c - 编写我自己的 shell，未声明的标识符 "output"

Question

这是我的代码，用于我自己的 C shell。编译时出现错误:使用未声明的标识符“输出”。以下是编译时的一些错误示例:

error: use of undeclared identifier 'output' char input[100];output[100];

test3.c:53:15: error: use of undeclared identifier 'output' strcpy(output,args[i+1]); ^

test3.c:53:15: error: use of undeclared identifier 'output'

test3.c:60:8: warning: implicit declaration of function 'open' is invalid in C99 [-Wimplicit-function-declaration] j = open(input, O_RDONLY, 0); ^

test3.c:60:20: error: use of undeclared identifier 'O_RDONLY' j = open(input, O_RDONLY, 0); ^

test3.c:61:29: error: use of undeclared identifier 'O_RDONLY' if ((j = open(input, O_RDONLY, 0)) < 0) {

test3.c:70:12: warning: implicit declaration of function 'creat' is invalid in C99 [-Wimplicit-function-declaration] if ((i= creat(output , 0644)) < 0) {

test3.c:70:18: error: use of undeclared identifier 'output' if ((i= creat(output , 0644)) < 0) {

这是我的代码:

#include "stdio.h"
#include "stdlib.h"
#include "string.h"
#include "signal.h"
#include "unistd.h"

void prompt(char*);
void execute( char* );
char** parse( char* );

int main( int ac, char* av[] )
{
  char input[255]; // buffer for supporting command

  signal( SIGINT, SIG_IGN ); // ignore ctrl-c

  while(1)
  {
    prompt(input);
    execute( input );
  }
};

void execute( char* str)
{
  int fork_result, status, i = 0,j=0,in=0,out=0;
  char input[100];output[100];
  char** args = parse( str ); // splits the user command into arguments

  fork_result = fork(); // attempt to fork
  if ( fork_result == -1 ) // failure
  {
    perror("Failed to fork\n");
    exit(1);
  }
  else if ( fork_result == 0 ) // I'm the child
  {
    for(i=0;args[i]!='\0';i++)
    {
      if(strcmp(args[i],"<")==0)
      {   
        args[i]=NULL;
        strcpy(input,args[i+1]);
        in=2;   
      }   
      if(strcmp(args[i],">")==0)
      {
        args[i]=NULL;
        strcpy(output,args[i+1]);
        out=2;
      }   
    }

    if (in) 
    {
      j = open(input, O_RDONLY, 0);
      if ((j = open(input, O_RDONLY, 0)) < 0) 
      {
        perror("Couldn't open input file");
        exit(0);
      }
      dup2(j, 0);
      close(j);
    }

    if (out) 
    {
      if ((i= creat(output , 0644)) < 0) 
      {
        perror("Couldn't open the output file");
        exit(0);
      }
      dup2(i, STDOUT_FILENO);
      close(i);
    }

    execvp( args[0], args );
    perror("failed to exec\n");
    exit(2);
  }
  else // I'm the parent
  {
    // wait here
    wait(&status); // wait for child to finish
    free( args ); // free dynamic memory
  }
}

char** parse( char* str )
{
  char** args = malloc( 256 );
  int i = 0;

  args[i] = strtok( str, " " );

  while( args[i] )
  {
    i++;
    args[i] = strtok( NULL, " " );
  }

  return args;
}

void prompt(char* input)
{
  printf("$ "); // print prompt
  fgets( input, 255, stdin );

  input[strlen(input)-1] = '\0'; // overwrite \n with \0

  if ( strcmp( input, "exit" ) == 0 ) // shell command
    exit(0);
}

Answer 1

char input[100];output[100];

你想要:

char input[100], output[100];

同时添加:#include <fcntl.h>

一般来说，man open (以及您使用的其他功能)是您的 friend ——它告诉您什么 #include s 来补充。

您的代码中存在更多潜在错误和任意限制。一些例子:

void execute( char* str)
{
  char input[100], output[100];
  ...
  if(strcmp(args[i],"<")==0)
  {   
    args[i]=NULL;
    strcpy(input,args[i+1]);  // possible stack buffer overflow.

  if(strcmp(args[i],">")==0)
  {
    args[i]=NULL;
    strcpy(output,args[i+1]);  // possible stack buffer overflow


char** parse( char* str )
{
  char** args = malloc( 256 );  // limit of 256/sizeof(char*) parameters.
  // on a 64-bit system, if more than 32 parameters are supplied ...

         args[i] = strtok( NULL, " " );  // ... possible heap buffer overflow.


  fgets( input, 255, stdin );  // arbitrary limit of 254 characters on command line.

不能保证字符串以 \n 结尾:

  input[strlen(input)-1] = '\0'; // overwrite \n with \0

如果我给这个“shell ”打分，我会给它一个“F”。