Question

所以，我正在为LC3b机器编写一个解释器。我需要编写一个函数来检查输入是否是有效的操作码。为此，我编写了如下函数：

int isOpcode(char* oput)//oput=Opcode under test
{
  FILE* isa_file = NULL;
  int i,j;
  char * line = NULL;
  size_t len = 0;
  ssize_t nread;

  //printf("Entered opcode checking function \n");
  /*convert to lower case */
  //for( i = 0; i < strlen( oput ); i++ )
  //    oput[i] = tolower( oput[i] );
  //printf("Entered opcode checking function1 \n");

  /*for modularity, using the file */
   isa_file = fopen("isa", "r");
   if(isa_file == NULL)
       printf("Error in file opening\n");
   int flag;
   while ((nread = getline(&line, &len, isa_file)) != -1)
   {
       printf("Retrieved line of length %zu :\n", nread);
       printf("%s", line);  
       //Solving the new line character problem which resulted in the improper difference of strcmp ..now it is proper...
       //char final[strlen(line)-1];
       //strcpy(final,line);
       //final[strlen(line)-1]='\0'; 
       //  printf("Entered opcode checking function2 \n");
       //  printf("%s \n",oput);
       // overwrite \n with nul-terminator
       line[strcspn(line, "\n")] = 0;
       char final[strlen(line)];
       strcpy(final,line);
       flag=strcmp(oput,final);
       printf("%s \n",oput);
       printf("flag: %d \n",flag ); 
       if(flag < 0 || flag >0)
       {
           flag=-1;
           //printf("entered 1st condition \n");
           printf("final flag: %d \n", flag);
        }
       else 
       {
           flag=1;
           //printf("entered negative \n");
           printf("final flag: %d \n", flag);
           break;
       }

   }
   printf("end of loop \n");//this is not getting printed
   fclose(isa_file);
   if (line)
       free(line);
   return flag;
}

现在，当我使用其中一个有效操作码调用该函数时，该函数会打印最终标志（= 1），然后进入分段错误。它甚至不打印循环结束语句。有人能告诉我这是什么问题吗？我尝试将部分功能部分作为单独的文件执行，看起来效果很好。

更新：当我注释掉toLower代码时，循环语句的结尾现在被打印在屏幕上。但是，即使然后代码也会抛出一个seg错误。有趣的是，它只在最终标志设置为1时才会抛出错误。否则，它运行正常！我无法理解这个奇怪的问题！

Answer 1

发表评论后我可以说：

strlen已经为您提供了阅读字符-1的长度，因为nul-terminator不计算在内。此外，您无法使用strcpy来执行您需要执行的操作，因为它会将源字符串的所有字符复制到一个较小的字符串中，从而导致UB。

您可以做的是删除源字符串中的'\n'，然后将其复制，如：

// overwrite \n with nul-terminator
line[strcspn(line, "\n")] = 0;
// allocate new string less by one
char final[strlen(line)+1];
// allocate new string less by one
strcpy(final,line);

Answer 2

以下提议的代码：

干净地编译
不会链接，因为没有main()功能
适用所有评论

现在建议的代码：

#include <stdio.h>  // printf(), FILE, fopen(), fclose(), perror()
#include <string.h> // strcpy(), strchr()
#include <stdlib.h> // malloc(), free(), exit(), EXIT_FAILURE
#include <ctype.h>  // tolower()

int isOpcode(char* oput);

int isOpcode(char* oput)//oput=Opcode under test
{
    FILE* isa_file = NULL;
    //int i;  <-- unused
    //int j;  <-- unused



    //printf("Entered opcode checking function \n");
    /*convert to lower case */
    for( size_t i = 0; i < strlen( oput ); i++ )
        oput[i] = (char)tolower( oput[i] );
    //printf("Entered opcode checking function1 \n");

    /*for modularity, using the file */
    isa_file = fopen("isa", "r");

    if(isa_file == NULL)
    {
        perror( "fopen to open 'isa' for read, failed" );
        exit( EXIT_FAILURE );
    }

    // implied else, fopen successful

    int flag;
    char * line = NULL;
    size_t len = 0;
    ssize_t nread;

    while ((nread = getline(&line, &len, isa_file)) != -1)
    {
       printf("Retrieved line of length %zu :\n", nread);

       printf("%s", line);

       //Solving the new line character problem which resulted in the improper difference of strcmp ..now it is proper...
       char final[ nread+1 ];
       strcpy(final, line);

       //  printf("Entered opcode checking function2 \n");
       //  printf("%s \n",oput);
       // overwrite \n with nul-terminator
       char *newline = NULL;
       if( NULL != (newline = strchr( final, '\n' ) ) )
           *newline = '\0';

       printf("%s \n",oput);
       printf("flag: %d \n",flag );

       strcmp(oput,final);
       switch( strcmp(oput,final) )
       {
            case -1:
            case 1:
               flag=-1;
               //printf("entered 1st condition \n");
               printf("final flag: %d \n", flag);
               break;

            case 0:
               flag=1;
               //printf("entered negative \n");
               printf("final flag: %d \n", flag);
               break;

           default:
               break;
        }

        free( line );
        line = NULL;
        len = 0;
    }

    if( !feof( isa_file ) )
        perror( "getline failed" );

    printf("end of loop \n");//this is not getting printed
    fclose(isa_file);


    free(line);

    return flag;
}

我不确定这是否是您正在寻找的功能。

返回函数时出现分段错误错误

2 个答案: