C中strcat函數的遞歸實現

Question

我正在執行中 ！ 在C. strcat的功能，我無法理解他們是如何與指針一起使用遞歸 。

以下是該來源的代碼段。

char dest[100] = "I love";
char *src = "food";`

/* my_strcat(dest, src) copies data of src to dest. */
void my_strcat(char *dest, char *src)
{
  (*dest)? my_strcat(++dest, src): (*dest++ = *src++)? my_strcat(dest, src): 0 ;
}

Answer 1

分成幾部分：

(*dest) /* Is *dest different than '\0' ? */
   ? my_strcat(++dest, src) /* *dest is different than '\0', so increment dest pointer so it'll point to the next character and call my_strcat() again. Here we're searching for the end of the string dest. */

   : (*dest++ = *src++) /* *dest is equal to '\0', so we're at the end of *dest... We start to assign *src to *dest and increment both pointers to point to the next character. The lvalue of this assignment is also a comparison (is it different than '\0' ?). */

     ? my_strcat(dest, src) /* The previous comparison is different than '\0', so we'll call my_strcat() again (pointers have already been incremented and they now point to the next character) */

     : 0; /* The previous comparison is '\0', so we've reached the end of the src, so we're done. */

用if / else代替三元運算符：

/* Is *dest different than '\0' ? */
if (*dest != '\0') {
  /* *dest is different than '\0', so increment dest pointer so it'll point to the next character and call my_strcat() again. Here we're searching for the end of the string dest. */
  my_strcat(++dest, src);
} else {
  /* *dest is equal to '\0', so we're at the end of *dest... We start to assign *src to *dest and increment both pointers to point to the next character. The lvalue of this assignment is also a comparison (is it different than '\0' ?). */
  if ((*dest = *src) != '\0') {
    /* The previous comparison is different than '\0', so we'll call my_strcat() again (pointers have already been incremented and they now point to the next character) */
    my_strcat(++ dest, ++ src); /* Moved increments down for readability */
  } else {
     /* The previous comparison is '\0', so we've reached the end of the src, so we're done. */
    return; 
  }
}

如果/沒有其他注釋（也許更具可讀性）：

if (*dest != '\0') {
  my_strcat(++dest, src);
} else {
  if ((*dest = *src) != '\0') {
    my_strcat(++ dest, ++ src);
  } else {
    return; 
  }
}

Answer 2

這是一個很差的實現，但是我想它確實起作用。 第一個?:測試目標字符串是否在結尾。 如果不是，它將碰撞目標指針，然后遞歸調用自身。 如果目標已經在末尾，則它將復制一個字符，同時碰觸兩個指針並測試零。 如果那不是來自src的尾隨NUL ，則它將使用更新的指針遞歸調用自身。

哦，等等，有一個錯誤（或者可能是一個“功能”）。 假定dest末尾的所有字符最初都用NUL填充。 我猜您實際上可以利用這一點，如果您可以繼續依賴該實現並擁有一個dest字符串，該字符串的文本中插入NUL字符，那么它將一次用一個字符填充src中的NUL 。

更不用說過度使用遞歸了，這意味着堆棧中的調用框架與結果字符串中包含的字符一樣多。 您從哪里獲得了這種愚蠢的strcat實現？ 當然，沒有真正的庫會使用此實現，迭代變體既易於理解，又快得多。