Why does my program display the Array address instead of its contents? - c

Question

My C program consists of an array called 'test_var'. It has another integer array 'arr_int' that consists of a set of integer numbers. My code looks something like this:
#include <stdlib.h>
#include <stddef.h>
#include <stdio.h>
int State(var);
int main()
{
int arr_int[3] ={1000, 1001, 1002, 1003};
int var;
int *test_var[4]={0};
State(var)
{
int i;
for(i=0; i<4; i++){
test_var[i] = arr_int[i];
i++;
}
return test_var[var];
}
printf("Enter a number between 0 and 3\n");
scanf("%d",&var);
State(var);
printf ("The array structure is %d", test_var[var]);
return 0;
}
However now when I try to print the returned value array test_var for the user input var=0 instead of the whole array(1000) I just get 0. What am I doing wrong here ? COuld somebody please tell me? Am I dereferencing the array in a wrong way?
EDIT : the code without the typo:
#include <stdlib.h>
#include <stddef.h>
#include <stdio.h>
int State(var);
int main()
{
int *arr_int[3] ={1000, 1001, 1002, 1003};
int var;
int *test_var[4]={0};
State(var)
{
int i;
for(i=0; i<4; i++){
test_var[i] = arr_int[i];
i++;
}
return test_var[var];
}
printf("Enter a number between 0 and 3\n");
scanf("%d",&var);
State(var);
printf ("The array structure is %d", test_var[var]);
return 0;
}

Answer 1

In your case,
the statement
test_var[i] = arr_int[i];
is wrong. test_var[i] is of type int *, and arr_int[i] is of type int, and they are not really compatible. To quote the standard, chapter §6.3.2.3
An integer may be converted to any pointer type. Except as previously specified, the result is implementation-defined, might not be correctly aligned, might not point to an entity of the referenced type, and might be a trap representation.
Then, you're using another statement like
printf ("The array structure is %d", test_var[var]);
%d expects an int argument, but you're supplying an int *. Pure undefined behaviour.
After that, int State(var) relies upon the older concept of default-to-int, which has been made invalid in C99 onwards.
In an array of 3 elements, you're supplying 4-member initializer list. Useless and confusing.
That said, FWIW, you're using a nested function, which is also not a part of standard C. It is supported as a GCC extension.
EDIT:
As per the comments, even if your arr_int is an array of of type int *, the initializers do not provide a seemingly valid value for the pointers to be dereferenced. That means, you cannot access the memory pointed to by those pointers, they are most likely to reside outside of the allocated memory area of your process and hence, invalid.
Nevertheless, for the point 2, mentioned earlier, the code is UB.

Answer 2

int arr_int[3] ={1000, 1001, 1002, 1003};
Array is of size 3 you initialize it with 4 elements.
Also the nested function is supposed to create a problem.They are supported as an extension in GNU C.
Note-Though you can declare a function inside of a function, but it's not a nested function.
EDIT
Your second code when you give input as 0 and 2 it seems to give correct output but when input is 1 or 3 it gives as 0. Surely it doesn't behave as intended.

Answer 3

int *arr_int[3] ={1000, 1001, 1002, 1003};
int *test_var[4]={0};
Why did you declare pointers instead of arrays? And why arr_int[3] is initialized with 4 integers?
It should be like this.
int arr_int[3] ={1000, 1001, 1002};
int test_var[4]={0}