地址0x0未堆叠,malloc'd或(最近)free'd

Question

我是C的新手,似乎无法弄清楚以下代码有什么问题.

int main() {
    char filen[] = "file.txt";
    FILE *file = fopen ( filen, "r" );
    if ( file != NULL )
    {
        char line [ 128 ];
        while ( fgets ( line, sizeof line, file ) != NULL ) /* read a line */
        {
            int i;
            char *result;
            for(i=0; i< NUM;i++)
            {
                char *rep;
                rep = (char *) malloc (sizeof(mychars[i][0]));
                strcpy(rep, mychars[i][0]);
                char *with;
                with = (char *) malloc (sizeof(mychars[i][1]));
                strcpy(with, cgichars[i][1]);
                result = (char *) malloc (sizeof(char) * 128);
                result = str_replace(line, rep, with);
            }


            fputs(result, stdout);
        }
    }
    fclose ( file );


    return 0;
}

Valgrind给了我这个错误:

==4266== Invalid read of size 1
==4266==    at 0x4C286D2: __GI_strlen (mc_replace_strmem.c:284)
==4266==    by 0x5118A8D: fputs (iofputs.c:37)
==4266==    by 0x400A0F: main (repl.c:35)
==4266==  Address 0x0 is not stack'd, malloc'd or (recently) free'd

repl.c对应于以此代码末尾的fput开头的行.

此外,mychars是一个二维数组,如下所示:

char *mychars[NUM][2] = {
  "a", "97",
  "b", "98",
  ....

有人可以告诉我如何解决这个问题吗？此外,任何关于如何改进我的当前代码(特别是使用malloc)的指针将非常感激.

编辑: str_replace的代码

char *str_replace(char *str, char *orig, char *rep) {
  char buffer[4096];
  char *p;

  if(!(p = strstr(str, orig)))
    return NULL;

  strncpy(buffer, str, p-str);
  buffer[p-str] = '\0';
  sprintf(buffer+(p-str), "%s%s", rep, p+strlen(orig));

  return buffer;

}

编辑2 str_replace和main的新代码

出于测试目的,我用我在这里找到的方法替换了我的str_replace方法:

在C中替换字符串的功能是什么？

我的主要内容略有改变:

int main() {
    static const char filen[] = "file.txt";
    FILE *file = fopen ( filen, "r" );
    if ( file != NULL )
    {
        char line [ 128 ];
        while ( fgets ( line, sizeof line, file ) != NULL ) /* read a line */
        {
            int i;
            char *result;
            for(i=0; i< NUM;i++)
            {
                char *rep;
                rep = (char *) malloc (sizeof(mychars[i][0]));
                strcpy(rep, mychars[i][0]);
                char *with;
                with = (char *) malloc (sizeof(mychars[i][1]));
                strcpy(with, mychars[i][1]);
                result = str_replace(line, rep, with);
            }


            fputs(result, stdout);
        }
    }
    fclose ( file );


    return 0;
}

但我还是得到了

==6730== Invalid read of size 1
==6730==    at 0x4C286D2: __GI_strlen (mc_replace_strmem.c:284)
==6730==    by 0x5118A8D: fputs (iofputs.c:37)
==6730==    by 0x400995: main (repl.c:29)
==6730==  Address 0x0 is not stack'd, malloc'd or (recently) free'd

也许最令人沮丧的部分是不知道这些无效的读错误是什么.

编辑3 我已经更新了for循环中心的代码:

        int i;
        char* result;
        result = &line[0];
        for(i=0; i< NUM_CGICHARS;i++)
        {
            char *rep;
            rep = (char *) malloc (sizeof(char));
            strcpy(rep, cgichars[i][1]);
            char *with;
            with = (char *) malloc (sizeof(char)*3);
            strcpy(with, cgichars[i][0]);
            result = str_replace(result, rep, with);
            fputs(result, stdout);
            free(rep);
            free(with);
        }

现在我开始输出了!但是,经过两次迭代后,我得到了一个分段错误,valgrind给了我一大堆这个:

==9130== Invalid read of size 1
==9130==    at 0x4C286D2: __GI_strlen (mc_replace_strmem.c:284)
==9130==    by 0x5118A8D: fputs (iofputs.c:37)
==9130==    by 0x4009DF: main (teststep1.c:27)
==9130==  Address 0x0 is not stack'd, malloc'd or (recently) free'd

Answer 1

在这两行中

            result = (char *) malloc (sizeof(char) * 128);
            result = str_replace(line, rep, with);

你首先分配空间result,然后通过返回覆盖它然后立即松开str_replace.该功能可能会返回0,因此您fputs失败了.

顺便说一下,不要把它归还malloc,在C中这是多余的,可能会隐藏你忘记包含原型的事实.

编辑:您的str_replace功能在内存处理方面完全错误.永远不要将指针返回到局部变量,在离开函数后空间无效.