Thread: basic input and output with printf and fgets

>Should i use scanf?
Not if you can avoid it without bending over backwards.

>I am also trying to learn how to write portable code.
Bravo! This should be everyone's goal. I've worked with some people who couldn't care less about portability. This usually translates into the quality of their code as well, so you can tell who is good and who isn't by their attitude on portability. (From my experience)

You really don't need two arrays, just have the input formatted into one array and use sscanf to read the values into integer variables:

Code:

#include <stdio.h>
/*
** Macros are traditionally upper case so we
** can tell them apart from variables and functions.
*/
#define MAX 100 /* Some arbitrary value */

int main ( void )
{
  char a[MAX];
  int ia, ib;

  printf ( "Enter two numbers:\n" );
  /*
  ** Check all return values in case something ugly happened.
  */
  if ( fgets ( a, MAX, stdin ) != NULL ) {
    /*
    ** Ditto, especially with user input. :)
    */
    if ( sscanf ( a, "%d %d", &ia, &ib ) != 2 )
      printf ( "Hmm, did you enter it correctly?" );
    else
      printf ( "Your numbers were: %d and %d\n", ia, ib );
  }
  return 0;
}

[suggestion]
Also take note of whitespace in between tokens, such as your line here:

>fgets(b,max,stdin);

It is widely accepted that no whitespace between tokens is harder to read than well placed spacing. Some of the more popular styles for this are:

/* Separate the arguments and the argument list */
fgets (b, max, stdin);

/* Separate everything */
fgets ( b, max, stdin );

/* Separate only the arguments */
fgets( b, max, stdin );

C is cryptic enough without our adding difficult to read code through bad formatting. Look into some style guides that can be found online for further information. The best style is the one that is easy to read and consistent, despite what some people will say.
[/suggestion]

-Prelude

[EDIT]
ROFL ... Prelude got it
[/EDIT]

fgets() gets a string. your code gets two strings and prints them back. if you want to get two numbers and print them back, then you have to use sscanf and sprintf like shown (there are alternatives for sscanf though)

you have to pass the addresses of ia and ib to sscanf() because it is programmed to expect pointers, so you need to give it the memory address which points to the memory location containing the values of ia and ib.

i'm sure that wasnt very helpful, maybe someone else can elaborate.

>And why cant i just assign values to ia and ib.
All function arguments are passed by value, this means that a copy is made of the original and you change the copy in the function:

Code:

#include <stdio.h>

int f ( int copy )
{
  copy++;
}

int main ( void )
{
  int i = 5;
  f ( i );
  printf ( "%d\n", i );
  return 0;
}

Since i is passed by value, the copy is incremented by one, but i remains unchanged. The output of this program is 5.

Now, one can simulate pass by reference and modify the original object by passing it as a pointer:

Code:

#include <stdio.h>

int f ( int *copy )
{
  *copy++;
}

int main ( void )
{
  int i = 5;
  f ( &i );
  printf ( "%d\n", i );
  return 0;
}

The output of this program is 6 because the address of i was passed by value. The pointer in f was a copy of the address of i, so when it increments the object it points to, the value of i changes. This change is obvious when f returns and i has the value of 6.

The scanf family does something similar, to read from input is simple, but if you pass scanf the copy of a variable it will not place the input in the correct place and your copy of i will remain empty:

Code:

#include <stdio.h>

void f ( int copy )
{
  copy = 10;
}

int main ( void )
{
  int i;
  f ( i )
  printf ( "%d\n", i );
  return 0;
}

The output of this program is undefined, it is dependent on the random value placed in i when you declared it. Let's try this again with the address of i:

Code:

#include <stdio.h>

void f ( int *copy )
{
  *copy = 10;
}

int main ( void )
{
  int i;
  f ( &i )
  printf ( "%d\n", i );
  return 0;
}

Since you pass the address of i to f, the value of 10 is placed at the address that i rests in. So the output of this program is 10. The scanf family works the same way and for the same reasons.

-Prelude