inputting integer array in a row recursevly..

i need to enter an array in a row number so they will become an array
when i enter
1 2 3 4 5 6 7 8
it says debug assertion failed

why its not working??

Code:

#include <stdio.h>
int read_array(int input[],int i,int size);
 
int main()
{
    int input[40];
    printf("%d",read_array(input,0,8));
}
int read_array(int input[],int i,int size)
{
	int flag;
       
	if (i==size-1)
	{
     
	   return 1;
	}
     flag=scanf("%d",&input[i]);
	 if (flag==0)
	 {
		 return 0;
	 }
	 else
	 {
	 }
     read(input,i+1,size); 
}

I don't see any values being passed into input... But I don't understand all of that code eather, so I could be wrong, but I still don't see anywhere where anything could be assigned to input :P

did you try
1234567

oh another recursion topic, I'm surprised

Code:

read(input,i+1,size);

did you mean read_array(input,i+1,size)?

PS: what tools are you using? any basic debugger should allow to spot the error, did you try?

PS2: once again, why open a new thread while the very same thing is discussed in another one? you won't get more answers that way...

i enter 1 2 3 4 5 6 7 8
ok it works
but i get a warning that not all path return value
i cant see what case i missed

Code:

 
#include <stdio.h>
int read_array(int input[],int i,int size);
 
int main()
{
    int input[40];
    printf("%d",read_array(input,0,8));
}
int read_array(int input[],int i,int size)
{
	int flag;
       
	if (i==size-1)
	{
     
	   return 1;
	}
     flag=scanf("%d",&input[i]);
	 if (flag==0)
	 {
		 return 0;
	 }
	 else
	 {
	 }
     read_array(input,i+1,size); 
}

but i get a warning that not all path return value
i cant see what case i missed

Are you joking? your function is 10 lines long and you can't see that?

Code:

     read_array(input,i+1,size); 
     // hint-of-the-day: function ends here
}

hint 2:
int

it need to end when i==size
or when the input is wrong

i dont need to return anything in the end except the recursive call
i changed the code so the recursive call will go to some variable
but the warning keeps showing
what case i missed??

Code:

 
#include <stdio.h>
int read_array(int input[],int i,int size);
 
int main()
{
    int input[40];
    printf("%d",read_array(input,0,8));
}
int read_array(int input[],int i,int size)
{
	int flag,rt;
       
	if (i==size-1)
	{
     
	   return 1;
	}
     flag=scanf("%d",&input[i]);
	 if (flag==0)
	 {
		 return 0;
	 }
	 else
	 {
	 }
     rt=read_array(input,i+1,size); 
}

i fixed it thanks

Code:

     rt=read_array(input,i+1,size); 
     return 0;
}

Sigh.

i put return rt in the end
and it worked fine
is there a problem with how i did it??

I suggest you get back to C programming basics before resuming this.

i dont need to return anything in the end

Sure. But your program tells otherwise and the compiler works on your program not on what you wish it should do.

i put return rt in the end instead of return 0
and it worked fine
is there a problem with how i did it??

as long as you return something, the compiler will be happy.

Originally Posted by transgalactic2

i put return rt in the end instead of return 0
and it worked fine
is there a problem with how i did it??

Yes, that is okay. Your compiler or debugger is just being picky because it cannot be sure that control will never reach the end of the function, even if

Code:

int func() {
     while (1) return 0;
     return 0;    // still need this
}