how to stop this proccess prematurely..

i added an "if" command that if ch<0 return 0
but its still doesnt work

??

Code:

#include <stdio.h>
int read_array(int input[],int i,int size);

int main()
{
	int i;
    int input[40];
    printf("%d\n",read_array(input,0,8));
  for(i=0;i<8;i++)
  {
    printf("%d ",input[i]);
  }
   printf("\n");
   return 0;
}
int read_array(int input[],int i,int size)
{
	int flag,rt,ch;
	if (i==size)
	{

	   return 1;
	}
    
    
     flag=scanf("%d",&input[i]);
	 ch=getchar();
	 if ((ch=='\n')&&(i!=size))
    {
       return 0;
    }

	 if (ch<0)
    {
		return 0;
	}

	 if (flag==0)
	 {
		 return 0;
	 }
	 else
	 {
	 }
     rt=read_array(input,i+1,size);
	 return rt;
}

Originally Posted by transgalactic2

i added an "if" command that if ch<0 return 0
but its still doesnt work

??

Like I said, yes you do need to deal with EOF at some point - but at this point in time, you should concentrate on getting your logic right. Perhaps adding a printf or two in the function would help - for example printing "i" at the entry of the function, and the number you received from scanf?

--
Mats

Thinking about the problem further, I think that the scanf() + getchar() combination is problematic. Before we continue, what are your exact and complete requirements?

Honestly, just write out the exact assignment that you have been given, word for word, with every comma and full stop. You have a history of leaving out important details which in turn frustrates those who suggest things to you that are forbidden, but you conveniently did not say so.

my whole program is almost complete i just need to do this array input part
i need to enter numbers in a row like +001 -007 8 9 6 and the size of the array
should be 5 for this example to be a legal input
also it should ignore spaces and tabs.

i need to do this recursively
no loops allowed
i can use only
scanf getchar gets

if my size is 6
and i enter 1 2 3 enter
then it should stop and return 0

if my size is 3
and i enter 1 2 3 4 5 enter
then it should stop and return 0

So, are you allowed any of these:
ungetc()
atoi()/strtol()
strtok()
isdigit()/isspace()

Are you allowed to write other functions.

I don't think it's possible to solve this case RELIABLY with getchar() and scanf() alone - and gets() is really no help here if you can't use some other function to translate the numeric string into an integer.

--
Mats

gets scanf and getchar is the only things allowed

Originally Posted by transgalactic2

gets scanf and getchar is the only things allowed

We have to assume that the problem is solvable, right? Your teacher seems to be a sadist, but I doubt that it's so sadistic that he/she is giving you tasks that are impossible to solve.

Let's forget about gets(), as it makes no sense for the purpose here (unless you fancy translating a long string into a set of numbers, which is a possibility, but a rather poor one).

That leaves getchar() and scanf(). getchar() on it's own is no better - it still leaves you doing the "ascii to integer" translation yourself.

Calling getchar() after scanf() should be fine, as whatever you enter that is a number will have to end with either a non-numeric character or EOF. In valid input [given the format string you have at the moment], the non-numeric character is a whitespace (space, tab or newline). If the input is invalid, then you may have letters or other symbols there - your description doesn't say what to do about that, but I'd say you should probably ignore that case altogether.

Now, given the above post-conditions of scanf, we have a pretty easy solution for validating the end of input: use getchar after scanf, if it's a whitespace other than newline, there is more input.

--
Mats

i did call getchar after scanf
and i checked for EOF
and its not working

Code:

#include <stdio.h>
int read_array(int input[],int i,int size);

int main()
{
	int i;
    int input[40];
    printf("%d\n",read_array(input,0,8));
  for(i=0;i<8;i++)
  {
    printf("%d ",input[i]);
  }
   printf("\n");
   return 0;
}
int read_array(int input[],int i,int size)
{
	int flag,rt,ch;
	if (i==size)
	{

	   return 1;
	}
    
    
     flag=scanf("%d",&input[i]);
	 ch=getchar();
	 if ((ch=='\n')&&(i!=size))
    {
       return 0;
    }

	 if (ch<0)
    {
		return 0;
	}

	 if (flag==0)
	 {
		 return 0;
	 }
	 else
	 {
	 }
     rt=read_array(input,i+1,size);
	 return rt;
}

And you didn't read my post prior to the last one:
add some tracing to see what your function is actually doing.

--
Mats

you said that its not in the right order
but i dont know whats the right order

Originally Posted by transgalactic2

you said that its not in the right order
but i dont know whats the right order

Neither do I, unless I spend some time tracing through the exact logic of your code - and that's part of what you need to learn how to do - you can't just post the same non-functioning code over and over again until someone posts the right answer for you.

I guess it would take about 3 different (correctly placed) printf's added to your code to see what is happening, and understand the difference between what actually happens and what you want to have happen. If in doubt, add a few extra printf's. Make sure you can tell WHICH printf is which one -- so mark each one with a letter (A, B, C, etc) at the start. For example:

Code:

printf("A i = %d\n", i);
...
printf("B x = %d\n", x);   // No, you don't use x in your code, this is just an example!

--
Mats

i did debugging and i can see the values of the variables without
the printf
but its doesnt enter
this part

Code:

if (i==size)
	{

	   return 1;
	}

Ok, so why not? Is i never equal to size? Why not? [No, I don't know the answer, but I can guarantee that it's not because the compiler is broken].

--
Mats

i dont know why
??

Originally Posted by transgalactic2

i dont know why
??

I understand that, and I'm trying to help you FIND out why. You have to learn to debug your own code - it is a VERY important part of programming. So me telling you what is wrong isn't going to teach you anything useful.

I still think you have a "one off" problem.

--
Mats