Thread: Dynamic allocation for array of pointers to char

Enter a number of students in a group as a command-line argument. Allocate memory for two arrays: one is the array of pointers to char to store students’ names, and another - an array of integers to store the results (from 0 to 100). Read the names and results from the keyboard and fill in the arrays. Find and display on the screen a name and a mark for students who have the lowest and the highest mark.

Well, I've been trying to do this for the last couple of days and I just can't figure it out.

Here's what I have so far:

Code:

int main(int argc, char *argv[])
{
    char * students;            // ? - 'char *students[]'
    int * scores;
    int num, i;
    int max = 0, min = 0, min_i, max_i;
    
    num = atoi( argv[1] );
    
    if ( num <= 0 )
    {
         printf("Please enter a number of students.\n");
         exit(0);
    }
    
    students = (char *) malloc( num * 15 * sizeof( char ) );
    scores = (int *) malloc( num * sizeof( int ) );
    
    for ( i = 0; i < num; i++ )
    {
        printf("Enter the student's name: ");
        gets( &students[i] );            // I know this is wrong, due to the declaration
        
        printf("Enter this student's score: ");
        scanf("%d", &scores[i] );
        fflush( stdin );
        
        printf("\n");
    }
    
    free(students);
    free(scores);
    
    
    system("PAUSE");	
    return 0;
}

* I am just making the assumption that each name will be no more than 14 characters long

As you can probably see from my highlighting, I can tell where my problems are - but it's a catch 22. If I declare an array of pointers to char in the beginning, I need to define a number of pointers the array will hold. But if I just declare a char pointer, when I malloc - it's just one chuck of memory. When indexed as an array, it will not be able to store a string, only a single char (per index).

Basically, my question is; is there any way to do this, other than reading in each character using getchar() and then appending my own null character at the end of each name?

Hopefully that all made sense.

Code:

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

int main(int argc, char *argv[])
{
  char  **students; /* one is the array of pointers to char to store students’ names */
  int * scores;     /* an array of integers to store the results */
  int num, i;

  if ( argc > 1 ) {
    num = atoi( argv[1] );
  } else {
    printf("Please enter a number of students.\n");
    exit(0);
  }

  if ( num <= 0 )
  {
    printf("Please enter a number of students.\n");
    exit(0);
  }

  students = malloc( num * sizeof *students );
  scores =   malloc( num * sizeof *scores );

  for ( i = 0; i < num; i++ )
  {
    char buff[BUFSIZ];
    printf("Enter the student's name: ");
    fgets( buff, BUFSIZ, stdin ); /* NEVER use gets */

    students[i] = malloc ( strlen(buff) + 1 );
    strcpy( students[i], buff );

    printf("Enter this student's score: ");
    fgets( buff, BUFSIZ, stdin );
    sscanf( buff, "%d", &scores[i] );
  }

  for ( i = 0 ; i < num ; i++ ) {
    free( students[i] );
  }
  free(students);
  free(scores);

  return 0;
}

And read the FAQ on
- casting malloc
- using gets
- using fflush(stdin)

I'll have a read through the faqs, but this has me a bit confused:

Code:

char buff[BUFSIZ];
printf("Enter the student's name: ");
fgets( buff, BUFSIZ, stdin ); /* NEVER use gets */

I assume you just forgot to declare BUFSIZE?

Oh, and what should I read to learn more about this line:

Code:

char  **students; /* one is the array of pointers to char to store students’ names */

BUFSIZ is a constant in stdio.h
Unless you have some brain-damaged fossil compiler.

> Oh, and what should I read to learn more about this line:
Maybe this
http://home.netcom.com/~tjensen/ptr/pointers.htm

This pointer business is starting to do my head in. I was fine with the basics, but it's starting to get a bit hairy.

Cheers mate.

Should I point out that strcopy() is bad (yes, I know it won't overflow in this case), or just go back to my learning?

Oops, I always pronounce it copy - hence the error. *paranoid*

And yeh, I was just being facetious.