Passing my array to function

I am having a problem where I do not understand the syntax on how to pass my array to a function and then access each argument. As it is now, while my array is in main, I can print each argument like:

Code:

---

printf("%s\n", args[0]);
printf("%s\n", args[1]);
etc...

---

However when I try to access it like above when I try to pass the args array to a function, I get a segmentation fault, I can only print the first element using:

Code:

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printf("%s\n", args)

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Could somebody please tell me if I'm passing the array incorrectly to the translate function and then subsequent functions, or how to acces each element beyond the first one.

Code:

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void translate(char *input);
void list();
void changedir(char *list);
void lsdir(char *list);

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

main()
{
  char *cptr, *csh;
  char buffer[256];
  char **args = NULL;
  int nargs = 0;


  printf("sh%% ");
  cptr = fgets(buffer, 256, stdin);
  csh = strtok(buffer, " ,\n");
 
  while(strcmp(csh, "bye") != 0)
    {   
      while(csh != NULL)
        {
          args = realloc(args, sizeof(char *)*(nargs + 1));
          args[nargs] = malloc(strlen(csh) + 1);
          strcpy(args[nargs], csh);
          nargs++;

          csh = strtok(NULL, " ,\n");
        }
 
     
      translate(*args);
     
      nargs = 0;
      printf("sh%% ");
      cptr = fgets(buffer, 256, stdin);
      csh = strtok(buffer, " ,\n");
    }
}


void translate(char *input)
{/*This function translates what command is entered and calls
  *appropriate function to execute command
  */
 
if(strcmp(input, "?") == 0)
    {
      list();
    }
 else if(strcmp(input, "cd") == 0)
  {
    changedir(input);
  }
 else if(strcmp(input, "ls") == 0)
  {
    lsdir(input);
  }
 else return;
}


void list()
{/*This function prints a list of all available commands when
  *? is entered
  */
 
  printf("\n***List of Available Commands***\n");
  printf("?  - Displays this message\n");
  printf("bye - Exits UNIX shell\n");
  printf("cd  - changes current directory\n");
  printf("ls  - lists files in current directory\n");
  printf("\n");
}
       
 
void changedir(char *input)
{
}

void lsdir(char *input)
{
  int pid;
  int status = 0;

  pid = fork();
 
  if(pid == -1)
    printf("error!!!\n");

  else if(pid == 0)
    {
      execvp(list, list);
      exit(1);
    }
  else if(pid > 0)
    {
      waitpid(pid, &status, 0);
    }

  printf("%d\n", pid);
}

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Passing functions utilize the unary operator “&”. To do this properly, you would send the address of a single row of your two-dimensional pointer:

Code:

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translate(&args[n]);

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Instead of:

Code:

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translate(*args);

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*args will result in an error as it may already since args is a char **, and you would have to send the memory address of args[] to translate since it only recognizes a single string, not multi-dimensional.


char **args is in the same dimension as:

Code:

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int myValue[2][3] = { {5, -3, 0}, {10, 17, -25} };

---

Which consists of rows and columns, instead of just a single row and column. To get a better look, this initialization can be viewed as:

Code:

---

Rows/Columns        Column 0        Column 1        Column 2
Row 0                5                -3                0
Row 1                10                17                -25

---

It is unlikely your function translate() will take all the rows and n amount of columns by sending it to a single char * called input. That's why sending the memory address of args[] is beneficial. args[] already holds a single row and column per array index.

On further note, the unary operator “&” gives the memory address of an existing object. This assigns the address of one variable and “points to” or “passes to” another.

I can further expound if neccessary.

Edit: You many not need to send the memory address of args[n] to translate() if not needed.


- Stack Overflow

I'm afraid I must disagree.

Code:

---

#include <stdio.h>

void argtest( char * arg )
{
        printf("arg is %s\n", arg );
}

int main( void )
{
        char **args;
        char *s[] = { "Hello", "World", "!", 0 };
        args = s;
        while( *args )
        {
                argtest( *args++ );
        }
        return 0;
}
/*
arg is Hello
arg is World
arg is !
*/

---

Dereferencing a pointer to a pointer to a type, gives you a pointer to a type, which is exactly what the function calls for.

Quzah.

Ok I understand how u are printing out each element out a time by passing one by one. How could I pass the whole thing to the function so that I could print it out like this:

Code:

---

printf("%s\n", args[0]);
printf("%s\n", args[1]);
printf("%s\n", args[2]);
etc...

---

I hope that makes sense.

How do you think you'd past a char ** to a function?
If you can pass a char * to this prototype:

Code:

---

void foo( char * );

---

Then what would you need for a char **? Take a second and think about it. It's not hard.

Quzah.

Ok I figure if I want to pass **args to translate I would do it like this:

Code:

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translate(**args);

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with this prototype:

Code:

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void  translate(char **input);

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but I get an error:

Code:

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project1a.c:42: warning: passing arg 1 of `translate' makes pointer from integer without a cast

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Close, but you're passing it wrong.

Q) How do you pass a character pointer to a function that wants one?
A) By passing the name of a variable:

Code:

---

void foo( char * );

int main( void )
{
    char *s = "Hello World!";
    foo( s );
    return 0;
}
...

---

The same holds true for a pointer to a pointer. If the variable you have to pass is exactly what the function expects, you simply pass it's name over.

Quzah.

Quzah, I love you, if I were a woman and I assume you are a man, I would marry you :D . You have been a great help to me, thank you very much.

here's the first major problem:

Code:

---

while(csh != NULL)
        {
          args = realloc(args, sizeof(char *)*(nargs + 1));
          args[nargs] = malloc(strlen(csh) + 1);
          strcpy(args[nargs], csh);
          nargs++;

          csh = strtok(NULL, " ,\n");
        }

---

each iteration of the loop you reallocate the char ** meaning when the loop finishes, only the last char * in the array points to valid memory. it doesn't help that you don't even free the memory when you're done with it, either!

>> translate(*args);

the way you have it here only one command gets executed - you'll need to either set up a loop to call the function with or else redesign the function to process a char ** internally.

this is a good example of why it is so important to modularize code as much as possible. just start with the obvious - you need an array of char * and also a means for safely destroying the array when you're done with it. so start with a set of functions:

Code:

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/*
 DeleteMatrix() takes the address of a char **
 so that it can set the pointer to NULL internally.
*/
void DeleteMatrix(char *** array, int rows) {
 if(*array != NULL) {
  int index;
  for(index = 0; index < rows; ++index) {
  free((*array)[index]);
  }
  free(*array);
  *array = 0;
  }
 return;
 }
/*
 NewMatrix() allocates an extra row which is set to
 NULL - this makes traversing the array easier...
 also, each column gets an extra char for posterity.
*/
char ** NewMatrix(int rows, int columns) {
 char ** array = malloc(rows+1);
 if(array != NULL) {
  array[rows] = NULL;
  int index;
  for(index = 0; index < rows; ++index) {
  array[index] = malloc(columns+1);
  if(array[index] == NULL) {
    DeleteMatrix(&array, index);
    }
  }
  }
 return array;
 }

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putting all the 'guts' of the code within functions really can make coding a lot easier to develop and debug. make a list of repetitive tasks and start building a toolkit of useful functions - you won't regret it. ;)

[edit]
foiled!
[/edit]