c - 查找第一次出现的未转义字符

Question

查找给定字符串中第一个未转义字符出现的最佳方法是什么？

我就是这样做的，但我感觉它过于复杂。

/*
 * Just like strchr, but find first -unescaped- occurrence of c in s.
 */
char *
strchr_unescaped(char *s, char c) 
{
  int i, escaped;
  char *p;

  /* Search for c until an unescaped occurrence is found or end of string is
     reached. */
  for (p=s; p=strchr(p, c); p++) {
    escaped = -1;
    /* We found a c. Backtrace from it's location to determine if it is
       escaped. */
    for (i=1; i<=p-s; i++) {
      if (*(p-i) == '\\') {
        /* Switch escaped flag every time a \ is found. */
        escaped *= -1;
        continue;
      }
      /* Stop backtracking when something other than a \ is found. */
      break;
    }
    /* If an odd number of escapes were found, c is indeed escaped. Keep 
       looking. */
    if (escaped == 1) 
      continue;
    /* We found an unescaped c! */
    return p;
  }
  return NULL;
}

Answer 1

如果搜索字符相当罕见，那么您的方法是合理的。一般来说，像 strchr 这样的 C 库例程是用严格的机器语言编码的，并且比用 C 语言编码的几乎任何循环都运行得更快。某些型号的硬件具有用于搜索内存块的机器指令；而某些型号的硬件具有用于搜索内存块的机器指令。使用它的 C 库例程的运行速度比您用 C 编写的任何循环都要快得多。

为了稍微加强你的方法，这样如何:

#define isEven(a) ((a) & 1) == 0)

char* p = strchr( s, c );
while (p != NULL) {   /* loop through all the c's */
    char* q = p;   /* scan backwards through preceding escapes */
    while (q > s && *(q-1) == '\\')
        --q;
    if (isEven( p - q ))   /* even number of esc's => c is good */
        return p;
    p = strchr( p+1, c );   /* else odd escapes => c is escaped, keep going */
}
return null;