Why compiler is showing unexpected result? - c

Question

I'm new in programming. At current learning C language.I'm confused right now in a problem which is showing unexpected result after running.
The problem is-
I want to make a program which take input one more than when the user input 42 and stop taking input after it and print all the numbers user had typed before 42 has come.
For example-Input:
1
2
88
42
99
Output:
1
2
88
My Program-
#include <stdio.h>
int main()
{
int i = 1, j, temp;
printf("enter numbers:\n");
while (i)
{
scanf("%d", &temp);
i++;
int a[i];
a[i - 2] = temp;
if (temp == 42)
{
scanf("%d", &a[i - 1]);
for (j = 0 ; j < i - 1 ; j++)
{
if (a[j] == 42)
break;
else
printf("%d\n",a[j]);
}
}
if (temp == 42)
break;
}
return 0;
}
When i input the same value as in example i get unexpected result such as-
4196363
0
-1
Please help.And sorry for my bad english.Thankyou in advance.

Answer 1

You are printing uninitialized values because if i's value becomes 3 and greater values after that you never set a[0] as well as all the values before a[i - 2].
Thus causing Undefined Behavior at the following line
printf("%d\n", a[j]);
some values are uninitialized and you still try to print them.
You should notice that a is being redeclared at each iteration and you are not initializing the values because the values in the previous a where only valid in the previous iteration.
The a variable is valid only within the scope of the while loop but not only because it's not declared outside, also because it's deallocated when it becomes out of scope. Perhaps an example will make it clearer.
Suppose you had the following program
int main(void)
{
int *pointer = NULL;
if (pointer == NULL)
{
int array[5] = {1, 2, 3, 4, 5};
pointer = array;
}
/* `poitner' points to `array` but array was deallocated */
for (int i = 0 ; i < 5 ; ++i)
printf("%d\n", pointer[i]); /* This is UB */
return 0;
}
this would compile, however as you can see it will invokes undefined behavior.
Since it's not possible to understand what your code does, I suggest the following
#include <stdio.h>
int main()
{
int i = 1;
int a[100];
printf("enter numbers:\n");
while (i++ < 100)
{
if (scanf("%d%*c", &a[i - 2]) != 1)
return -1; /* this was invalid input, it will cause UB */
if (a[i - 2] == 42)
{
if (scanf("%d%*c", &a[i - 1]) != 1)
return -1; /* this was invalid input, it will cause UB */
for (int j = 0 ; j < i - 1 ; j++)
{
if (a[j] == 42)
break;
else
printf("%d\n", a[j]);
}
}
if (a[i - 2] == 42)
break;
}
return 0;
}
This at least removes undefiend behavior and meakes teh code more readable, but it's still not clear what is this program for or what it should do.

Answer 2

In general case because the number of entered values is unknown you need to allocate dynamically storage for a new entered value.
So I would build a list that contains entered values. The other approach as to use malloc and realloc for the whole array when a new value is entered. In my opinion it is a worse solution than to build a list.
The program can look the following way
#include <stdio.h>
#include <stdlib.h>
int main( void )
{
struct node
{
int value;
struct node *next;
} * head = NULL;
struct node **current = &head;
int value;
int next = -1;
while ( next != 1 && scanf( "%d", &value ) == 1 )
{
if ( next == 0 || value == 42 ) ++next;
if ( next == -1 )
{
*current = malloc( sizeof( struct node ) );
( *current )->value = value;
( *current )->next = NULL;
current = &( *current )->next;
}
}
while ( head )
{
printf( "%d\n", head->value );
struct node *tmp = head;
head = head->next;
free( tmp );
}
}
If to enter for example the following values
1 2 3 4 5 6 7 8 9 42 10
then the program output will be
1
2
3
4
5
6
7
8
9
The list stores neither 42 nor the next value after 42 though you can modify the program such a way that it would store also these values.
As for your code then at least this statement
a[i - 2] = temp;
is wrong when i is equal to 1 and results in undefined behaviour of the program.
And I do not see that i would be changed.

Answer 3

I think your problem is trying to declare(and redeclare) a[i] multiple times with different values. The compiler needs to know the size of an array at compile time. So the compiler is setting aside memory for an integer array with one element and then the dynamic part of your code is attempting to redeclare it a run-time. Try moving a outside the while loop and declaring it with a well defined maximum number of integer values like:
#define MAX_NUMS 1000
int a[MAX_NUMS];
There are some other problems as listed above, but I tested this with GCC 4.9 compiler and it seems to give the desired result.

Answer 4

You are doing some unnecessary matter.
You have to check just this--
while(input_number)
{
if(input_number==42)
break;
else
printf("%d\n",input_number);
}
In my thought I had edited your code to store all number before 42 based on your code.
#include<stdio.h>
int main()
{
int i=1,j,temp;
int a[10010];
printf("enter numbers:\n");
while(i)
{
scanf("%d",&temp);
i++;
a[i-2]=temp;
if(temp==42)
{
for(j=0; j<i-2; j++)
{
if(a[j]==42)
break;
else
printf("%d\n",a[j]);
}
}
if (temp == 42)
break;
}
return 0;
}