如何替换字符串在C？ [英] how to replace substring in c?

问题描述

本示例工作，但我认为内存泄漏。简单的Web服务器模块中使用的功能，如果你使用此功能从而共享内存增长。

This example works but I think that the memory leaks. Function used in the simple web server module and thus shared memory grows if you use this function.

    char *str_replace ( const char *string, const char *substr, const char *replacement ){
      char *tok = NULL;
      char *newstr = NULL;
      char *oldstr = NULL;
      if ( substr == NULL || replacement == NULL ) return strdup (string);
      newstr = strdup (string);
      while ( (tok = strstr ( newstr, substr ))){
        oldstr = newstr;
        newstr = malloc ( strlen ( oldstr ) - strlen ( substr ) + strlen ( replacement ) + 1 );
        memset(newstr,0,strlen ( oldstr ) - strlen ( substr ) + strlen ( replacement ) + 1);
        if ( newstr == NULL ){
          free (oldstr);
          return NULL;
        }
        memcpy ( newstr, oldstr, tok - oldstr );
        memcpy ( newstr + (tok - oldstr), replacement, strlen ( replacement ) );
        memcpy ( newstr + (tok - oldstr) + strlen( replacement ), tok + strlen ( substr ), strlen ( oldstr ) - strlen ( substr ) - ( tok - oldstr ) );
        memset ( newstr + strlen ( oldstr ) - strlen ( substr ) + strlen ( replacement ) , 0, 1 );
        free (oldstr);
      }
      return newstr;
    }

推荐答案

有一个问题，我可以看到的是，如果替换字符串包含搜索字符串，你永远循环下去（直到耗尽内存）。

One problem I can see is that if the replacement string contains the search string, you'll loop forever (until you run out of memory).

例如：

char *result = str_replace("abc", "a", "aa");

此外，每更换一个实例的时间做其他的malloc / free是pretty昂贵。

Also, doing another malloc/free every time you replace one instance is pretty expensive.

有一个更好的办法是在输入字符串做的正是通过2次：

A better approach would be to do exactly 2 passes over the input string:

• 第一遍，算上搜索字符串的许多情况下是如何present

• the first pass, count how many instances of the search string are present

现在你知道很多比赛怎么样，你的计算结果与放大器的长度;曾经的malloc：

now that you know how many matches, compute the length of your result & malloc once:

的strlen（字符串）+匹配*（strlen的（替代）-strlen（SUBSTR））+ 1

strlen(string) + matches*(strlen(replacement)-strlen(substr)) + 1

进行第二次通过源字符串，复制/替换

make a second pass through the source string, copying/replacing

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