Thread: Wrong access for char *string[]

Your function doesn't need to take a pointer to an array. Just pass it the array. What you've got as an argument is: "An array, with 255 members, of pointers to type char." Read it from right to left.

Just pass the array name to the function, and lose the * in the prototype, and try there.


Quzah.

I have followed your recommendation then this is the new code
cleaner. I remove the tolower() function to simplify but the problem remains:
as result special/unreadable character stored in kbufft .

Why the pointer "point" give wrong data ?

Code:

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

int main(void)
{
 char kbuff[15];
 parser( &kbuff );
}

int parser( char *kbuff )

{
  char i = 0;  // This  fill the kbufft[i]
  cahr j;        // This is the range 0 to xxxxx to read the source kbuff
  char kbufft[15]; // This contains the result: string reversed of kbuff
  char *kbuffz[15];  // This is an array pointed to kbuff[0] ,
		   // I need this to obtain the length of kbuff
  char *point;	   // This a char pointed to end of kbuff (just before the '\0')
  
  kbufft[0] = '\0';
  kbuffz[0]=kbuff;
  j = strlen (*kbuffz) -1;
  point= kbuff[j];

  while (  ( (i<=14) && (j>=0) ) )
        {
   	  kbufft[i] = *point;
	  kbufft[i+1]= '\0';
	  point--;
	  i++;
	  j--;
	}
  printf("\nMy reversed, string is:%s \n", kbufft);
}

>csisz3r

kbuff is not a pointer in the main program.
kbuff is not the same as *kbuff and &kbuff.
The name are identical but *kbuff is pointer only available inside he function.
They are treated differently even the name are same.

You're missing the point, clearly. Go read this FAQ.

1) An array name, 'kbuff' is treated as a pointer to its first element. As such, all you should be doing is passing 'kbuff' to the function, as you've been told multiple times now.

2) You're right. It's not, so stop doing it wrong, and start listening. Just pass 'kbuff' to the function. Nothing else.

3) Wrong. Arrays are passed as pointers to their first element. Any change to a passed array that happen inside a function will affect the array outside the function. Don't believe us still? Test it yourself:

Code:

#include<stdio.h>
void fun( int array[3] )
{
    array[0] = 10;
    array[1] = 21;
    array[2] = 32;
}

int main( void )
{
    int array[3] = { 1, 2, 3 };
    int x;

    printf("Displaying array contents:\n");
    for( x = 0; x < 3; x++ )
        printf("array[ %d ] is %d\n", x, array[ x ] );

    printf("Calling 'fun( array )'\n" );
    fun( array );

    printf("Displaying array contents:\n");
    for( x = 0; x < 3; x++ )
        printf("array[ %d ] is %d\n", x, array[ x ] );

    return 0;
}

/*
    Output for the lazy, or compiler deprived:

Displaying array contents:
array[ 0 ] is 1
array[ 1 ] is 2
array[ 2 ] is 3
Calling 'fun( array )'
Displaying array contents:
array[ 0 ] is 10
array[ 1 ] is 21
array[ 2 ] is 32

*/

4) No, not really they aren't. Not in this case.


Quzah.