= ((uint64_t)tail[ 1]) << 8;
  case  1: k1 ^= ((uint64_t)tail[ 0]) << 0;
           k1 *= c1; k1  = rotl64(k1,31); k1 *= c2; h1 ^= k1;
  };

  //----------
  // finalization

  h1 ^= len; h2 ^= len;

  h1 += h2;
  h2 += h1;

  h1 = fmix(h1);
  h2 = fmix(h2);

  h1 += h2;
  h2 += h1;

  ((uint64_t*)out)[0] = h1;
  ((uint64_t*)out)[1] = h2;

}

rotl64 和 fmix 都定义在同一个文件中，并且在函数 MurmurHash3_x64_128 之上。

Answer 1

如果您正在编译为 C99，那么编译器不必使用 inline 函数定义，但 GCC 会在优化时使用。当不进行优化时，它假设程序中某处有一个正常的 extern 定义，这就是它试图链接到的内容。

通过定义它 extern inline ，您可以将其定义为 extern 定义，因此在未优化时将通过相同或不同翻译单元中的调用使用它。

通过定义它 static inline 你使它成为一个 static 定义，所以它会在未优化时被使用，仅在同一个翻译单元中调用。该 inline 定义不能用于解析其他翻译单元中的调用。

参见 http://www.greenend.org.uk/rjk/tech/inline.html了解更多详情。