c - 将 c 代码片段翻译为伪代码

Question

我刚刚学习 Java，我有一个作业需要将 C 代码片段翻译为 Java。有人可以帮助我将代码片段翻译为伪代码吗？我想自己编写 Java 编码，但我不懂 C，也无法从代码片段中获得太多意义。这是作业:

<小时/>

您正在寻找一种简单的模式匹配方法。类似于C中的strstr(...)，它应该在字符串中搜索搜索字符串。搜索字符串应包含“*”(替换多个字符)和“?”。您有一个示例，但它位于 C:

int match ( char *pat, char *str ) {
 switch ( *pat ) {
  case '\0' : return !*str;
  case '*'  : return match( pat+1, str ) || *str && match( pat, str+1 );
  case '?'  : return *str && match( pat+1, str+1 );
  default   : return *pat == *str && match( pat+1, str+1 );
 }
}

将其翻译为 Java。

<小时/>

我知道尝试一项需要从您不知道的语言进行翻译的作业是愚蠢的，而且我无法理解为什么该作业包含在 Java 学习任务列表中，但我必须解决它，这会非常困难。如果有人愿意帮助我，那就太好了。

Answer 1

我已尝试为您发表评论。阅读一下，看看是否有助于您理解:)。

/* Returns an integer (nonzero if the strings match, zero if they don't).
 * - pat: A string (char *) which is your pattern.
 * - str: A string (char *) which is your source string.
 * Note: All strings end in the null-character ('\0') which has integer value zero. 
*/
int match ( char *pat, char *str ) {

 /* The switch extracts the first character of the string "pat" (*pat).
  * Then, it will run the code in the case to which that character matches:
 */
 switch ( *pat ) {

  /* If you match null-character, then return nonzero only if the value 
   * of the leading character in str (*str) is zero (!*str) This means
   * that it returns a match if the leading character in str is also 
   * the null character
  */ 
  case '\0' : return !*str;

  /* If you match an asterisk '*', then return nonzero in two cases:
   * (1) Calling your own function recursively having dropped the asterisk from
   *     the pattern returns nonzero (match(pat+1, str)).
   * ... OR ... 
   * (2) The leading character of str is nonzero (*str) AND calling
   *     this very function having dropped the leading character of str returns
   *     nonzero (match(pat, str + 1)).
  */
  case '*'  : return match( pat+1, str ) || *str && match( pat, str+1 );

  /* If you match '?', then return nonzero if both cases are true:
   * (1) *str is not the null-char (it is nonzero).
   * (2) Calling match recursively having skipped the current character
   * in both the pattern AND the string returns nonzero (match(pat+1, str+1)).
  */                             
  case '?'  : return *str && match( pat+1, str+1 );


  /* Otherwise, if you didn't match on any of the above patterns, return
   * nonzero only if both the following conditions are true:
   * (1) The current character at the head of pat is the same as that of str
   *      (*pat == *str)
   * (2) calling match recursively having skipped both the current character
   *     at the head of pattern AND the string also returns nonzero. 
   *     match(pat + 1, str + 1)
  */
  default   : return *pat == *str && match( pat+1, str+1 );
 }
}