什么是 - 运营商做的char *？

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问题描述

我刚接触C.我读对C一个查找替换算法和我有点困惑是什么 - ＆安培; + 运营商在这种code做的：

 的char *取代（的char * SRC，为const char *搜索，为const char *代替）{
   字符*缓冲=的malloc（4096）; //在内存中新字符串分配4096个字节
   字符* P; //我在src字符串搜索的字符串
   INT I;   P =的strstr（SRC，搜索）;
   如果（NULL == p）的回报SRC;如果//'搜索'对'src'中的回报SRC未找到
      我= P  -  SRC; //我的子指数   函数strncpy（缓冲液，SRC，ⅰ）; //复制子值缓冲
   sprintf的（缓冲+ I，％s％S，更换，
   P + strlen的（搜索））; //？   返回缓冲区;
}

解决方案

由于 P 是在你的字符数组（字符串）和 SRC的位置是它的开始，

  I = P  -  SRC;

将设置 I 来的索引， P 分。

例如，请考虑以下的内存布局：

  [0] [1] [2] [3] [4] [5] [6] [7] [8] [9]＆下;  - 索引
 123 124 125 126 127 128 129 130 131 132 LT;  - 地址
+ --- + --- + --- + --- + --- + --- + --- + --- + --- + ---- +
| H |我| ，| | W | Ø| -  [R |升|开发| \\ 0 |
+ --- + --- + --- + --- + --- + --- + --- + --- + --- + ---- +
  ^^
  | |
 SRC p

在这种情况下，对 - SRC 会给你 127 - 123 或 4 ，这是中的是W 指数你好，世界。

这就是所谓的指针运算在覆盖加法运算符 ISO标准（ C99 6.5.6 / 9 ）：

It provides a scaled way of working out differences within the same array or with one pointing just beyond the end of the array (all else is undefined).

By that I mean doing pointer arithmetic with (for example) four-byte integers, will give you a difference of one between the addresses of arr[7] and arr[8], not four as some may think.

The buffer + i construct is simply another way of saying &(buffer[i]), the address of the ith element of the array buffer. I actually prefer the latter method since it seems more explicit in what I'm trying to represent.

For what it's worth, that's not actually a very good string replacement code. It has numerous problems:

if no replacements are made, you have a 4K memory leak with buffer.
in any case, you should always check to ensure malloc hasn't failed.
you have a possibility of buffer overflow the way the new string is allocated, you should really allocate based on the lengths of src search and replace.
you could create the new string with a single sprintf ("%*.*s%s%s", i, i, src, replace, &(src[i + strlen (search)])); or a strcpy and two strcat operations. Mixing the two seems incongruous to me.

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