C / C ++指针和数组帮助

Question

I have the following C++ code and I can't seem to get it working. What I am trying to do is read numerous entries from the command line, separated by ('|') pipe characters, and then splitting the resulting strings by spaces.

eg. 

mkdir C:/unixcode/shells|cd D:/margins/code | pwd| finger kobojunkie | last -l kobojunkie

but so far, I get errors, something about declaring the size of the pointer:

Initializer fails to determine the size of argv2 cannot convert char** to char* for argument 1 to char strtok(char*, const char*)

int main(int argc, char *argv[])
{
    char * pch;
    pch = strtok (argv,"|");
    //parse the contents of the generated arrays
    while (pch != NULL)
    {
        printf ("%s\n",pch);
        char * argv2[] = pch;
        char * subpch = strtok(argv2," ");

        while (subpch !=NULL)
        {
            printf ("%s\n",subpch);
            subpch = strtok (NULL, " ");
        }
        pch = strtok (NULL, " ");
    }
    return 0;
}

Answer 1

the type of argv is char** , not char* hence you cannot pass it to strtok. Use argv[ 1 ] instead, but check that argc >= 2 first.

Or, since this is tagged c++, use stl to split the string, eg

void split( const std::string& s, char delim, std::vector<std::string>& elems )
{
  std::stringstream ss( s );
  std::string item;
  while( std::getline( ss, item, delim ) )
    if( !item.empty() )
      elems.push_back( item );
}

int main( int argc, char *argv[] )
{
  if( argc == 2 )
  {
    std::vector< std::string > elements;
    split( argv[ 1 ], '|', elements );
    //elements now contains all items..
  }
}

Answer 2

argv is an array of arrays pointers. You cannot pass it as is to strtok: you need to pass its elements in a loop

for (k = 1; k < argc; k++) {
    pch = strtok(argv[k], "|");
    /* ... */
}

Also: are you sure you want to delimit with "|" ? That character has a special meaning to shells and, usually, does not make it to your program. Unless you call your program with them escaped, eg

bash$ ./a.out 'one|two|three' 'four|five|six'

Answer 3

The command-line is managed by a program: the shell (probably cmd.exe in Windows or bash in Linux). That shell gets all the stuff written in the command line and parses it and executes the commands specified.

Unless you are writing a shell, you cannot ever see the "|" of your example command-line inside the programs you write. They are effectively processed from the shell and removed from the parameters sent to the programs.

In

mkdir C:/unixcode/shells|cd D:/margins/code | pwd| finger kobojunkie | last -l kobojunkie

the shell calls the 5 following commands, each with the parameters specified

• mkdir C:/unixcode/shells
• cd D:/margins/code
• pwd
• finger kobojunkie
• last -l kobojunkie

Note none of the programs receive a "|" .

---

If you are indeed writing a shell, the command-line is not available in the argv array. It depends on the way you manage input inside your shell.

Answer 4

argv is not a string. argv is an array of strings. strtok takes a string, so you cannot pass it an array of strings and expect it to do something meaningful.

Each string element of the argv array is a separate command line parameter, except for the first which is the name of the executable. So what you should be doing is looking through each string entry for "|", and acting accordingly.