检测到C堆粉碎

Question

我有一个函数可以计算单词出现在文件上的次数。 现在由于某种原因它得到堆栈粉碎检测到的错误，我没有看到错误。

这是代码：

int contar_palabra(const char *nombre_file, const char *palabra){

/*variables locales*/
FILE *file; 
char buffer[50]; 
int bytes_leidos, contador = 0, comparar, cerrar, i;    

/*Abrimos el archivo para lectura*/
file = fopen(nombre_file, "r");

/*Verificamos que se haya abierto correctamente*/
if (file == NULL){
  printf("No se pudo abrir el archivo \n");
  perror(nombre_file);
  exit(EXIT_FAILURE);
}

/*Procedemos a contar cuantas veces aparece una palabra*/
while (!feof(file)){

  bytes_leidos = fscanf(file, "%s", buffer);

  if (bytes_leidos > 0){

    /*Hacemos la comparacion*/
    comparar = strcmp(buffer, palabra);
    if (comparar == 0)
      contador++;
  }
  else if(errno == EOF)     
        printf("Error al leer alguna palabra de %s \n", nombre_file);
  else if (bytes_leidos == EOF)
    break;
}

cerrar = fclose(file);

if(cerrar == EOF){
  printf("Error: no se pudo cerra el archivo.");
}

printf("antes de retornar contador \n");
return contador; 

}

我用valgrind试图识别错误，日志文件给了我这个：

  ==2252== Memcheck, a memory error detector
  ==2252== Copyright (C) 2002-2010, and GNU GPL'd, by Julian Seward et al.
  ==2252== Using Valgrind-3.6.1 and LibVEX; rerun with -h for copyright info
  ==2252== Command: ./pargrep viejo.txt
  ==2252== Parent PID: 1756
  ==2252== 
  ==2252== 
  ==2252== HEAP SUMMARY:
  ==2252==     in use at exit: 55 bytes in 1 blocks
  ==2252==   total heap usage: 7 allocs, 6 frees, 1,389 bytes allocated
  ==2252== 
  ==2252== 55 bytes in 1 blocks are still reachable in loss record 1 of 1
  ==2252==    at 0x4026864: malloc (vg_replace_malloc.c:236)
  ==2252==    by 0x40B878B: __libc_message (libc_fatal.c:138)
  ==2252==    by 0x413D09F: __fortify_fail (fortify_fail.c:32)
  ==2252==    by 0x413D049: __stack_chk_fail (stack_chk_fail.c:29)
  ==2252==    by 0x8049142: contar_palabra (in /home/alessandro/OS/Proyecto2/prueba1.0/pargrep)
  ==2252==    by 0x80489D4: main (in /home/alessandro/OS/Proyecto2/prueba1.0/pargrep)
  ==2252== 
  ==2252== LEAK SUMMARY:
  ==2252==    definitely lost: 0 bytes in 0 blocks
  ==2252==    indirectly lost: 0 bytes in 0 blocks
  ==2252==      possibly lost: 0 bytes in 0 blocks
  ==2252==    still reachable: 55 bytes in 1 blocks
  ==2252==         suppressed: 0 bytes in 0 blocks
  ==2252== 
  ==2252== For counts of detected and suppressed errors, rerun with: -v
  ==2252== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 15 from 8)

奇怪的是它在返回之前打印了一条消息。 我真的没有看到错误，感谢帮助。 提前致谢

Answer 1

使用普通的%s作为fscanf的格式说明符是非常危险的，因为你没有防止写入超出缓冲区末尾的保护，在你的情况下只有50个字节。 考虑为格式说明符提供宽度修饰符，但是当宽度限制达到更复杂时，这将处理准确的计数。

您可能会发现逐字符读取（ fgetc ）或读取固定缓冲区（ fread ）并手动检测单词分隔符会产生更简单的代码。