Behavior of int x[size]={0} - c

Question

I am reading through K&R 2nd Edition and C by Dissection 4th Edition, and came to arrays where they both didn't explicitly state it but I remember from a class that we want to always initialize an array to zero to circumvent garbage being in the indices from previous memory allocations.
I looked online and similar suggestions were made, but I saw an array declaration/initialization I hadn't seen before so I wanted to test it: int x[size]={0}; found here
I wrote a little c program to check things out, and I'm currently going down the rabbit hole it seems.
Can someone please help explain the output?
gcc is compiling to c99, -std=c89 won't compile (because of // type comments so I know it isn't c89 (ansi) ), but looking at the gcc flags I cannot find STDC_VERSION flag to tell me which is; therefore, I've manually compiled like: gcc -Wall -std=c99 -o arrayTest arrayTest.c
The little program is:
#include <stdio.h>
typedef int bool;
#define true 1
#define false 2
#define DEBUG_ true
int main()
{
int x[5] ={0};// what does this do? Guess: Initialize x[0]-x[4] values of 0?
int i=0;// for loop
printf("Hello, World!\n");
if(DEBUG_){
printf("TESTING FOR NULLITY OF INDICES\n");
for(i=0; i<5; i++){
if(x[i]==NULL) printf("The Value at %d is NULL\n", i);
}// If each is null, thats what int x[5]={0} decl/initl does..
// Is 0 NULL in C Language (I wouldn't have thought so..)
}
printf("assigning 0 to all indices of the array..\n");
for(i=0; i<5; i++){ x[i]=0; }
// assigning the array to zeros..
if(DEBUG_){
printf("printing contents of array after filling with zero's\n");
for(i=0; i<5; i++){ printf(x[i]+"\n"); }
// hopefully print out 5 zeros..
printf("SUSPECTING THAT ZERO FILL DIDN'T REPLACE NULL VALUES IN ARRAY\n");
for(i=0; i<5; i++){
if(x[i]==NULL) printf("The Value at %d is NULL\n", i);
}
}
printf("Filling the array, each index has value i*2\n");
for(i=0; i<5; i++){ x[i]= i*2; }
// initialize the array to i*2.
// Expected Values 0, 2, 4, 6, 8
for(i=0; i<5; i++){ printf(x[i]+"\n"); }
// hopefully print out Expected Values
return 0;
}
The output from the program:
-bash-4.3$
-bash-4.3$ arrayTest
Hello, World!
TESTING FOR NULLITY OF INDICES
The Value at 0 is NULL
The Value at 1 is NULL
The Value at 2 is NULL
The Value at 3 is NULL
The Value at 4 is NULL
assigning 0 to all indices of the array..
printing contents of array after filling with zero's
SUSPECTING THAT ZERO FILL DIDN'T REPLACE NULL VALUES IN ARRAY
The Value at 0 is NULL
The Value at 1 is NULL
The Value at 2 is NULL
The Value at 3 is NULL
The Value at 4 is NULL
Filling the array, each index has value i*2
USPECTING THAT ZERO FILL DIDN'T REPLACE NULL VALUES IN ARRAYPECTING THAT ZERO FILL DIDN'T REPLACE NULL VALUES IN ARRAYCTING THAT ZERO FILL DIDN'T REPLACE NULL VALUES IN ARRAY
From the looks of that last line there is a repeating sequence occurring:
USPECTING THAT ZERO FILL DIDN'T REPLACE NULL VALUES IN ARRAY
PECTING THAT ZERO FILL DIDN'T REPLACE NULL VALUES IN ARRAY
CTING THAT ZERO FILL DIDN'T REPLACE NULL VALUES IN ARRAY
I am not sure why this is occurring.
Specifically can someone explain what the array declaration/initialization actually does, why assigning zeros to the array isn't working in this case, why trying to set each index to i*2 doesn't work, and what is going on with the print pattern at the end of the output?

Answer 1

The printf() usage is wrong here. Change
printf(x[i]+"\n");
to
printf("%d\n", x[i]);
See the man page for details.
Also, regarding the initializaion, if you need any reference, see this.

Answer 2

According to the C Standard (6.7.9 Initialization)
21 If there are fewer initializers in a brace-enclosed list than there
are elements or members of an aggregate, or fewer characters in a
string literal used to initialize an array of known size than there
are elements in the array, the remainder of the aggregate shall be
initialized implicitly the same as objects that have static storage
duration.
This means that in this declaration
int x[5] ={0};
the first element of the array is explicitly initialized by 0 and all other elements of the array are implicitly initialized the same way as objects with static storage duration that is they are also initialized by 0.
If you would wriite for example
int x[5] ={1};
then the elements of the array would be initialized like { 1, 0, 0, 0, 0 }
Also take into account that this statement
printf(x[i]+"\n");
does not do what you expect. In this expression x[i]+"\n" there is used the pointer arithmetic becuase string literal "\n" is converted to pointer to its first character and after adding to the pointer value x[i] can result in undefined behaviour. I think you mean
printf( "%d\n", x[i] );
Here is a more interesting example of initializing odd elements of an array with 1 and even elements with 0 which you do not know yet..
#include <stdio.h>
int main(void)
{
int a[] = { [1] = 1, [3] = 1, [5] = 1 };
printf( "The number of elements in the array is %zu\n",
sizeof( a ) / sizeof( *a) );
for ( size_t i = 0; i < sizeof( a ) / sizeof( *a ); i++ )
{
printf( "%d ", a[i] );
}
printf( "\n" );
return 0;
}
The program output is
The number of elements in the array is 6
0 1 0 1 0 1
Thus if for example you want to initialize only the last element of an array with 1 and all other elements with 0 you can write
int a[] = { [4] = 1 };